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---
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comments: true
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---
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# 2.3 時間複雜度
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執行時間可以直觀且準確地反映演算法的效率。如果我們想準確預估一段程式碼的執行時間，應該如何操作呢？
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1. **確定執行平臺**，包括硬體配置、程式語言、系統環境等，這些因素都會影響程式碼的執行效率。
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2. **評估各種計算操作所需的執行時間**，例如加法操作 `+` 需要 1 ns ，乘法操作 `*` 需要 10 ns ，列印操作 `print()` 需要 5 ns 等。
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3. **統計程式碼中所有的計算操作**，並將所有操作的執行時間求和，從而得到執行時間。
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例如在以下程式碼中，輸入資料大小為 $n$ ：
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=== "Python"
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```python title=""
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# 在某執行平臺下
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def algorithm(n: int):
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a = 2 # 1 ns
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a = a + 1 # 1 ns
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a = a * 2 # 10 ns
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# 迴圈 n 次
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for _ in range(n): # 1 ns
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print(0) # 5 ns
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```
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=== "C++"
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```cpp title=""
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// 在某執行平臺下
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void algorithm(int n) {
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int a = 2; // 1 ns
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a = a + 1; // 1 ns
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a = a * 2; // 10 ns
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// 迴圈 n 次
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for (int i = 0; i < n; i++) { // 1 ns ，每輪都要執行 i++
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cout << 0 << endl; // 5 ns
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}
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}
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```
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=== "Java"
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```java title=""
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// 在某執行平臺下
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void algorithm(int n) {
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int a = 2; // 1 ns
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a = a + 1; // 1 ns
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a = a * 2; // 10 ns
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// 迴圈 n 次
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for (int i = 0; i < n; i++) { // 1 ns ，每輪都要執行 i++
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System.out.println(0); // 5 ns
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}
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}
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```
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=== "C#"
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```csharp title=""
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// 在某執行平臺下
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void Algorithm(int n) {
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int a = 2; // 1 ns
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a = a + 1; // 1 ns
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a = a * 2; // 10 ns
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// 迴圈 n 次
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for (int i = 0; i < n; i++) { // 1 ns ，每輪都要執行 i++
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Console.WriteLine(0); // 5 ns
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}
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}
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```
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=== "Go"
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```go title=""
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// 在某執行平臺下
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func algorithm(n int) {
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a := 2 // 1 ns
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a = a + 1 // 1 ns
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a = a * 2 // 10 ns
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// 迴圈 n 次
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for i := 0; i < n; i++ { // 1 ns
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fmt.Println(a) // 5 ns
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}
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}
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```
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=== "Swift"
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```swift title=""
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// 在某執行平臺下
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func algorithm(n: Int) {
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var a = 2 // 1 ns
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a = a + 1 // 1 ns
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a = a * 2 // 10 ns
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// 迴圈 n 次
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for _ in 0 ..< n { // 1 ns
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print(0) // 5 ns
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}
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}
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```
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=== "JS"
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```javascript title=""
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// 在某執行平臺下
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function algorithm(n) {
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var a = 2; // 1 ns
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a = a + 1; // 1 ns
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a = a * 2; // 10 ns
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// 迴圈 n 次
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for(let i = 0; i < n; i++) { // 1 ns ，每輪都要執行 i++
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console.log(0); // 5 ns
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}
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}
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```
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=== "TS"
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```typescript title=""
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// 在某執行平臺下
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function algorithm(n: number): void {
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var a: number = 2; // 1 ns
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a = a + 1; // 1 ns
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a = a * 2; // 10 ns
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// 迴圈 n 次
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for(let i = 0; i < n; i++) { // 1 ns ，每輪都要執行 i++
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console.log(0); // 5 ns
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}
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}
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```
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=== "Dart"
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```dart title=""
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// 在某執行平臺下
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void algorithm(int n) {
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int a = 2; // 1 ns
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a = a + 1; // 1 ns
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a = a * 2; // 10 ns
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// 迴圈 n 次
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for (int i = 0; i < n; i++) { // 1 ns ，每輪都要執行 i++
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print(0); // 5 ns
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}
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}
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```
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=== "Rust"
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```rust title=""
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// 在某執行平臺下
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fn algorithm(n: i32) {
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let mut a = 2; // 1 ns
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a = a + 1; // 1 ns
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a = a * 2; // 10 ns
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// 迴圈 n 次
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for _ in 0..n { // 1 ns ，每輪都要執行 i++
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println!("{}", 0); // 5 ns
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}
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}
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```
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=== "C"
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```c title=""
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// 在某執行平臺下
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void algorithm(int n) {
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int a = 2; // 1 ns
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a = a + 1; // 1 ns
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a = a * 2; // 10 ns
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// 迴圈 n 次
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for (int i = 0; i < n; i++) { // 1 ns ，每輪都要執行 i++
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printf("%d", 0); // 5 ns
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}
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}
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```
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=== "Kotlin"
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```kotlin title=""
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// 在某執行平臺下
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fun algorithm(n: Int) {
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var a = 2 // 1 ns
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a = a + 1 // 1 ns
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a = a * 2 // 10 ns
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// 迴圈 n 次
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for (i in 0..<n) { // 1 ns ，每輪都要執行 i++
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println(0) // 5 ns
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}
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}
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```
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=== "Ruby"
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```ruby title=""
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# 在某執行平臺下
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def algorithm(n)
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a = 2 # 1 ns
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a = a + 1 # 1 ns
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a = a * 2 # 10 ns
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# 迴圈 n 次
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(n...0).each do # 1 ns
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puts 0 # 5 ns
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end
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end
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```
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=== "Zig"
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```zig title=""
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// 在某執行平臺下
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fn algorithm(n: usize) void {
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var a: i32 = 2; // 1 ns
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a += 1; // 1 ns
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a *= 2; // 10 ns
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// 迴圈 n 次
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for (0..n) |_| { // 1 ns
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std.debug.print("{}\n", .{0}); // 5 ns
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}
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}
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```
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根據以上方法，可以得到演算法的執行時間為 $(6n + 12)$ ns ：
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$$
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1 + 1 + 10 + (1 + 5) \times n = 6n + 12
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$$
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但實際上，**統計演算法的執行時間既不合理也不現實**。首先，我們不希望將預估時間和執行平臺繫結，因為演算法需要在各種不同的平臺上執行。其次，我們很難獲知每種操作的執行時間，這給預估過程帶來了極大的難度。
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## 2.3.1 統計時間增長趨勢
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時間複雜度分析統計的不是演算法執行時間，**而是演算法執行時間隨著資料量變大時的增長趨勢**。
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“時間增長趨勢”這個概念比較抽象，我們透過一個例子來加以理解。假設輸入資料大小為 $n$ ，給定三個演算法 `A`、`B` 和 `C` ：
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=== "Python"
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```python title=""
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# 演算法 A 的時間複雜度：常數階
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def algorithm_A(n: int):
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print(0)
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# 演算法 B 的時間複雜度：線性階
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def algorithm_B(n: int):
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for _ in range(n):
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print(0)
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# 演算法 C 的時間複雜度：常數階
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def algorithm_C(n: int):
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for _ in range(1000000):
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print(0)
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```
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=== "C++"
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```cpp title=""
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// 演算法 A 的時間複雜度：常數階
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void algorithm_A(int n) {
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cout << 0 << endl;
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}
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// 演算法 B 的時間複雜度：線性階
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void algorithm_B(int n) {
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for (int i = 0; i < n; i++) {
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cout << 0 << endl;
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}
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}
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// 演算法 C 的時間複雜度：常數階
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void algorithm_C(int n) {
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for (int i = 0; i < 1000000; i++) {
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cout << 0 << endl;
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}
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}
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```
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=== "Java"
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```java title=""
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// 演算法 A 的時間複雜度：常數階
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void algorithm_A(int n) {
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System.out.println(0);
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}
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// 演算法 B 的時間複雜度：線性階
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void algorithm_B(int n) {
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for (int i = 0; i < n; i++) {
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System.out.println(0);
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}
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}
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// 演算法 C 的時間複雜度：常數階
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void algorithm_C(int n) {
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for (int i = 0; i < 1000000; i++) {
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System.out.println(0);
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}
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}
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```
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=== "C#"
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```csharp title=""
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// 演算法 A 的時間複雜度：常數階
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void AlgorithmA(int n) {
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Console.WriteLine(0);
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}
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// 演算法 B 的時間複雜度：線性階
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void AlgorithmB(int n) {
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for (int i = 0; i < n; i++) {
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Console.WriteLine(0);
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}
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}
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// 演算法 C 的時間複雜度：常數階
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void AlgorithmC(int n) {
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for (int i = 0; i < 1000000; i++) {
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Console.WriteLine(0);
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}
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}
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```
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=== "Go"
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```go title=""
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// 演算法 A 的時間複雜度：常數階
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func algorithm_A(n int) {
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fmt.Println(0)
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}
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// 演算法 B 的時間複雜度：線性階
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func algorithm_B(n int) {
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for i := 0; i < n; i++ {
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fmt.Println(0)
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}
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}
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// 演算法 C 的時間複雜度：常數階
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func algorithm_C(n int) {
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for i := 0; i < 1000000; i++ {
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fmt.Println(0)
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}
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}
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```
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=== "Swift"
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```swift title=""
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// 演算法 A 的時間複雜度：常數階
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func algorithmA(n: Int) {
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print(0)
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}
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// 演算法 B 的時間複雜度：線性階
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func algorithmB(n: Int) {
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for _ in 0 ..< n {
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print(0)
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}
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}
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// 演算法 C 的時間複雜度：常數階
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func algorithmC(n: Int) {
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for _ in 0 ..< 1_000_000 {
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print(0)
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}
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}
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```
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=== "JS"
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```javascript title=""
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// 演算法 A 的時間複雜度：常數階
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function algorithm_A(n) {
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console.log(0);
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}
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// 演算法 B 的時間複雜度：線性階
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function algorithm_B(n) {
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for (let i = 0; i < n; i++) {
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console.log(0);
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}
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}
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// 演算法 C 的時間複雜度：常數階
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function algorithm_C(n) {
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for (let i = 0; i < 1000000; i++) {
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console.log(0);
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}
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}
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```
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=== "TS"
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```typescript title=""
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// 演算法 A 的時間複雜度：常數階
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function algorithm_A(n: number): void {
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console.log(0);
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}
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// 演算法 B 的時間複雜度：線性階
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function algorithm_B(n: number): void {
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for (let i = 0; i < n; i++) {
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console.log(0);
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}
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}
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// 演算法 C 的時間複雜度：常數階
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function algorithm_C(n: number): void {
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for (let i = 0; i < 1000000; i++) {
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console.log(0);
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}
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}
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```
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=== "Dart"
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```dart title=""
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// 演算法 A 的時間複雜度：常數階
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void algorithmA(int n) {
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print(0);
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}
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// 演算法 B 的時間複雜度：線性階
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void algorithmB(int n) {
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for (int i = 0; i < n; i++) {
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print(0);
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}
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}
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// 演算法 C 的時間複雜度：常數階
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void algorithmC(int n) {
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for (int i = 0; i < 1000000; i++) {
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print(0);
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}
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}
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```
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=== "Rust"
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```rust title=""
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// 演算法 A 的時間複雜度：常數階
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fn algorithm_A(n: i32) {
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println!("{}", 0);
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}
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// 演算法 B 的時間複雜度：線性階
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fn algorithm_B(n: i32) {
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for _ in 0..n {
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println!("{}", 0);
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}
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}
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|
|
// 演算法 C 的時間複雜度：常數階
|
|
|
fn algorithm_C(n: i32) {
|
|
|
for _ in 0..1000000 {
|
|
|
println!("{}", 0);
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
```c title=""
|
|
|
// 演算法 A 的時間複雜度：常數階
|
|
|
void algorithm_A(int n) {
|
|
|
printf("%d", 0);
|
|
|
}
|
|
|
// 演算法 B 的時間複雜度：線性階
|
|
|
void algorithm_B(int n) {
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
printf("%d", 0);
|
|
|
}
|
|
|
}
|
|
|
// 演算法 C 的時間複雜度：常數階
|
|
|
void algorithm_C(int n) {
|
|
|
for (int i = 0; i < 1000000; i++) {
|
|
|
printf("%d", 0);
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Kotlin"
|
|
|
|
|
|
```kotlin title=""
|
|
|
// 演算法 A 的時間複雜度：常數階
|
|
|
fun algoritm_A(n: Int) {
|
|
|
println(0)
|
|
|
}
|
|
|
// 演算法 B 的時間複雜度：線性階
|
|
|
fun algorithm_B(n: Int) {
|
|
|
for (i in 0..<n){
|
|
|
println(0)
|
|
|
}
|
|
|
}
|
|
|
// 演算法 C 的時間複雜度：常數階
|
|
|
fun algorithm_C(n: Int) {
|
|
|
for (i in 0..<1000000) {
|
|
|
println(0)
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Ruby"
|
|
|
|
|
|
```ruby title=""
|
|
|
# 演算法 A 的時間複雜度：常數階
|
|
|
def algorithm_A(n)
|
|
|
puts 0
|
|
|
end
|
|
|
|
|
|
# 演算法 B 的時間複雜度：線性階
|
|
|
def algorithm_B(n)
|
|
|
(0...n).each { puts 0 }
|
|
|
end
|
|
|
|
|
|
# 演算法 C 的時間複雜度：常數階
|
|
|
def algorithm_C(n)
|
|
|
(0...1_000_000).each { puts 0 }
|
|
|
end
|
|
|
```
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
```zig title=""
|
|
|
// 演算法 A 的時間複雜度：常數階
|
|
|
fn algorithm_A(n: usize) void {
|
|
|
_ = n;
|
|
|
std.debug.print("{}\n", .{0});
|
|
|
}
|
|
|
// 演算法 B 的時間複雜度：線性階
|
|
|
fn algorithm_B(n: i32) void {
|
|
|
for (0..n) |_| {
|
|
|
std.debug.print("{}\n", .{0});
|
|
|
}
|
|
|
}
|
|
|
// 演算法 C 的時間複雜度：常數階
|
|
|
fn algorithm_C(n: i32) void {
|
|
|
_ = n;
|
|
|
for (0..1000000) |_| {
|
|
|
std.debug.print("{}\n", .{0});
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
圖 2-7 展示了以上三個演算法函式的時間複雜度。
|
|
|
|
|
|
- 演算法 `A` 只有 $1$ 個列印操作，演算法執行時間不隨著 $n$ 增大而增長。我們稱此演算法的時間複雜度為“常數階”。
|
|
|
- 演算法 `B` 中的列印操作需要迴圈 $n$ 次，演算法執行時間隨著 $n$ 增大呈線性增長。此演算法的時間複雜度被稱為“線性階”。
|
|
|
- 演算法 `C` 中的列印操作需要迴圈 $1000000$ 次，雖然執行時間很長，但它與輸入資料大小 $n$ 無關。因此 `C` 的時間複雜度和 `A` 相同，仍為“常數階”。
|
|
|
|
|
|
{ class="animation-figure" }
|
|
|
|
|
|
<p align="center"> 圖 2-7 演算法 A、B 和 C 的時間增長趨勢 </p>
|
|
|
|
|
|
相較於直接統計演算法的執行時間，時間複雜度分析有哪些特點呢？
|
|
|
|
|
|
- **時間複雜度能夠有效評估演算法效率**。例如，演算法 `B` 的執行時間呈線性增長，在 $n > 1$ 時比演算法 `A` 更慢，在 $n > 1000000$ 時比演算法 `C` 更慢。事實上，只要輸入資料大小 $n$ 足夠大，複雜度為“常數階”的演算法一定優於“線性階”的演算法，這正是時間增長趨勢的含義。
|
|
|
- **時間複雜度的推算方法更簡便**。顯然，執行平臺和計算操作型別都與演算法執行時間的增長趨勢無關。因此在時間複雜度分析中，我們可以簡單地將所有計算操作的執行時間視為相同的“單位時間”，從而將“計算操作執行時間統計”簡化為“計算操作數量統計”，這樣一來估算難度就大大降低了。
|
|
|
- **時間複雜度也存在一定的侷限性**。例如，儘管演算法 `A` 和 `C` 的時間複雜度相同，但實際執行時間差別很大。同樣，儘管演算法 `B` 的時間複雜度比 `C` 高，但在輸入資料大小 $n$ 較小時，演算法 `B` 明顯優於演算法 `C` 。在這些情況下，我們很難僅憑時間複雜度判斷演算法效率的高低。當然，儘管存在上述問題，複雜度分析仍然是評判演算法效率最有效且常用的方法。
|
|
|
|
|
|
## 2.3.2 函式漸近上界
|
|
|
|
|
|
給定一個輸入大小為 $n$ 的函式：
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
```python title=""
|
|
|
def algorithm(n: int):
|
|
|
a = 1 # +1
|
|
|
a = a + 1 # +1
|
|
|
a = a * 2 # +1
|
|
|
# 迴圈 n 次
|
|
|
for i in range(n): # +1
|
|
|
print(0) # +1
|
|
|
```
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
```cpp title=""
|
|
|
void algorithm(int n) {
|
|
|
int a = 1; // +1
|
|
|
a = a + 1; // +1
|
|
|
a = a * 2; // +1
|
|
|
// 迴圈 n 次
|
|
|
for (int i = 0; i < n; i++) { // +1（每輪都執行 i ++）
|
|
|
cout << 0 << endl; // +1
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
```java title=""
|
|
|
void algorithm(int n) {
|
|
|
int a = 1; // +1
|
|
|
a = a + 1; // +1
|
|
|
a = a * 2; // +1
|
|
|
// 迴圈 n 次
|
|
|
for (int i = 0; i < n; i++) { // +1（每輪都執行 i ++）
|
|
|
System.out.println(0); // +1
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
```csharp title=""
|
|
|
void Algorithm(int n) {
|
|
|
int a = 1; // +1
|
|
|
a = a + 1; // +1
|
|
|
a = a * 2; // +1
|
|
|
// 迴圈 n 次
|
|
|
for (int i = 0; i < n; i++) { // +1（每輪都執行 i ++）
|
|
|
Console.WriteLine(0); // +1
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
```go title=""
|
|
|
func algorithm(n int) {
|
|
|
a := 1 // +1
|
|
|
a = a + 1 // +1
|
|
|
a = a * 2 // +1
|
|
|
// 迴圈 n 次
|
|
|
for i := 0; i < n; i++ { // +1
|
|
|
fmt.Println(a) // +1
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
```swift title=""
|
|
|
func algorithm(n: Int) {
|
|
|
var a = 1 // +1
|
|
|
a = a + 1 // +1
|
|
|
a = a * 2 // +1
|
|
|
// 迴圈 n 次
|
|
|
for _ in 0 ..< n { // +1
|
|
|
print(0) // +1
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
```javascript title=""
|
|
|
function algorithm(n) {
|
|
|
var a = 1; // +1
|
|
|
a += 1; // +1
|
|
|
a *= 2; // +1
|
|
|
// 迴圈 n 次
|
|
|
for(let i = 0; i < n; i++){ // +1（每輪都執行 i ++）
|
|
|
console.log(0); // +1
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
```typescript title=""
|
|
|
function algorithm(n: number): void{
|
|
|
var a: number = 1; // +1
|
|
|
a += 1; // +1
|
|
|
a *= 2; // +1
|
|
|
// 迴圈 n 次
|
|
|
for(let i = 0; i < n; i++){ // +1（每輪都執行 i ++）
|
|
|
console.log(0); // +1
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
```dart title=""
|
|
|
void algorithm(int n) {
|
|
|
int a = 1; // +1
|
|
|
a = a + 1; // +1
|
|
|
a = a * 2; // +1
|
|
|
// 迴圈 n 次
|
|
|
for (int i = 0; i < n; i++) { // +1（每輪都執行 i ++）
|
|
|
print(0); // +1
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
```rust title=""
|
|
|
fn algorithm(n: i32) {
|
|
|
let mut a = 1; // +1
|
|
|
a = a + 1; // +1
|
|
|
a = a * 2; // +1
|
|
|
|
|
|
// 迴圈 n 次
|
|
|
for _ in 0..n { // +1（每輪都執行 i ++）
|
|
|
println!("{}", 0); // +1
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
```c title=""
|
|
|
void algorithm(int n) {
|
|
|
int a = 1; // +1
|
|
|
a = a + 1; // +1
|
|
|
a = a * 2; // +1
|
|
|
// 迴圈 n 次
|
|
|
for (int i = 0; i < n; i++) { // +1（每輪都執行 i ++）
|
|
|
printf("%d", 0); // +1
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Kotlin"
|
|
|
|
|
|
```kotlin title=""
|
|
|
fun algorithm(n: Int) {
|
|
|
var a = 1 // +1
|
|
|
a = a + 1 // +1
|
|
|
a = a * 2 // +1
|
|
|
// 迴圈 n 次
|
|
|
for (i in 0..<n) { // +1（每輪都執行 i ++）
|
|
|
println(0) // +1
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Ruby"
|
|
|
|
|
|
```ruby title=""
|
|
|
def algorithm(n)
|
|
|
a = 1 # +1
|
|
|
a = a + 1 # +1
|
|
|
a = a * 2 # +1
|
|
|
# 迴圈 n 次
|
|
|
(0...n).each do # +1
|
|
|
puts 0 # +1
|
|
|
end
|
|
|
end
|
|
|
```
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
```zig title=""
|
|
|
fn algorithm(n: usize) void {
|
|
|
var a: i32 = 1; // +1
|
|
|
a += 1; // +1
|
|
|
a *= 2; // +1
|
|
|
// 迴圈 n 次
|
|
|
for (0..n) |_| { // +1（每輪都執行 i ++）
|
|
|
std.debug.print("{}\n", .{0}); // +1
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
設演算法的操作數量是一個關於輸入資料大小 $n$ 的函式，記為 $T(n)$ ，則以上函式的操作數量為：
|
|
|
|
|
|
$$
|
|
|
T(n) = 3 + 2n
|
|
|
$$
|
|
|
|
|
|
$T(n)$ 是一次函式，說明其執行時間的增長趨勢是線性的，因此它的時間複雜度是線性階。
|
|
|
|
|
|
我們將線性階的時間複雜度記為 $O(n)$ ，這個數學符號稱為<u>大 $O$ 記號（big-$O$ notation）</u>，表示函式 $T(n)$ 的<u>漸近上界（asymptotic upper bound）</u>。
|
|
|
|
|
|
時間複雜度分析本質上是計算“操作數量 $T(n)$”的漸近上界，它具有明確的數學定義。
|
|
|
|
|
|
!!! abstract "函式漸近上界"
|
|
|
|
|
|
若存在正實數 $c$ 和實數 $n_0$ ，使得對於所有的 $n > n_0$ ，均有 $T(n) \leq c \cdot f(n)$ ，則可認為 $f(n)$ 給出了 $T(n)$ 的一個漸近上界，記為 $T(n) = O(f(n))$ 。
|
|
|
|
|
|
如圖 2-8 所示，計算漸近上界就是尋找一個函式 $f(n)$ ，使得當 $n$ 趨向於無窮大時，$T(n)$ 和 $f(n)$ 處於相同的增長級別，僅相差一個常數項 $c$ 的倍數。
|
|
|
|
|
|
{ class="animation-figure" }
|
|
|
|
|
|
<p align="center"> 圖 2-8 函式的漸近上界 </p>
|
|
|
|
|
|
## 2.3.3 推算方法
|
|
|
|
|
|
漸近上界的數學味兒有點重，如果你感覺沒有完全理解，也無須擔心。我們可以先掌握推算方法，在不斷的實踐中，就可以逐漸領悟其數學意義。
|
|
|
|
|
|
根據定義，確定 $f(n)$ 之後，我們便可得到時間複雜度 $O(f(n))$ 。那麼如何確定漸近上界 $f(n)$ 呢？總體分為兩步：首先統計操作數量，然後判斷漸近上界。
|
|
|
|
|
|
### 1. 第一步：統計操作數量
|
|
|
|
|
|
針對程式碼，逐行從上到下計算即可。然而，由於上述 $c \cdot f(n)$ 中的常數項 $c$ 可以取任意大小，**因此操作數量 $T(n)$ 中的各種係數、常數項都可以忽略**。根據此原則，可以總結出以下計數簡化技巧。
|
|
|
|
|
|
1. **忽略 $T(n)$ 中的常數項**。因為它們都與 $n$ 無關，所以對時間複雜度不產生影響。
|
|
|
2. **省略所有係數**。例如，迴圈 $2n$ 次、$5n + 1$ 次等，都可以簡化記為 $n$ 次，因為 $n$ 前面的係數對時間複雜度沒有影響。
|
|
|
3. **迴圈巢狀時使用乘法**。總操作數量等於外層迴圈和內層迴圈操作數量之積，每一層迴圈依然可以分別套用第 `1.` 點和第 `2.` 點的技巧。
|
|
|
|
|
|
給定一個函式，我們可以用上述技巧來統計操作數量：
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
```python title=""
|
|
|
def algorithm(n: int):
|
|
|
a = 1 # +0（技巧 1）
|
|
|
a = a + n # +0（技巧 1）
|
|
|
# +n（技巧 2）
|
|
|
for i in range(5 * n + 1):
|
|
|
print(0)
|
|
|
# +n*n（技巧 3）
|
|
|
for i in range(2 * n):
|
|
|
for j in range(n + 1):
|
|
|
print(0)
|
|
|
```
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
```cpp title=""
|
|
|
void algorithm(int n) {
|
|
|
int a = 1; // +0（技巧 1）
|
|
|
a = a + n; // +0（技巧 1）
|
|
|
// +n（技巧 2）
|
|
|
for (int i = 0; i < 5 * n + 1; i++) {
|
|
|
cout << 0 << endl;
|
|
|
}
|
|
|
// +n*n（技巧 3）
|
|
|
for (int i = 0; i < 2 * n; i++) {
|
|
|
for (int j = 0; j < n + 1; j++) {
|
|
|
cout << 0 << endl;
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
```java title=""
|
|
|
void algorithm(int n) {
|
|
|
int a = 1; // +0（技巧 1）
|
|
|
a = a + n; // +0（技巧 1）
|
|
|
// +n（技巧 2）
|
|
|
for (int i = 0; i < 5 * n + 1; i++) {
|
|
|
System.out.println(0);
|
|
|
}
|
|
|
// +n*n（技巧 3）
|
|
|
for (int i = 0; i < 2 * n; i++) {
|
|
|
for (int j = 0; j < n + 1; j++) {
|
|
|
System.out.println(0);
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
```csharp title=""
|
|
|
void Algorithm(int n) {
|
|
|
int a = 1; // +0（技巧 1）
|
|
|
a = a + n; // +0（技巧 1）
|
|
|
// +n（技巧 2）
|
|
|
for (int i = 0; i < 5 * n + 1; i++) {
|
|
|
Console.WriteLine(0);
|
|
|
}
|
|
|
// +n*n（技巧 3）
|
|
|
for (int i = 0; i < 2 * n; i++) {
|
|
|
for (int j = 0; j < n + 1; j++) {
|
|
|
Console.WriteLine(0);
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
```go title=""
|
|
|
func algorithm(n int) {
|
|
|
a := 1 // +0（技巧 1）
|
|
|
a = a + n // +0（技巧 1）
|
|
|
// +n（技巧 2）
|
|
|
for i := 0; i < 5 * n + 1; i++ {
|
|
|
fmt.Println(0)
|
|
|
}
|
|
|
// +n*n（技巧 3）
|
|
|
for i := 0; i < 2 * n; i++ {
|
|
|
for j := 0; j < n + 1; j++ {
|
|
|
fmt.Println(0)
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
```swift title=""
|
|
|
func algorithm(n: Int) {
|
|
|
var a = 1 // +0（技巧 1）
|
|
|
a = a + n // +0（技巧 1）
|
|
|
// +n（技巧 2）
|
|
|
for _ in 0 ..< (5 * n + 1) {
|
|
|
print(0)
|
|
|
}
|
|
|
// +n*n（技巧 3）
|
|
|
for _ in 0 ..< (2 * n) {
|
|
|
for _ in 0 ..< (n + 1) {
|
|
|
print(0)
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
```javascript title=""
|
|
|
function algorithm(n) {
|
|
|
let a = 1; // +0（技巧 1）
|
|
|
a = a + n; // +0（技巧 1）
|
|
|
// +n（技巧 2）
|
|
|
for (let i = 0; i < 5 * n + 1; i++) {
|
|
|
console.log(0);
|
|
|
}
|
|
|
// +n*n（技巧 3）
|
|
|
for (let i = 0; i < 2 * n; i++) {
|
|
|
for (let j = 0; j < n + 1; j++) {
|
|
|
console.log(0);
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
```typescript title=""
|
|
|
function algorithm(n: number): void {
|
|
|
let a = 1; // +0（技巧 1）
|
|
|
a = a + n; // +0（技巧 1）
|
|
|
// +n（技巧 2）
|
|
|
for (let i = 0; i < 5 * n + 1; i++) {
|
|
|
console.log(0);
|
|
|
}
|
|
|
// +n*n（技巧 3）
|
|
|
for (let i = 0; i < 2 * n; i++) {
|
|
|
for (let j = 0; j < n + 1; j++) {
|
|
|
console.log(0);
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
```dart title=""
|
|
|
void algorithm(int n) {
|
|
|
int a = 1; // +0（技巧 1）
|
|
|
a = a + n; // +0（技巧 1）
|
|
|
// +n（技巧 2）
|
|
|
for (int i = 0; i < 5 * n + 1; i++) {
|
|
|
print(0);
|
|
|
}
|
|
|
// +n*n（技巧 3）
|
|
|
for (int i = 0; i < 2 * n; i++) {
|
|
|
for (int j = 0; j < n + 1; j++) {
|
|
|
print(0);
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
```rust title=""
|
|
|
fn algorithm(n: i32) {
|
|
|
let mut a = 1; // +0（技巧 1）
|
|
|
a = a + n; // +0（技巧 1）
|
|
|
|
|
|
// +n（技巧 2）
|
|
|
for i in 0..(5 * n + 1) {
|
|
|
println!("{}", 0);
|
|
|
}
|
|
|
|
|
|
// +n*n（技巧 3）
|
|
|
for i in 0..(2 * n) {
|
|
|
for j in 0..(n + 1) {
|
|
|
println!("{}", 0);
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
```c title=""
|
|
|
void algorithm(int n) {
|
|
|
int a = 1; // +0（技巧 1）
|
|
|
a = a + n; // +0（技巧 1）
|
|
|
// +n（技巧 2）
|
|
|
for (int i = 0; i < 5 * n + 1; i++) {
|
|
|
printf("%d", 0);
|
|
|
}
|
|
|
// +n*n（技巧 3）
|
|
|
for (int i = 0; i < 2 * n; i++) {
|
|
|
for (int j = 0; j < n + 1; j++) {
|
|
|
printf("%d", 0);
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Kotlin"
|
|
|
|
|
|
```kotlin title=""
|
|
|
fun algorithm(n: Int) {
|
|
|
var a = 1 // +0（技巧 1）
|
|
|
a = a + n // +0（技巧 1）
|
|
|
// +n（技巧 2）
|
|
|
for (i in 0..<5 * n + 1) {
|
|
|
println(0)
|
|
|
}
|
|
|
// +n*n（技巧 3）
|
|
|
for (i in 0..<2 * n) {
|
|
|
for (j in 0..<n + 1) {
|
|
|
println(0)
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Ruby"
|
|
|
|
|
|
```ruby title=""
|
|
|
def algorithm(n)
|
|
|
a = 1 # +0（技巧 1）
|
|
|
a = a + n # +0（技巧 1）
|
|
|
# +n（技巧 2）
|
|
|
(0...(5 * n + 1)).each do { puts 0 }
|
|
|
# +n*n（技巧 3）
|
|
|
(0...(2 * n)).each do
|
|
|
(0...(n + 1)).each do { puts 0 }
|
|
|
end
|
|
|
end
|
|
|
```
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
```zig title=""
|
|
|
fn algorithm(n: usize) void {
|
|
|
var a: i32 = 1; // +0（技巧 1）
|
|
|
a = a + @as(i32, @intCast(n)); // +0（技巧 1）
|
|
|
|
|
|
// +n（技巧 2）
|
|
|
for(0..(5 * n + 1)) |_| {
|
|
|
std.debug.print("{}\n", .{0});
|
|
|
}
|
|
|
|
|
|
// +n*n（技巧 3）
|
|
|
for(0..(2 * n)) |_| {
|
|
|
for(0..(n + 1)) |_| {
|
|
|
std.debug.print("{}\n", .{0});
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
```
|
|
|
|
|
|
以下公式展示了使用上述技巧前後的統計結果，兩者推算出的時間複雜度都為 $O(n^2)$ 。
|
|
|
|
|
|
$$
|
|
|
\begin{aligned}
|
|
|
T(n) & = 2n(n + 1) + (5n + 1) + 2 & \text{完整統計 (-.-|||)} \newline
|
|
|
& = 2n^2 + 7n + 3 \newline
|
|
|
T(n) & = n^2 + n & \text{偷懶統計 (o.O)}
|
|
|
\end{aligned}
|
|
|
$$
|
|
|
|
|
|
### 2. 第二步：判斷漸近上界
|
|
|
|
|
|
**時間複雜度由 $T(n)$ 中最高階的項來決定**。這是因為在 $n$ 趨於無窮大時，最高階的項將發揮主導作用，其他項的影響都可以忽略。
|
|
|
|
|
|
表 2-2 展示了一些例子，其中一些誇張的值是為了強調“係數無法撼動階數”這一結論。當 $n$ 趨於無窮大時，這些常數變得無足輕重。
|
|
|
|
|
|
<p align="center"> 表 2-2 不同操作數量對應的時間複雜度 </p>
|
|
|
|
|
|
<div class="center-table" markdown>
|
|
|
|
|
|
| 操作數量 $T(n)$ | 時間複雜度 $O(f(n))$ |
|
|
|
| ---------------------- | -------------------- |
|
|
|
| $100000$ | $O(1)$ |
|
|
|
| $3n + 2$ | $O(n)$ |
|
|
|
| $2n^2 + 3n + 2$ | $O(n^2)$ |
|
|
|
| $n^3 + 10000n^2$ | $O(n^3)$ |
|
|
|
| $2^n + 10000n^{10000}$ | $O(2^n)$ |
|
|
|
|
|
|
</div>
|
|
|
|
|
|
## 2.3.4 常見型別
|
|
|
|
|
|
設輸入資料大小為 $n$ ，常見的時間複雜度型別如圖 2-9 所示（按照從低到高的順序排列）。
|
|
|
|
|
|
$$
|
|
|
\begin{aligned}
|
|
|
O(1) < O(\log n) < O(n) < O(n \log n) < O(n^2) < O(2^n) < O(n!) \newline
|
|
|
\text{常數階} < \text{對數階} < \text{線性階} < \text{線性對數階} < \text{平方階} < \text{指數階} < \text{階乘階}
|
|
|
\end{aligned}
|
|
|
$$
|
|
|
|
|
|
{ class="animation-figure" }
|
|
|
|
|
|
<p align="center"> 圖 2-9 常見的時間複雜度型別 </p>
|
|
|
|
|
|
### 1. 常數階 $O(1)$ {data-toc-label="1. 常數階"}
|
|
|
|
|
|
常數階的操作數量與輸入資料大小 $n$ 無關，即不隨著 $n$ 的變化而變化。
|
|
|
|
|
|
在以下函式中，儘管操作數量 `size` 可能很大，但由於其與輸入資料大小 $n$ 無關，因此時間複雜度仍為 $O(1)$ ：
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
```python title="time_complexity.py"
|
|
|
def constant(n: int) -> int:
|
|
|
"""常數階"""
|
|
|
count = 0
|
|
|
size = 100000
|
|
|
for _ in range(size):
|
|
|
count += 1
|
|
|
return count
|
|
|
```
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
```cpp title="time_complexity.cpp"
|
|
|
/* 常數階 */
|
|
|
int constant(int n) {
|
|
|
int count = 0;
|
|
|
int size = 100000;
|
|
|
for (int i = 0; i < size; i++)
|
|
|
count++;
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
```java title="time_complexity.java"
|
|
|
/* 常數階 */
|
|
|
int constant(int n) {
|
|
|
int count = 0;
|
|
|
int size = 100000;
|
|
|
for (int i = 0; i < size; i++)
|
|
|
count++;
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
```csharp title="time_complexity.cs"
|
|
|
/* 常數階 */
|
|
|
int Constant(int n) {
|
|
|
int count = 0;
|
|
|
int size = 100000;
|
|
|
for (int i = 0; i < size; i++)
|
|
|
count++;
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
```go title="time_complexity.go"
|
|
|
/* 常數階 */
|
|
|
func constant(n int) int {
|
|
|
count := 0
|
|
|
size := 100000
|
|
|
for i := 0; i < size; i++ {
|
|
|
count++
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
```swift title="time_complexity.swift"
|
|
|
/* 常數階 */
|
|
|
func constant(n: Int) -> Int {
|
|
|
var count = 0
|
|
|
let size = 100_000
|
|
|
for _ in 0 ..< size {
|
|
|
count += 1
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
```javascript title="time_complexity.js"
|
|
|
/* 常數階 */
|
|
|
function constant(n) {
|
|
|
let count = 0;
|
|
|
const size = 100000;
|
|
|
for (let i = 0; i < size; i++) count++;
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
```typescript title="time_complexity.ts"
|
|
|
/* 常數階 */
|
|
|
function constant(n: number): number {
|
|
|
let count = 0;
|
|
|
const size = 100000;
|
|
|
for (let i = 0; i < size; i++) count++;
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
```dart title="time_complexity.dart"
|
|
|
/* 常數階 */
|
|
|
int constant(int n) {
|
|
|
int count = 0;
|
|
|
int size = 100000;
|
|
|
for (var i = 0; i < size; i++) {
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
```rust title="time_complexity.rs"
|
|
|
/* 常數階 */
|
|
|
fn constant(n: i32) -> i32 {
|
|
|
_ = n;
|
|
|
let mut count = 0;
|
|
|
let size = 100_000;
|
|
|
for _ in 0..size {
|
|
|
count += 1;
|
|
|
}
|
|
|
count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
```c title="time_complexity.c"
|
|
|
/* 常數階 */
|
|
|
int constant(int n) {
|
|
|
int count = 0;
|
|
|
int size = 100000;
|
|
|
int i = 0;
|
|
|
for (int i = 0; i < size; i++) {
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Kotlin"
|
|
|
|
|
|
```kotlin title="time_complexity.kt"
|
|
|
/* 常數階 */
|
|
|
fun constant(n: Int): Int {
|
|
|
var count = 0
|
|
|
val size = 10_0000
|
|
|
for (i in 0..<size)
|
|
|
count++
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Ruby"
|
|
|
|
|
|
```ruby title="time_complexity.rb"
|
|
|
### 常數階 ###
|
|
|
def constant(n)
|
|
|
count = 0
|
|
|
size = 100000
|
|
|
|
|
|
(0...size).each { count += 1 }
|
|
|
|
|
|
count
|
|
|
end
|
|
|
```
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
```zig title="time_complexity.zig"
|
|
|
// 常數階
|
|
|
fn constant(n: i32) i32 {
|
|
|
_ = n;
|
|
|
var count: i32 = 0;
|
|
|
const size: i32 = 100_000;
|
|
|
var i: i32 = 0;
|
|
|
while(i<size) : (i += 1) {
|
|
|
count += 1;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
??? pythontutor "視覺化執行"
|
|
|
|
|
|
<div style="height: 459px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20constant%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%B8%B8%E6%95%B0%E9%98%B6%22%22%22%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20size%20%3D%2010%0A%20%20%20%20for%20_%20in%20range%28size%29%3A%0A%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20constant%28n%29%0A%20%20%20%20print%28%22%E5%B8%B8%E6%95%B0%E9%98%B6%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
|
|
|
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20constant%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%B8%B8%E6%95%B0%E9%98%B6%22%22%22%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20size%20%3D%2010%0A%20%20%20%20for%20_%20in%20range%28size%29%3A%0A%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20constant%28n%29%0A%20%20%20%20print%28%22%E5%B8%B8%E6%95%B0%E9%98%B6%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全螢幕觀看 ></a></div>
|
|
|
|
|
|
### 2. 線性階 $O(n)$ {data-toc-label="2. 線性階"}
|
|
|
|
|
|
線性階的操作數量相對於輸入資料大小 $n$ 以線性級別增長。線性階通常出現在單層迴圈中：
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
```python title="time_complexity.py"
|
|
|
def linear(n: int) -> int:
|
|
|
"""線性階"""
|
|
|
count = 0
|
|
|
for _ in range(n):
|
|
|
count += 1
|
|
|
return count
|
|
|
```
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
```cpp title="time_complexity.cpp"
|
|
|
/* 線性階 */
|
|
|
int linear(int n) {
|
|
|
int count = 0;
|
|
|
for (int i = 0; i < n; i++)
|
|
|
count++;
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
```java title="time_complexity.java"
|
|
|
/* 線性階 */
|
|
|
int linear(int n) {
|
|
|
int count = 0;
|
|
|
for (int i = 0; i < n; i++)
|
|
|
count++;
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
```csharp title="time_complexity.cs"
|
|
|
/* 線性階 */
|
|
|
int Linear(int n) {
|
|
|
int count = 0;
|
|
|
for (int i = 0; i < n; i++)
|
|
|
count++;
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
```go title="time_complexity.go"
|
|
|
/* 線性階 */
|
|
|
func linear(n int) int {
|
|
|
count := 0
|
|
|
for i := 0; i < n; i++ {
|
|
|
count++
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
```swift title="time_complexity.swift"
|
|
|
/* 線性階 */
|
|
|
func linear(n: Int) -> Int {
|
|
|
var count = 0
|
|
|
for _ in 0 ..< n {
|
|
|
count += 1
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
```javascript title="time_complexity.js"
|
|
|
/* 線性階 */
|
|
|
function linear(n) {
|
|
|
let count = 0;
|
|
|
for (let i = 0; i < n; i++) count++;
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
```typescript title="time_complexity.ts"
|
|
|
/* 線性階 */
|
|
|
function linear(n: number): number {
|
|
|
let count = 0;
|
|
|
for (let i = 0; i < n; i++) count++;
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
```dart title="time_complexity.dart"
|
|
|
/* 線性階 */
|
|
|
int linear(int n) {
|
|
|
int count = 0;
|
|
|
for (var i = 0; i < n; i++) {
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
```rust title="time_complexity.rs"
|
|
|
/* 線性階 */
|
|
|
fn linear(n: i32) -> i32 {
|
|
|
let mut count = 0;
|
|
|
for _ in 0..n {
|
|
|
count += 1;
|
|
|
}
|
|
|
count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
```c title="time_complexity.c"
|
|
|
/* 線性階 */
|
|
|
int linear(int n) {
|
|
|
int count = 0;
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Kotlin"
|
|
|
|
|
|
```kotlin title="time_complexity.kt"
|
|
|
/* 線性階 */
|
|
|
fun linear(n: Int): Int {
|
|
|
var count = 0
|
|
|
// 迴圈次數與陣列長度成正比
|
|
|
for (i in 0..<n)
|
|
|
count++
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Ruby"
|
|
|
|
|
|
```ruby title="time_complexity.rb"
|
|
|
### 線性階 ###
|
|
|
def linear(n)
|
|
|
count = 0
|
|
|
(0...n).each { count += 1 }
|
|
|
count
|
|
|
end
|
|
|
```
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
```zig title="time_complexity.zig"
|
|
|
// 線性階
|
|
|
fn linear(n: i32) i32 {
|
|
|
var count: i32 = 0;
|
|
|
var i: i32 = 0;
|
|
|
while (i < n) : (i += 1) {
|
|
|
count += 1;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
??? pythontutor "視覺化執行"
|
|
|
|
|
|
<div style="height: 441px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20linear%28n%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%BA%BF%E6%80%A7%E9%98%B6%22%22%22%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20for%20_%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20linear%28n%29%0A%20%20%20%20print%28%22%E7%BA%BF%E6%80%A7%E9%98%B6%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
|
|
|
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20linear%28n%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%BA%BF%E6%80%A7%E9%98%B6%22%22%22%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20for%20_%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20linear%28n%29%0A%20%20%20%20print%28%22%E7%BA%BF%E6%80%A7%E9%98%B6%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全螢幕觀看 ></a></div>
|
|
|
|
|
|
走訪陣列和走訪鏈結串列等操作的時間複雜度均為 $O(n)$ ，其中 $n$ 為陣列或鏈結串列的長度：
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
```python title="time_complexity.py"
|
|
|
def array_traversal(nums: list[int]) -> int:
|
|
|
"""線性階（走訪陣列）"""
|
|
|
count = 0
|
|
|
# 迴圈次數與陣列長度成正比
|
|
|
for num in nums:
|
|
|
count += 1
|
|
|
return count
|
|
|
```
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
```cpp title="time_complexity.cpp"
|
|
|
/* 線性階（走訪陣列） */
|
|
|
int arrayTraversal(vector<int> &nums) {
|
|
|
int count = 0;
|
|
|
// 迴圈次數與陣列長度成正比
|
|
|
for (int num : nums) {
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
```java title="time_complexity.java"
|
|
|
/* 線性階（走訪陣列） */
|
|
|
int arrayTraversal(int[] nums) {
|
|
|
int count = 0;
|
|
|
// 迴圈次數與陣列長度成正比
|
|
|
for (int num : nums) {
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
```csharp title="time_complexity.cs"
|
|
|
/* 線性階（走訪陣列） */
|
|
|
int ArrayTraversal(int[] nums) {
|
|
|
int count = 0;
|
|
|
// 迴圈次數與陣列長度成正比
|
|
|
foreach (int num in nums) {
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
```go title="time_complexity.go"
|
|
|
/* 線性階（走訪陣列） */
|
|
|
func arrayTraversal(nums []int) int {
|
|
|
count := 0
|
|
|
// 迴圈次數與陣列長度成正比
|
|
|
for range nums {
|
|
|
count++
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
```swift title="time_complexity.swift"
|
|
|
/* 線性階（走訪陣列） */
|
|
|
func arrayTraversal(nums: [Int]) -> Int {
|
|
|
var count = 0
|
|
|
// 迴圈次數與陣列長度成正比
|
|
|
for _ in nums {
|
|
|
count += 1
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
```javascript title="time_complexity.js"
|
|
|
/* 線性階（走訪陣列） */
|
|
|
function arrayTraversal(nums) {
|
|
|
let count = 0;
|
|
|
// 迴圈次數與陣列長度成正比
|
|
|
for (let i = 0; i < nums.length; i++) {
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
```typescript title="time_complexity.ts"
|
|
|
/* 線性階（走訪陣列） */
|
|
|
function arrayTraversal(nums: number[]): number {
|
|
|
let count = 0;
|
|
|
// 迴圈次數與陣列長度成正比
|
|
|
for (let i = 0; i < nums.length; i++) {
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
```dart title="time_complexity.dart"
|
|
|
/* 線性階（走訪陣列） */
|
|
|
int arrayTraversal(List<int> nums) {
|
|
|
int count = 0;
|
|
|
// 迴圈次數與陣列長度成正比
|
|
|
for (var _num in nums) {
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
```rust title="time_complexity.rs"
|
|
|
/* 線性階（走訪陣列） */
|
|
|
fn array_traversal(nums: &[i32]) -> i32 {
|
|
|
let mut count = 0;
|
|
|
// 迴圈次數與陣列長度成正比
|
|
|
for _ in nums {
|
|
|
count += 1;
|
|
|
}
|
|
|
count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
```c title="time_complexity.c"
|
|
|
/* 線性階（走訪陣列） */
|
|
|
int arrayTraversal(int *nums, int n) {
|
|
|
int count = 0;
|
|
|
// 迴圈次數與陣列長度成正比
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Kotlin"
|
|
|
|
|
|
```kotlin title="time_complexity.kt"
|
|
|
/* 線性階（走訪陣列） */
|
|
|
fun arrayTraversal(nums: IntArray): Int {
|
|
|
var count = 0
|
|
|
// 迴圈次數與陣列長度成正比
|
|
|
for (num in nums) {
|
|
|
count++
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Ruby"
|
|
|
|
|
|
```ruby title="time_complexity.rb"
|
|
|
### 線性階（走訪陣列）###
|
|
|
def array_traversal(nums)
|
|
|
count = 0
|
|
|
|
|
|
# 迴圈次數與陣列長度成正比
|
|
|
for num in nums
|
|
|
count += 1
|
|
|
end
|
|
|
|
|
|
count
|
|
|
end
|
|
|
```
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
```zig title="time_complexity.zig"
|
|
|
// 線性階（走訪陣列）
|
|
|
fn arrayTraversal(nums: []i32) i32 {
|
|
|
var count: i32 = 0;
|
|
|
// 迴圈次數與陣列長度成正比
|
|
|
for (nums) |_| {
|
|
|
count += 1;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
??? pythontutor "視覺化執行"
|
|
|
|
|
|
<div style="height: 459px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20array_traversal%28nums%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%BA%BF%E6%80%A7%E9%98%B6%EF%BC%88%E9%81%8D%E5%8E%86%E6%95%B0%E7%BB%84%EF%BC%89%22%22%22%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E6%AC%A1%E6%95%B0%E4%B8%8E%E6%95%B0%E7%BB%84%E9%95%BF%E5%BA%A6%E6%88%90%E6%AD%A3%E6%AF%94%0A%20%20%20%20for%20num%20in%20nums%3A%0A%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20array_traversal%28%5B0%5D%20*%20n%29%0A%20%20%20%20print%28%22%E7%BA%BF%E6%80%A7%E9%98%B6%EF%BC%88%E9%81%8D%E5%8E%86%E6%95%B0%E7%BB%84%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
|
|
|
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20array_traversal%28nums%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%BA%BF%E6%80%A7%E9%98%B6%EF%BC%88%E9%81%8D%E5%8E%86%E6%95%B0%E7%BB%84%EF%BC%89%22%22%22%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E6%AC%A1%E6%95%B0%E4%B8%8E%E6%95%B0%E7%BB%84%E9%95%BF%E5%BA%A6%E6%88%90%E6%AD%A3%E6%AF%94%0A%20%20%20%20for%20num%20in%20nums%3A%0A%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20array_traversal%28%5B0%5D%20*%20n%29%0A%20%20%20%20print%28%22%E7%BA%BF%E6%80%A7%E9%98%B6%EF%BC%88%E9%81%8D%E5%8E%86%E6%95%B0%E7%BB%84%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全螢幕觀看 ></a></div>
|
|
|
|
|
|
值得注意的是，**輸入資料大小 $n$ 需根據輸入資料的型別來具體確定**。比如在第一個示例中，變數 $n$ 為輸入資料大小；在第二個示例中，陣列長度 $n$ 為資料大小。
|
|
|
|
|
|
### 3. 平方階 $O(n^2)$ {data-toc-label="3. 平方階"}
|
|
|
|
|
|
平方階的操作數量相對於輸入資料大小 $n$ 以平方級別增長。平方階通常出現在巢狀迴圈中，外層迴圈和內層迴圈的時間複雜度都為 $O(n)$ ，因此總體的時間複雜度為 $O(n^2)$ ：
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
```python title="time_complexity.py"
|
|
|
def quadratic(n: int) -> int:
|
|
|
"""平方階"""
|
|
|
count = 0
|
|
|
# 迴圈次數與資料大小 n 成平方關係
|
|
|
for i in range(n):
|
|
|
for j in range(n):
|
|
|
count += 1
|
|
|
return count
|
|
|
```
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
```cpp title="time_complexity.cpp"
|
|
|
/* 平方階 */
|
|
|
int quadratic(int n) {
|
|
|
int count = 0;
|
|
|
// 迴圈次數與資料大小 n 成平方關係
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
for (int j = 0; j < n; j++) {
|
|
|
count++;
|
|
|
}
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
```java title="time_complexity.java"
|
|
|
/* 平方階 */
|
|
|
int quadratic(int n) {
|
|
|
int count = 0;
|
|
|
// 迴圈次數與資料大小 n 成平方關係
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
for (int j = 0; j < n; j++) {
|
|
|
count++;
|
|
|
}
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
```csharp title="time_complexity.cs"
|
|
|
/* 平方階 */
|
|
|
int Quadratic(int n) {
|
|
|
int count = 0;
|
|
|
// 迴圈次數與資料大小 n 成平方關係
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
for (int j = 0; j < n; j++) {
|
|
|
count++;
|
|
|
}
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
```go title="time_complexity.go"
|
|
|
/* 平方階 */
|
|
|
func quadratic(n int) int {
|
|
|
count := 0
|
|
|
// 迴圈次數與資料大小 n 成平方關係
|
|
|
for i := 0; i < n; i++ {
|
|
|
for j := 0; j < n; j++ {
|
|
|
count++
|
|
|
}
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
```swift title="time_complexity.swift"
|
|
|
/* 平方階 */
|
|
|
func quadratic(n: Int) -> Int {
|
|
|
var count = 0
|
|
|
// 迴圈次數與資料大小 n 成平方關係
|
|
|
for _ in 0 ..< n {
|
|
|
for _ in 0 ..< n {
|
|
|
count += 1
|
|
|
}
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
```javascript title="time_complexity.js"
|
|
|
/* 平方階 */
|
|
|
function quadratic(n) {
|
|
|
let count = 0;
|
|
|
// 迴圈次數與資料大小 n 成平方關係
|
|
|
for (let i = 0; i < n; i++) {
|
|
|
for (let j = 0; j < n; j++) {
|
|
|
count++;
|
|
|
}
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
```typescript title="time_complexity.ts"
|
|
|
/* 平方階 */
|
|
|
function quadratic(n: number): number {
|
|
|
let count = 0;
|
|
|
// 迴圈次數與資料大小 n 成平方關係
|
|
|
for (let i = 0; i < n; i++) {
|
|
|
for (let j = 0; j < n; j++) {
|
|
|
count++;
|
|
|
}
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
```dart title="time_complexity.dart"
|
|
|
/* 平方階 */
|
|
|
int quadratic(int n) {
|
|
|
int count = 0;
|
|
|
// 迴圈次數與資料大小 n 成平方關係
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
for (int j = 0; j < n; j++) {
|
|
|
count++;
|
|
|
}
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
```rust title="time_complexity.rs"
|
|
|
/* 平方階 */
|
|
|
fn quadratic(n: i32) -> i32 {
|
|
|
let mut count = 0;
|
|
|
// 迴圈次數與資料大小 n 成平方關係
|
|
|
for _ in 0..n {
|
|
|
for _ in 0..n {
|
|
|
count += 1;
|
|
|
}
|
|
|
}
|
|
|
count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
```c title="time_complexity.c"
|
|
|
/* 平方階 */
|
|
|
int quadratic(int n) {
|
|
|
int count = 0;
|
|
|
// 迴圈次數與資料大小 n 成平方關係
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
for (int j = 0; j < n; j++) {
|
|
|
count++;
|
|
|
}
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Kotlin"
|
|
|
|
|
|
```kotlin title="time_complexity.kt"
|
|
|
/* 平方階 */
|
|
|
fun quadratic(n: Int): Int {
|
|
|
var count = 0
|
|
|
// 迴圈次數與資料大小 n 成平方關係
|
|
|
for (i in 0..<n) {
|
|
|
for (j in 0..<n) {
|
|
|
count++
|
|
|
}
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Ruby"
|
|
|
|
|
|
```ruby title="time_complexity.rb"
|
|
|
### 平方階 ###
|
|
|
def quadratic(n)
|
|
|
count = 0
|
|
|
|
|
|
# 迴圈次數與資料大小 n 成平方關係
|
|
|
for i in 0...n
|
|
|
for j in 0...n
|
|
|
count += 1
|
|
|
end
|
|
|
end
|
|
|
|
|
|
count
|
|
|
end
|
|
|
```
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
```zig title="time_complexity.zig"
|
|
|
// 平方階
|
|
|
fn quadratic(n: i32) i32 {
|
|
|
var count: i32 = 0;
|
|
|
var i: i32 = 0;
|
|
|
// 迴圈次數與資料大小 n 成平方關係
|
|
|
while (i < n) : (i += 1) {
|
|
|
var j: i32 = 0;
|
|
|
while (j < n) : (j += 1) {
|
|
|
count += 1;
|
|
|
}
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
??? pythontutor "視覺化執行"
|
|
|
|
|
|
<div style="height: 477px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20quadratic%28n%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%B9%B3%E6%96%B9%E9%98%B6%22%22%22%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E6%AC%A1%E6%95%B0%E4%B8%8E%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%E6%88%90%E5%B9%B3%E6%96%B9%E5%85%B3%E7%B3%BB%0A%20%20%20%20for%20i%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20quadratic%28n%29%0A%20%20%20%20print%28%22%E5%B9%B3%E6%96%B9%E9%98%B6%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
|
|
|
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20quadratic%28n%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%B9%B3%E6%96%B9%E9%98%B6%22%22%22%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E6%AC%A1%E6%95%B0%E4%B8%8E%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%E6%88%90%E5%B9%B3%E6%96%B9%E5%85%B3%E7%B3%BB%0A%20%20%20%20for%20i%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20quadratic%28n%29%0A%20%20%20%20print%28%22%E5%B9%B3%E6%96%B9%E9%98%B6%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全螢幕觀看 ></a></div>
|
|
|
|
|
|
圖 2-10 對比了常數階、線性階和平方階三種時間複雜度。
|
|
|
|
|
|
{ class="animation-figure" }
|
|
|
|
|
|
<p align="center"> 圖 2-10 常數階、線性階和平方階的時間複雜度 </p>
|
|
|
|
|
|
以泡沫排序為例，外層迴圈執行 $n - 1$ 次，內層迴圈執行 $n-1$、$n-2$、$\dots$、$2$、$1$ 次，平均為 $n / 2$ 次，因此時間複雜度為 $O((n - 1) n / 2) = O(n^2)$ ：
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
```python title="time_complexity.py"
|
|
|
def bubble_sort(nums: list[int]) -> int:
|
|
|
"""平方階（泡沫排序）"""
|
|
|
count = 0 # 計數器
|
|
|
# 外迴圈：未排序區間為 [0, i]
|
|
|
for i in range(len(nums) - 1, 0, -1):
|
|
|
# 內迴圈：將未排序區間 [0, i] 中的最大元素交換至該區間的最右端
|
|
|
for j in range(i):
|
|
|
if nums[j] > nums[j + 1]:
|
|
|
# 交換 nums[j] 與 nums[j + 1]
|
|
|
tmp: int = nums[j]
|
|
|
nums[j] = nums[j + 1]
|
|
|
nums[j + 1] = tmp
|
|
|
count += 3 # 元素交換包含 3 個單元操作
|
|
|
return count
|
|
|
```
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
```cpp title="time_complexity.cpp"
|
|
|
/* 平方階（泡沫排序） */
|
|
|
int bubbleSort(vector<int> &nums) {
|
|
|
int count = 0; // 計數器
|
|
|
// 外迴圈：未排序區間為 [0, i]
|
|
|
for (int i = nums.size() - 1; i > 0; i--) {
|
|
|
// 內迴圈：將未排序區間 [0, i] 中的最大元素交換至該區間的最右端
|
|
|
for (int j = 0; j < i; j++) {
|
|
|
if (nums[j] > nums[j + 1]) {
|
|
|
// 交換 nums[j] 與 nums[j + 1]
|
|
|
int tmp = nums[j];
|
|
|
nums[j] = nums[j + 1];
|
|
|
nums[j + 1] = tmp;
|
|
|
count += 3; // 元素交換包含 3 個單元操作
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
```java title="time_complexity.java"
|
|
|
/* 平方階（泡沫排序） */
|
|
|
int bubbleSort(int[] nums) {
|
|
|
int count = 0; // 計數器
|
|
|
// 外迴圈：未排序區間為 [0, i]
|
|
|
for (int i = nums.length - 1; i > 0; i--) {
|
|
|
// 內迴圈：將未排序區間 [0, i] 中的最大元素交換至該區間的最右端
|
|
|
for (int j = 0; j < i; j++) {
|
|
|
if (nums[j] > nums[j + 1]) {
|
|
|
// 交換 nums[j] 與 nums[j + 1]
|
|
|
int tmp = nums[j];
|
|
|
nums[j] = nums[j + 1];
|
|
|
nums[j + 1] = tmp;
|
|
|
count += 3; // 元素交換包含 3 個單元操作
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
```csharp title="time_complexity.cs"
|
|
|
/* 平方階（泡沫排序） */
|
|
|
int BubbleSort(int[] nums) {
|
|
|
int count = 0; // 計數器
|
|
|
// 外迴圈：未排序區間為 [0, i]
|
|
|
for (int i = nums.Length - 1; i > 0; i--) {
|
|
|
// 內迴圈：將未排序區間 [0, i] 中的最大元素交換至該區間的最右端
|
|
|
for (int j = 0; j < i; j++) {
|
|
|
if (nums[j] > nums[j + 1]) {
|
|
|
// 交換 nums[j] 與 nums[j + 1]
|
|
|
(nums[j + 1], nums[j]) = (nums[j], nums[j + 1]);
|
|
|
count += 3; // 元素交換包含 3 個單元操作
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
```go title="time_complexity.go"
|
|
|
/* 平方階（泡沫排序） */
|
|
|
func bubbleSort(nums []int) int {
|
|
|
count := 0 // 計數器
|
|
|
// 外迴圈：未排序區間為 [0, i]
|
|
|
for i := len(nums) - 1; i > 0; i-- {
|
|
|
// 內迴圈：將未排序區間 [0, i] 中的最大元素交換至該區間的最右端
|
|
|
for j := 0; j < i; j++ {
|
|
|
if nums[j] > nums[j+1] {
|
|
|
// 交換 nums[j] 與 nums[j + 1]
|
|
|
tmp := nums[j]
|
|
|
nums[j] = nums[j+1]
|
|
|
nums[j+1] = tmp
|
|
|
count += 3 // 元素交換包含 3 個單元操作
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
```swift title="time_complexity.swift"
|
|
|
/* 平方階（泡沫排序） */
|
|
|
func bubbleSort(nums: inout [Int]) -> Int {
|
|
|
var count = 0 // 計數器
|
|
|
// 外迴圈：未排序區間為 [0, i]
|
|
|
for i in nums.indices.dropFirst().reversed() {
|
|
|
// 內迴圈：將未排序區間 [0, i] 中的最大元素交換至該區間的最右端
|
|
|
for j in 0 ..< i {
|
|
|
if nums[j] > nums[j + 1] {
|
|
|
// 交換 nums[j] 與 nums[j + 1]
|
|
|
let tmp = nums[j]
|
|
|
nums[j] = nums[j + 1]
|
|
|
nums[j + 1] = tmp
|
|
|
count += 3 // 元素交換包含 3 個單元操作
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
```javascript title="time_complexity.js"
|
|
|
/* 平方階（泡沫排序） */
|
|
|
function bubbleSort(nums) {
|
|
|
let count = 0; // 計數器
|
|
|
// 外迴圈：未排序區間為 [0, i]
|
|
|
for (let i = nums.length - 1; i > 0; i--) {
|
|
|
// 內迴圈：將未排序區間 [0, i] 中的最大元素交換至該區間的最右端
|
|
|
for (let j = 0; j < i; j++) {
|
|
|
if (nums[j] > nums[j + 1]) {
|
|
|
// 交換 nums[j] 與 nums[j + 1]
|
|
|
let tmp = nums[j];
|
|
|
nums[j] = nums[j + 1];
|
|
|
nums[j + 1] = tmp;
|
|
|
count += 3; // 元素交換包含 3 個單元操作
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
```typescript title="time_complexity.ts"
|
|
|
/* 平方階（泡沫排序） */
|
|
|
function bubbleSort(nums: number[]): number {
|
|
|
let count = 0; // 計數器
|
|
|
// 外迴圈：未排序區間為 [0, i]
|
|
|
for (let i = nums.length - 1; i > 0; i--) {
|
|
|
// 內迴圈：將未排序區間 [0, i] 中的最大元素交換至該區間的最右端
|
|
|
for (let j = 0; j < i; j++) {
|
|
|
if (nums[j] > nums[j + 1]) {
|
|
|
// 交換 nums[j] 與 nums[j + 1]
|
|
|
let tmp = nums[j];
|
|
|
nums[j] = nums[j + 1];
|
|
|
nums[j + 1] = tmp;
|
|
|
count += 3; // 元素交換包含 3 個單元操作
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
```dart title="time_complexity.dart"
|
|
|
/* 平方階（泡沫排序） */
|
|
|
int bubbleSort(List<int> nums) {
|
|
|
int count = 0; // 計數器
|
|
|
// 外迴圈：未排序區間為 [0, i]
|
|
|
for (var i = nums.length - 1; i > 0; i--) {
|
|
|
// 內迴圈：將未排序區間 [0, i] 中的最大元素交換至該區間的最右端
|
|
|
for (var j = 0; j < i; j++) {
|
|
|
if (nums[j] > nums[j + 1]) {
|
|
|
// 交換 nums[j] 與 nums[j + 1]
|
|
|
int tmp = nums[j];
|
|
|
nums[j] = nums[j + 1];
|
|
|
nums[j + 1] = tmp;
|
|
|
count += 3; // 元素交換包含 3 個單元操作
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
```rust title="time_complexity.rs"
|
|
|
/* 平方階（泡沫排序） */
|
|
|
fn bubble_sort(nums: &mut [i32]) -> i32 {
|
|
|
let mut count = 0; // 計數器
|
|
|
|
|
|
// 外迴圈：未排序區間為 [0, i]
|
|
|
for i in (1..nums.len()).rev() {
|
|
|
// 內迴圈：將未排序區間 [0, i] 中的最大元素交換至該區間的最右端
|
|
|
for j in 0..i {
|
|
|
if nums[j] > nums[j + 1] {
|
|
|
// 交換 nums[j] 與 nums[j + 1]
|
|
|
let tmp = nums[j];
|
|
|
nums[j] = nums[j + 1];
|
|
|
nums[j + 1] = tmp;
|
|
|
count += 3; // 元素交換包含 3 個單元操作
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
```c title="time_complexity.c"
|
|
|
/* 平方階（泡沫排序） */
|
|
|
int bubbleSort(int *nums, int n) {
|
|
|
int count = 0; // 計數器
|
|
|
// 外迴圈：未排序區間為 [0, i]
|
|
|
for (int i = n - 1; i > 0; i--) {
|
|
|
// 內迴圈：將未排序區間 [0, i] 中的最大元素交換至該區間的最右端
|
|
|
for (int j = 0; j < i; j++) {
|
|
|
if (nums[j] > nums[j + 1]) {
|
|
|
// 交換 nums[j] 與 nums[j + 1]
|
|
|
int tmp = nums[j];
|
|
|
nums[j] = nums[j + 1];
|
|
|
nums[j + 1] = tmp;
|
|
|
count += 3; // 元素交換包含 3 個單元操作
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Kotlin"
|
|
|
|
|
|
```kotlin title="time_complexity.kt"
|
|
|
/* 平方階（泡沫排序） */
|
|
|
fun bubbleSort(nums: IntArray): Int {
|
|
|
var count = 0
|
|
|
// 外迴圈：未排序區間為 [0, i]
|
|
|
for (i in nums.size - 1 downTo 1) {
|
|
|
// 內迴圈：將未排序區間 [0, i] 中的最大元素交換至該區間的最右端
|
|
|
for (j in 0..<i) {
|
|
|
if (nums[j] > nums[j + 1]) {
|
|
|
// 交換 nums[j] 與 nums[j + 1]
|
|
|
nums[j] = nums[j + 1].also { nums[j + 1] = nums[j] }
|
|
|
count += 3 // 元素交換包含 3 個單元操作
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Ruby"
|
|
|
|
|
|
```ruby title="time_complexity.rb"
|
|
|
### 平方階（泡沫排序）###
|
|
|
def bubble_sort(nums)
|
|
|
count = 0 # 計數器
|
|
|
|
|
|
# 外迴圈：未排序區間為 [0, i]
|
|
|
for i in (nums.length - 1).downto(0)
|
|
|
# 內迴圈：將未排序區間 [0, i] 中的最大元素交換至該區間的最右端
|
|
|
for j in 0...i
|
|
|
if nums[j] > nums[j + 1]
|
|
|
# 交換 nums[j] 與 nums[j + 1]
|
|
|
tmp = nums[j]
|
|
|
nums[j] = nums[j + 1]
|
|
|
nums[j + 1] = tmp
|
|
|
count += 3 # 元素交換包含 3 個單元操作
|
|
|
end
|
|
|
end
|
|
|
end
|
|
|
|
|
|
count
|
|
|
end
|
|
|
```
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
```zig title="time_complexity.zig"
|
|
|
// 平方階（泡沫排序）
|
|
|
fn bubbleSort(nums: []i32) i32 {
|
|
|
var count: i32 = 0; // 計數器
|
|
|
// 外迴圈：未排序區間為 [0, i]
|
|
|
var i: i32 = @as(i32, @intCast(nums.len)) - 1;
|
|
|
while (i > 0) : (i -= 1) {
|
|
|
var j: usize = 0;
|
|
|
// 內迴圈：將未排序區間 [0, i] 中的最大元素交換至該區間的最右端
|
|
|
while (j < i) : (j += 1) {
|
|
|
if (nums[j] > nums[j + 1]) {
|
|
|
// 交換 nums[j] 與 nums[j + 1]
|
|
|
var tmp = nums[j];
|
|
|
nums[j] = nums[j + 1];
|
|
|
nums[j + 1] = tmp;
|
|
|
count += 3; // 元素交換包含 3 個單元操作
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
??? pythontutor "視覺化執行"
|
|
|
|
|
|
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20bubble_sort%28nums%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%B9%B3%E6%96%B9%E9%98%B6%EF%BC%88%E5%86%92%E6%B3%A1%E6%8E%92%E5%BA%8F%EF%BC%89%22%22%22%0A%20%20%20%20count%20%3D%200%20%20%23%20%E8%AE%A1%E6%95%B0%E5%99%A8%0A%20%20%20%20%23%20%E5%A4%96%E5%BE%AA%E7%8E%AF%EF%BC%9A%E6%9C%AA%E6%8E%92%E5%BA%8F%E5%8C%BA%E9%97%B4%E4%B8%BA%20%5B0,%20i%5D%0A%20%20%20%20for%20i%20in%20range%28len%28nums%29%20-%201,%200,%20-1%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%86%85%E5%BE%AA%E7%8E%AF%EF%BC%9A%E5%B0%86%E6%9C%AA%E6%8E%92%E5%BA%8F%E5%8C%BA%E9%97%B4%20%5B0,%20i%5D%20%E4%B8%AD%E7%9A%84%E6%9C%80%E5%A4%A7%E5%85%83%E7%B4%A0%E4%BA%A4%E6%8D%A2%E8%87%B3%E8%AF%A5%E5%8C%BA%E9%97%B4%E7%9A%84%E6%9C%80%E5%8F%B3%E7%AB%AF%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%28i%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20nums%5Bj%5D%20%3E%20nums%5Bj%20%2B%201%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E4%BA%A4%E6%8D%A2%20nums%5Bj%5D%20%E4%B8%8E%20nums%5Bj%20%2B%201%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20tmp%20%3D%20nums%5Bj%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20nums%5Bj%5D%20%3D%20nums%5Bj%20%2B%201%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20nums%5Bj%20%2B%201%5D%20%3D%20tmp%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20count%20%2B%3D%203%20%20%23%20%E5%85%83%E7%B4%A0%E4%BA%A4%E6%8D%A2%E5%8C%85%E5%90%AB%203%20%E4%B8%AA%E5%8D%95%E5%85%83%E6%93%8D%E4%BD%9C%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20nums%20%3D%20%5Bi%20for%20i%20in%20range%28n,%200,%20-1%29%5D%20%20%23%20%5Bn,%20n-1,%20...,%202,%201%5D%0A%20%20%20%20count%20%3D%20bubble_sort%28nums%29%0A%20%20%20%20print%28%22%E5%B9%B3%E6%96%B9%E9%98%B6%EF%BC%88%E5%86%92%E6%B3%A1%E6%8E%92%E5%BA%8F%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
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<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20bubble_sort%28nums%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%B9%B3%E6%96%B9%E9%98%B6%EF%BC%88%E5%86%92%E6%B3%A1%E6%8E%92%E5%BA%8F%EF%BC%89%22%22%22%0A%20%20%20%20count%20%3D%200%20%20%23%20%E8%AE%A1%E6%95%B0%E5%99%A8%0A%20%20%20%20%23%20%E5%A4%96%E5%BE%AA%E7%8E%AF%EF%BC%9A%E6%9C%AA%E6%8E%92%E5%BA%8F%E5%8C%BA%E9%97%B4%E4%B8%BA%20%5B0,%20i%5D%0A%20%20%20%20for%20i%20in%20range%28len%28nums%29%20-%201,%200,%20-1%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%86%85%E5%BE%AA%E7%8E%AF%EF%BC%9A%E5%B0%86%E6%9C%AA%E6%8E%92%E5%BA%8F%E5%8C%BA%E9%97%B4%20%5B0,%20i%5D%20%E4%B8%AD%E7%9A%84%E6%9C%80%E5%A4%A7%E5%85%83%E7%B4%A0%E4%BA%A4%E6%8D%A2%E8%87%B3%E8%AF%A5%E5%8C%BA%E9%97%B4%E7%9A%84%E6%9C%80%E5%8F%B3%E7%AB%AF%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%28i%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20nums%5Bj%5D%20%3E%20nums%5Bj%20%2B%201%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E4%BA%A4%E6%8D%A2%20nums%5Bj%5D%20%E4%B8%8E%20nums%5Bj%20%2B%201%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20tmp%20%3D%20nums%5Bj%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20nums%5Bj%5D%20%3D%20nums%5Bj%20%2B%201%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20nums%5Bj%20%2B%201%5D%20%3D%20tmp%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20count%20%2B%3D%203%20%20%23%20%E5%85%83%E7%B4%A0%E4%BA%A4%E6%8D%A2%E5%8C%85%E5%90%AB%203%20%E4%B8%AA%E5%8D%95%E5%85%83%E6%93%8D%E4%BD%9C%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20nums%20%3D%20%5Bi%20for%20i%20in%20range%28n,%200,%20-1%29%5D%20%20%23%20%5Bn,%20n-1,%20...,%202,%201%5D%0A%20%20%20%20count%20%3D%20bubble_sort%28nums%29%0A%20%20%20%20print%28%22%E5%B9%B3%E6%96%B9%E9%98%B6%EF%BC%88%E5%86%92%E6%B3%A1%E6%8E%92%E5%BA%8F%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全螢幕觀看 ></a></div>
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### 4. 指數階 $O(2^n)$ {data-toc-label="4. 指數階"}
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生物學的“細胞分裂”是指數階增長的典型例子：初始狀態為 $1$ 個細胞，分裂一輪後變為 $2$ 個，分裂兩輪後變為 $4$ 個，以此類推，分裂 $n$ 輪後有 $2^n$ 個細胞。
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圖 2-11 和以下程式碼模擬了細胞分裂的過程，時間複雜度為 $O(2^n)$ ：
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=== "Python"
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```python title="time_complexity.py"
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def exponential(n: int) -> int:
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"""指數階（迴圈實現）"""
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count = 0
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base = 1
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# 細胞每輪一分為二，形成數列 1, 2, 4, 8, ..., 2^(n-1)
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for _ in range(n):
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for _ in range(base):
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count += 1
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base *= 2
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# count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
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return count
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```
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=== "C++"
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```cpp title="time_complexity.cpp"
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/* 指數階（迴圈實現） */
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int exponential(int n) {
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int count = 0, base = 1;
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// 細胞每輪一分為二，形成數列 1, 2, 4, 8, ..., 2^(n-1)
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < base; j++) {
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count++;
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}
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base *= 2;
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}
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// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
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return count;
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}
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```
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=== "Java"
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```java title="time_complexity.java"
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/* 指數階（迴圈實現） */
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int exponential(int n) {
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int count = 0, base = 1;
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// 細胞每輪一分為二，形成數列 1, 2, 4, 8, ..., 2^(n-1)
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < base; j++) {
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count++;
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}
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base *= 2;
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}
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// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
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return count;
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}
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```
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=== "C#"
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```csharp title="time_complexity.cs"
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/* 指數階（迴圈實現） */
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int Exponential(int n) {
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int count = 0, bas = 1;
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// 細胞每輪一分為二，形成數列 1, 2, 4, 8, ..., 2^(n-1)
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < bas; j++) {
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count++;
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}
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bas *= 2;
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}
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// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
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return count;
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}
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```
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=== "Go"
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```go title="time_complexity.go"
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/* 指數階（迴圈實現）*/
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func exponential(n int) int {
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count, base := 0, 1
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// 細胞每輪一分為二，形成數列 1, 2, 4, 8, ..., 2^(n-1)
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for i := 0; i < n; i++ {
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for j := 0; j < base; j++ {
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count++
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}
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base *= 2
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}
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// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
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return count
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}
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```
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=== "Swift"
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```swift title="time_complexity.swift"
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/* 指數階（迴圈實現） */
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func exponential(n: Int) -> Int {
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var count = 0
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var base = 1
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// 細胞每輪一分為二，形成數列 1, 2, 4, 8, ..., 2^(n-1)
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for _ in 0 ..< n {
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for _ in 0 ..< base {
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count += 1
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}
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base *= 2
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}
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// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
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return count
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}
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```
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=== "JS"
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```javascript title="time_complexity.js"
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/* 指數階（迴圈實現） */
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function exponential(n) {
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let count = 0,
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base = 1;
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// 細胞每輪一分為二，形成數列 1, 2, 4, 8, ..., 2^(n-1)
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for (let i = 0; i < n; i++) {
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for (let j = 0; j < base; j++) {
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count++;
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}
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base *= 2;
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}
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// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
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return count;
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}
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```
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=== "TS"
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```typescript title="time_complexity.ts"
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/* 指數階（迴圈實現） */
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function exponential(n: number): number {
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let count = 0,
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base = 1;
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// 細胞每輪一分為二，形成數列 1, 2, 4, 8, ..., 2^(n-1)
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for (let i = 0; i < n; i++) {
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for (let j = 0; j < base; j++) {
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count++;
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}
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base *= 2;
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}
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// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
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return count;
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}
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```
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=== "Dart"
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```dart title="time_complexity.dart"
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/* 指數階（迴圈實現） */
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int exponential(int n) {
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int count = 0, base = 1;
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// 細胞每輪一分為二，形成數列 1, 2, 4, 8, ..., 2^(n-1)
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for (var i = 0; i < n; i++) {
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for (var j = 0; j < base; j++) {
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count++;
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}
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base *= 2;
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}
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// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
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return count;
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}
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```
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|
=== "Rust"
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|
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|
|
```rust title="time_complexity.rs"
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|
/* 指數階（迴圈實現） */
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|
|
fn exponential(n: i32) -> i32 {
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|
let mut count = 0;
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|
let mut base = 1;
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|
// 細胞每輪一分為二，形成數列 1, 2, 4, 8, ..., 2^(n-1)
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for _ in 0..n {
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for _ in 0..base {
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count += 1
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}
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|
base *= 2;
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}
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|
// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
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|
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count
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}
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```
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|
=== "C"
|
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|
|
|
|
```c title="time_complexity.c"
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|
|
/* 指數階（迴圈實現） */
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|
|
int exponential(int n) {
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|
|
int count = 0;
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|
int bas = 1;
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|
|
// 細胞每輪一分為二，形成數列 1, 2, 4, 8, ..., 2^(n-1)
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|
for (int i = 0; i < n; i++) {
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|
|
for (int j = 0; j < bas; j++) {
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|
|
count++;
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|
}
|
|
|
bas *= 2;
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|
|
}
|
|
|
// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
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|
|
return count;
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|
|
}
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|
```
|
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|
|
=== "Kotlin"
|
|
|
|
|
|
```kotlin title="time_complexity.kt"
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|
|
/* 指數階（迴圈實現） */
|
|
|
fun exponential(n: Int): Int {
|
|
|
var count = 0
|
|
|
// 細胞每輪一分為二，形成數列 1, 2, 4, 8, ..., 2^(n-1)
|
|
|
var base = 1
|
|
|
for (i in 0..<n) {
|
|
|
for (j in 0..<base) {
|
|
|
count++
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|
|
}
|
|
|
base *= 2
|
|
|
}
|
|
|
// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
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|
|
return count
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|
|
}
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|
|
```
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|
=== "Ruby"
|
|
|
|
|
|
```ruby title="time_complexity.rb"
|
|
|
### 指數階（迴圈實現）###
|
|
|
def exponential(n)
|
|
|
count, base = 0, 1
|
|
|
|
|
|
# 細胞每輪一分為二，形成數列 1, 2, 4, 8, ..., 2^(n-1)
|
|
|
(0...n).each do
|
|
|
(0...base).each { count += 1 }
|
|
|
base *= 2
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|
|
end
|
|
|
|
|
|
# count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
|
|
|
count
|
|
|
end
|
|
|
```
|
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|
=== "Zig"
|
|
|
|
|
|
```zig title="time_complexity.zig"
|
|
|
// 指數階（迴圈實現）
|
|
|
fn exponential(n: i32) i32 {
|
|
|
var count: i32 = 0;
|
|
|
var bas: i32 = 1;
|
|
|
var i: i32 = 0;
|
|
|
// 細胞每輪一分為二，形成數列 1, 2, 4, 8, ..., 2^(n-1)
|
|
|
while (i < n) : (i += 1) {
|
|
|
var j: i32 = 0;
|
|
|
while (j < bas) : (j += 1) {
|
|
|
count += 1;
|
|
|
}
|
|
|
bas *= 2;
|
|
|
}
|
|
|
// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
??? pythontutor "視覺化執行"
|
|
|
|
|
|
<div style="height: 531px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20exponential%28n%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%8C%87%E6%95%B0%E9%98%B6%EF%BC%88%E5%BE%AA%E7%8E%AF%E5%AE%9E%E7%8E%B0%EF%BC%89%22%22%22%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20base%20%3D%201%0A%20%20%20%20%23%20%E7%BB%86%E8%83%9E%E6%AF%8F%E8%BD%AE%E4%B8%80%E5%88%86%E4%B8%BA%E4%BA%8C%EF%BC%8C%E5%BD%A2%E6%88%90%E6%95%B0%E5%88%97%201,%202,%204,%208,%20...,%202%5E%28n-1%29%0A%20%20%20%20for%20_%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20for%20_%20in%20range%28base%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20%20%20%20%20base%20*%3D%202%0A%20%20%20%20%23%20count%20%3D%201%20%2B%202%20%2B%204%20%2B%208%20%2B%20..%20%2B%202%5E%28n-1%29%20%3D%202%5En%20-%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20exponential%28n%29%0A%20%20%20%20print%28%22%E6%8C%87%E6%95%B0%E9%98%B6%EF%BC%88%E5%BE%AA%E7%8E%AF%E5%AE%9E%E7%8E%B0%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
|
|
|
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20exponential%28n%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%8C%87%E6%95%B0%E9%98%B6%EF%BC%88%E5%BE%AA%E7%8E%AF%E5%AE%9E%E7%8E%B0%EF%BC%89%22%22%22%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20base%20%3D%201%0A%20%20%20%20%23%20%E7%BB%86%E8%83%9E%E6%AF%8F%E8%BD%AE%E4%B8%80%E5%88%86%E4%B8%BA%E4%BA%8C%EF%BC%8C%E5%BD%A2%E6%88%90%E6%95%B0%E5%88%97%201,%202,%204,%208,%20...,%202%5E%28n-1%29%0A%20%20%20%20for%20_%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20for%20_%20in%20range%28base%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20%20%20%20%20base%20*%3D%202%0A%20%20%20%20%23%20count%20%3D%201%20%2B%202%20%2B%204%20%2B%208%20%2B%20..%20%2B%202%5E%28n-1%29%20%3D%202%5En%20-%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20exponential%28n%29%0A%20%20%20%20print%28%22%E6%8C%87%E6%95%B0%E9%98%B6%EF%BC%88%E5%BE%AA%E7%8E%AF%E5%AE%9E%E7%8E%B0%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全螢幕觀看 ></a></div>
|
|
|
|
|
|
{ class="animation-figure" }
|
|
|
|
|
|
<p align="center"> 圖 2-11 指數階的時間複雜度 </p>
|
|
|
|
|
|
在實際演算法中，指數階常出現於遞迴函式中。例如在以下程式碼中，其遞迴地一分為二，經過 $n$ 次分裂後停止：
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
```python title="time_complexity.py"
|
|
|
def exp_recur(n: int) -> int:
|
|
|
"""指數階（遞迴實現）"""
|
|
|
if n == 1:
|
|
|
return 1
|
|
|
return exp_recur(n - 1) + exp_recur(n - 1) + 1
|
|
|
```
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
```cpp title="time_complexity.cpp"
|
|
|
/* 指數階（遞迴實現） */
|
|
|
int expRecur(int n) {
|
|
|
if (n == 1)
|
|
|
return 1;
|
|
|
return expRecur(n - 1) + expRecur(n - 1) + 1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
```java title="time_complexity.java"
|
|
|
/* 指數階（遞迴實現） */
|
|
|
int expRecur(int n) {
|
|
|
if (n == 1)
|
|
|
return 1;
|
|
|
return expRecur(n - 1) + expRecur(n - 1) + 1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
```csharp title="time_complexity.cs"
|
|
|
/* 指數階（遞迴實現） */
|
|
|
int ExpRecur(int n) {
|
|
|
if (n == 1) return 1;
|
|
|
return ExpRecur(n - 1) + ExpRecur(n - 1) + 1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
```go title="time_complexity.go"
|
|
|
/* 指數階（遞迴實現）*/
|
|
|
func expRecur(n int) int {
|
|
|
if n == 1 {
|
|
|
return 1
|
|
|
}
|
|
|
return expRecur(n-1) + expRecur(n-1) + 1
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
```swift title="time_complexity.swift"
|
|
|
/* 指數階（遞迴實現） */
|
|
|
func expRecur(n: Int) -> Int {
|
|
|
if n == 1 {
|
|
|
return 1
|
|
|
}
|
|
|
return expRecur(n: n - 1) + expRecur(n: n - 1) + 1
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
```javascript title="time_complexity.js"
|
|
|
/* 指數階（遞迴實現） */
|
|
|
function expRecur(n) {
|
|
|
if (n === 1) return 1;
|
|
|
return expRecur(n - 1) + expRecur(n - 1) + 1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
```typescript title="time_complexity.ts"
|
|
|
/* 指數階（遞迴實現） */
|
|
|
function expRecur(n: number): number {
|
|
|
if (n === 1) return 1;
|
|
|
return expRecur(n - 1) + expRecur(n - 1) + 1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
```dart title="time_complexity.dart"
|
|
|
/* 指數階（遞迴實現） */
|
|
|
int expRecur(int n) {
|
|
|
if (n == 1) return 1;
|
|
|
return expRecur(n - 1) + expRecur(n - 1) + 1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
```rust title="time_complexity.rs"
|
|
|
/* 指數階（遞迴實現） */
|
|
|
fn exp_recur(n: i32) -> i32 {
|
|
|
if n == 1 {
|
|
|
return 1;
|
|
|
}
|
|
|
exp_recur(n - 1) + exp_recur(n - 1) + 1
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
```c title="time_complexity.c"
|
|
|
/* 指數階（遞迴實現） */
|
|
|
int expRecur(int n) {
|
|
|
if (n == 1)
|
|
|
return 1;
|
|
|
return expRecur(n - 1) + expRecur(n - 1) + 1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Kotlin"
|
|
|
|
|
|
```kotlin title="time_complexity.kt"
|
|
|
/* 指數階（遞迴實現） */
|
|
|
fun expRecur(n: Int): Int {
|
|
|
if (n == 1) {
|
|
|
return 1
|
|
|
}
|
|
|
return expRecur(n - 1) + expRecur(n - 1) + 1
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Ruby"
|
|
|
|
|
|
```ruby title="time_complexity.rb"
|
|
|
### 指數階（遞迴實現）###
|
|
|
def exp_recur(n)
|
|
|
return 1 if n == 1
|
|
|
exp_recur(n - 1) + exp_recur(n - 1) + 1
|
|
|
end
|
|
|
```
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
```zig title="time_complexity.zig"
|
|
|
// 指數階（遞迴實現）
|
|
|
fn expRecur(n: i32) i32 {
|
|
|
if (n == 1) return 1;
|
|
|
return expRecur(n - 1) + expRecur(n - 1) + 1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
??? pythontutor "視覺化執行"
|
|
|
|
|
|
<div style="height: 423px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20exp_recur%28n%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%8C%87%E6%95%B0%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%22%22%22%0A%20%20%20%20if%20n%20%3D%3D%201%3A%0A%20%20%20%20%20%20%20%20return%201%0A%20%20%20%20return%20exp_recur%28n%20-%201%29%20%2B%20exp_recur%28n%20-%201%29%20%2B%201%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%207%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20exp_recur%28n%29%0A%20%20%20%20print%28%22%E6%8C%87%E6%95%B0%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
|
|
|
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20exp_recur%28n%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%8C%87%E6%95%B0%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%22%22%22%0A%20%20%20%20if%20n%20%3D%3D%201%3A%0A%20%20%20%20%20%20%20%20return%201%0A%20%20%20%20return%20exp_recur%28n%20-%201%29%20%2B%20exp_recur%28n%20-%201%29%20%2B%201%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%207%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20exp_recur%28n%29%0A%20%20%20%20print%28%22%E6%8C%87%E6%95%B0%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全螢幕觀看 ></a></div>
|
|
|
|
|
|
指數階增長非常迅速，在窮舉法（暴力搜尋、回溯等）中比較常見。對於資料規模較大的問題，指數階是不可接受的，通常需要使用動態規劃或貪婪演算法等來解決。
|
|
|
|
|
|
### 5. 對數階 $O(\log n)$ {data-toc-label="5. 對數階"}
|
|
|
|
|
|
與指數階相反，對數階反映了“每輪縮減到一半”的情況。設輸入資料大小為 $n$ ，由於每輪縮減到一半，因此迴圈次數是 $\log_2 n$ ，即 $2^n$ 的反函式。
|
|
|
|
|
|
圖 2-12 和以下程式碼模擬了“每輪縮減到一半”的過程，時間複雜度為 $O(\log_2 n)$ ，簡記為 $O(\log n)$ ：
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
```python title="time_complexity.py"
|
|
|
def logarithmic(n: int) -> int:
|
|
|
"""對數階（迴圈實現）"""
|
|
|
count = 0
|
|
|
while n > 1:
|
|
|
n = n / 2
|
|
|
count += 1
|
|
|
return count
|
|
|
```
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
```cpp title="time_complexity.cpp"
|
|
|
/* 對數階（迴圈實現） */
|
|
|
int logarithmic(int n) {
|
|
|
int count = 0;
|
|
|
while (n > 1) {
|
|
|
n = n / 2;
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
```java title="time_complexity.java"
|
|
|
/* 對數階（迴圈實現） */
|
|
|
int logarithmic(int n) {
|
|
|
int count = 0;
|
|
|
while (n > 1) {
|
|
|
n = n / 2;
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
```csharp title="time_complexity.cs"
|
|
|
/* 對數階（迴圈實現） */
|
|
|
int Logarithmic(int n) {
|
|
|
int count = 0;
|
|
|
while (n > 1) {
|
|
|
n /= 2;
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
```go title="time_complexity.go"
|
|
|
/* 對數階（迴圈實現）*/
|
|
|
func logarithmic(n int) int {
|
|
|
count := 0
|
|
|
for n > 1 {
|
|
|
n = n / 2
|
|
|
count++
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
```swift title="time_complexity.swift"
|
|
|
/* 對數階（迴圈實現） */
|
|
|
func logarithmic(n: Int) -> Int {
|
|
|
var count = 0
|
|
|
var n = n
|
|
|
while n > 1 {
|
|
|
n = n / 2
|
|
|
count += 1
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
```javascript title="time_complexity.js"
|
|
|
/* 對數階（迴圈實現） */
|
|
|
function logarithmic(n) {
|
|
|
let count = 0;
|
|
|
while (n > 1) {
|
|
|
n = n / 2;
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
```typescript title="time_complexity.ts"
|
|
|
/* 對數階（迴圈實現） */
|
|
|
function logarithmic(n: number): number {
|
|
|
let count = 0;
|
|
|
while (n > 1) {
|
|
|
n = n / 2;
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
```dart title="time_complexity.dart"
|
|
|
/* 對數階（迴圈實現） */
|
|
|
int logarithmic(int n) {
|
|
|
int count = 0;
|
|
|
while (n > 1) {
|
|
|
n = n ~/ 2;
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
```rust title="time_complexity.rs"
|
|
|
/* 對數階（迴圈實現） */
|
|
|
fn logarithmic(mut n: i32) -> i32 {
|
|
|
let mut count = 0;
|
|
|
while n > 1 {
|
|
|
n = n / 2;
|
|
|
count += 1;
|
|
|
}
|
|
|
count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
```c title="time_complexity.c"
|
|
|
/* 對數階（迴圈實現） */
|
|
|
int logarithmic(int n) {
|
|
|
int count = 0;
|
|
|
while (n > 1) {
|
|
|
n = n / 2;
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Kotlin"
|
|
|
|
|
|
```kotlin title="time_complexity.kt"
|
|
|
/* 對數階（迴圈實現） */
|
|
|
fun logarithmic(n: Int): Int {
|
|
|
var n1 = n
|
|
|
var count = 0
|
|
|
while (n1 > 1) {
|
|
|
n1 /= 2
|
|
|
count++
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Ruby"
|
|
|
|
|
|
```ruby title="time_complexity.rb"
|
|
|
### 對數階（迴圈實現）###
|
|
|
def logarithmic(n)
|
|
|
count = 0
|
|
|
|
|
|
while n > 1
|
|
|
n /= 2
|
|
|
count += 1
|
|
|
end
|
|
|
|
|
|
count
|
|
|
end
|
|
|
```
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
```zig title="time_complexity.zig"
|
|
|
// 對數階（迴圈實現）
|
|
|
fn logarithmic(n: i32) i32 {
|
|
|
var count: i32 = 0;
|
|
|
var n_var = n;
|
|
|
while (n_var > 1)
|
|
|
{
|
|
|
n_var = n_var / 2;
|
|
|
count +=1;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
??? pythontutor "視覺化執行"
|
|
|
|
|
|
<div style="height: 459px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20logarithmic%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%AF%B9%E6%95%B0%E9%98%B6%EF%BC%88%E5%BE%AA%E7%8E%AF%E5%AE%9E%E7%8E%B0%EF%BC%89%22%22%22%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20while%20n%20%3E%201%3A%0A%20%20%20%20%20%20%20%20n%20%3D%20n%20/%202%0A%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20logarithmic%28n%29%0A%20%20%20%20print%28%22%E5%AF%B9%E6%95%B0%E9%98%B6%EF%BC%88%E5%BE%AA%E7%8E%AF%E5%AE%9E%E7%8E%B0%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
|
|
|
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20logarithmic%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%AF%B9%E6%95%B0%E9%98%B6%EF%BC%88%E5%BE%AA%E7%8E%AF%E5%AE%9E%E7%8E%B0%EF%BC%89%22%22%22%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20while%20n%20%3E%201%3A%0A%20%20%20%20%20%20%20%20n%20%3D%20n%20/%202%0A%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20logarithmic%28n%29%0A%20%20%20%20print%28%22%E5%AF%B9%E6%95%B0%E9%98%B6%EF%BC%88%E5%BE%AA%E7%8E%AF%E5%AE%9E%E7%8E%B0%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全螢幕觀看 ></a></div>
|
|
|
|
|
|
{ class="animation-figure" }
|
|
|
|
|
|
<p align="center"> 圖 2-12 對數階的時間複雜度 </p>
|
|
|
|
|
|
與指數階類似，對數階也常出現於遞迴函式中。以下程式碼形成了一棵高度為 $\log_2 n$ 的遞迴樹：
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
```python title="time_complexity.py"
|
|
|
def log_recur(n: int) -> int:
|
|
|
"""對數階（遞迴實現）"""
|
|
|
if n <= 1:
|
|
|
return 0
|
|
|
return log_recur(n / 2) + 1
|
|
|
```
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
```cpp title="time_complexity.cpp"
|
|
|
/* 對數階（遞迴實現） */
|
|
|
int logRecur(int n) {
|
|
|
if (n <= 1)
|
|
|
return 0;
|
|
|
return logRecur(n / 2) + 1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
```java title="time_complexity.java"
|
|
|
/* 對數階（遞迴實現） */
|
|
|
int logRecur(int n) {
|
|
|
if (n <= 1)
|
|
|
return 0;
|
|
|
return logRecur(n / 2) + 1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
```csharp title="time_complexity.cs"
|
|
|
/* 對數階（遞迴實現） */
|
|
|
int LogRecur(int n) {
|
|
|
if (n <= 1) return 0;
|
|
|
return LogRecur(n / 2) + 1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
```go title="time_complexity.go"
|
|
|
/* 對數階（遞迴實現）*/
|
|
|
func logRecur(n int) int {
|
|
|
if n <= 1 {
|
|
|
return 0
|
|
|
}
|
|
|
return logRecur(n/2) + 1
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
```swift title="time_complexity.swift"
|
|
|
/* 對數階（遞迴實現） */
|
|
|
func logRecur(n: Int) -> Int {
|
|
|
if n <= 1 {
|
|
|
return 0
|
|
|
}
|
|
|
return logRecur(n: n / 2) + 1
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
```javascript title="time_complexity.js"
|
|
|
/* 對數階（遞迴實現） */
|
|
|
function logRecur(n) {
|
|
|
if (n <= 1) return 0;
|
|
|
return logRecur(n / 2) + 1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
```typescript title="time_complexity.ts"
|
|
|
/* 對數階（遞迴實現） */
|
|
|
function logRecur(n: number): number {
|
|
|
if (n <= 1) return 0;
|
|
|
return logRecur(n / 2) + 1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
```dart title="time_complexity.dart"
|
|
|
/* 對數階（遞迴實現） */
|
|
|
int logRecur(int n) {
|
|
|
if (n <= 1) return 0;
|
|
|
return logRecur(n ~/ 2) + 1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
```rust title="time_complexity.rs"
|
|
|
/* 對數階（遞迴實現） */
|
|
|
fn log_recur(n: i32) -> i32 {
|
|
|
if n <= 1 {
|
|
|
return 0;
|
|
|
}
|
|
|
log_recur(n / 2) + 1
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
```c title="time_complexity.c"
|
|
|
/* 對數階（遞迴實現） */
|
|
|
int logRecur(int n) {
|
|
|
if (n <= 1)
|
|
|
return 0;
|
|
|
return logRecur(n / 2) + 1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Kotlin"
|
|
|
|
|
|
```kotlin title="time_complexity.kt"
|
|
|
/* 對數階（遞迴實現） */
|
|
|
fun logRecur(n: Int): Int {
|
|
|
if (n <= 1)
|
|
|
return 0
|
|
|
return logRecur(n / 2) + 1
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Ruby"
|
|
|
|
|
|
```ruby title="time_complexity.rb"
|
|
|
### 對數階（遞迴實現）###
|
|
|
def log_recur(n)
|
|
|
return 0 unless n > 1
|
|
|
log_recur(n / 2) + 1
|
|
|
end
|
|
|
```
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
```zig title="time_complexity.zig"
|
|
|
// 對數階（遞迴實現）
|
|
|
fn logRecur(n: i32) i32 {
|
|
|
if (n <= 1) return 0;
|
|
|
return logRecur(n / 2) + 1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
??? pythontutor "視覺化執行"
|
|
|
|
|
|
<div style="height: 423px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20log_recur%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%AF%B9%E6%95%B0%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%22%22%22%0A%20%20%20%20if%20n%20%3C%3D%201%3A%0A%20%20%20%20%20%20%20%20return%200%0A%20%20%20%20return%20log_recur%28n%20/%202%29%20%2B%201%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20log_recur%28n%29%0A%20%20%20%20print%28%22%E5%AF%B9%E6%95%B0%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
|
|
|
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20log_recur%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%AF%B9%E6%95%B0%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%22%22%22%0A%20%20%20%20if%20n%20%3C%3D%201%3A%0A%20%20%20%20%20%20%20%20return%200%0A%20%20%20%20return%20log_recur%28n%20/%202%29%20%2B%201%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20log_recur%28n%29%0A%20%20%20%20print%28%22%E5%AF%B9%E6%95%B0%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全螢幕觀看 ></a></div>
|
|
|
|
|
|
對數階常出現於基於分治策略的演算法中，體現了“一分為多”和“化繁為簡”的演算法思想。它增長緩慢，是僅次於常數階的理想的時間複雜度。
|
|
|
|
|
|
!!! tip "$O(\log n)$ 的底數是多少？"
|
|
|
|
|
|
準確來說，“一分為 $m$”對應的時間複雜度是 $O(\log_m n)$ 。而透過對數換底公式，我們可以得到具有不同底數、相等的時間複雜度：
|
|
|
|
|
|
$$
|
|
|
O(\log_m n) = O(\log_k n / \log_k m) = O(\log_k n)
|
|
|
$$
|
|
|
|
|
|
也就是說，底數 $m$ 可以在不影響複雜度的前提下轉換。因此我們通常會省略底數 $m$ ，將對數階直接記為 $O(\log n)$ 。
|
|
|
|
|
|
### 6. 線性對數階 $O(n \log n)$ {data-toc-label="6. 線性對數階"}
|
|
|
|
|
|
線性對數階常出現於巢狀迴圈中，兩層迴圈的時間複雜度分別為 $O(\log n)$ 和 $O(n)$ 。相關程式碼如下：
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
```python title="time_complexity.py"
|
|
|
def linear_log_recur(n: int) -> int:
|
|
|
"""線性對數階"""
|
|
|
if n <= 1:
|
|
|
return 1
|
|
|
count: int = linear_log_recur(n // 2) + linear_log_recur(n // 2)
|
|
|
for _ in range(n):
|
|
|
count += 1
|
|
|
return count
|
|
|
```
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
```cpp title="time_complexity.cpp"
|
|
|
/* 線性對數階 */
|
|
|
int linearLogRecur(int n) {
|
|
|
if (n <= 1)
|
|
|
return 1;
|
|
|
int count = linearLogRecur(n / 2) + linearLogRecur(n / 2);
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
```java title="time_complexity.java"
|
|
|
/* 線性對數階 */
|
|
|
int linearLogRecur(int n) {
|
|
|
if (n <= 1)
|
|
|
return 1;
|
|
|
int count = linearLogRecur(n / 2) + linearLogRecur(n / 2);
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
```csharp title="time_complexity.cs"
|
|
|
/* 線性對數階 */
|
|
|
int LinearLogRecur(int n) {
|
|
|
if (n <= 1) return 1;
|
|
|
int count = LinearLogRecur(n / 2) + LinearLogRecur(n / 2);
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
```go title="time_complexity.go"
|
|
|
/* 線性對數階 */
|
|
|
func linearLogRecur(n int) int {
|
|
|
if n <= 1 {
|
|
|
return 1
|
|
|
}
|
|
|
count := linearLogRecur(n/2) + linearLogRecur(n/2)
|
|
|
for i := 0; i < n; i++ {
|
|
|
count++
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
```swift title="time_complexity.swift"
|
|
|
/* 線性對數階 */
|
|
|
func linearLogRecur(n: Int) -> Int {
|
|
|
if n <= 1 {
|
|
|
return 1
|
|
|
}
|
|
|
var count = linearLogRecur(n: n / 2) + linearLogRecur(n: n / 2)
|
|
|
for _ in stride(from: 0, to: n, by: 1) {
|
|
|
count += 1
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
```javascript title="time_complexity.js"
|
|
|
/* 線性對數階 */
|
|
|
function linearLogRecur(n) {
|
|
|
if (n <= 1) return 1;
|
|
|
let count = linearLogRecur(n / 2) + linearLogRecur(n / 2);
|
|
|
for (let i = 0; i < n; i++) {
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
```typescript title="time_complexity.ts"
|
|
|
/* 線性對數階 */
|
|
|
function linearLogRecur(n: number): number {
|
|
|
if (n <= 1) return 1;
|
|
|
let count = linearLogRecur(n / 2) + linearLogRecur(n / 2);
|
|
|
for (let i = 0; i < n; i++) {
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
```dart title="time_complexity.dart"
|
|
|
/* 線性對數階 */
|
|
|
int linearLogRecur(int n) {
|
|
|
if (n <= 1) return 1;
|
|
|
int count = linearLogRecur(n ~/ 2) + linearLogRecur(n ~/ 2);
|
|
|
for (var i = 0; i < n; i++) {
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
```rust title="time_complexity.rs"
|
|
|
/* 線性對數階 */
|
|
|
fn linear_log_recur(n: i32) -> i32 {
|
|
|
if n <= 1 {
|
|
|
return 1;
|
|
|
}
|
|
|
let mut count = linear_log_recur(n / 2) + linear_log_recur(n / 2);
|
|
|
for _ in 0..n as i32 {
|
|
|
count += 1;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
```c title="time_complexity.c"
|
|
|
/* 線性對數階 */
|
|
|
int linearLogRecur(int n) {
|
|
|
if (n <= 1)
|
|
|
return 1;
|
|
|
int count = linearLogRecur(n / 2) + linearLogRecur(n / 2);
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
count++;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Kotlin"
|
|
|
|
|
|
```kotlin title="time_complexity.kt"
|
|
|
/* 線性對數階 */
|
|
|
fun linearLogRecur(n: Int): Int {
|
|
|
if (n <= 1)
|
|
|
return 1
|
|
|
var count = linearLogRecur(n / 2) + linearLogRecur(n / 2)
|
|
|
for (i in 0..<n.toInt()) {
|
|
|
count++
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Ruby"
|
|
|
|
|
|
```ruby title="time_complexity.rb"
|
|
|
### 線性對數階 ###
|
|
|
def linear_log_recur(n)
|
|
|
return 1 unless n > 1
|
|
|
|
|
|
count = linear_log_recur(n / 2) + linear_log_recur(n / 2)
|
|
|
(0...n).each { count += 1 }
|
|
|
|
|
|
count
|
|
|
end
|
|
|
```
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
```zig title="time_complexity.zig"
|
|
|
// 線性對數階
|
|
|
fn linearLogRecur(n: i32) i32 {
|
|
|
if (n <= 1) return 1;
|
|
|
var count: i32 = linearLogRecur(n / 2) + linearLogRecur(n / 2);
|
|
|
var i: i32 = 0;
|
|
|
while (i < n) : (i += 1) {
|
|
|
count += 1;
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
??? pythontutor "視覺化執行"
|
|
|
|
|
|
<div style="height: 477px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20linear_log_recur%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%BA%BF%E6%80%A7%E5%AF%B9%E6%95%B0%E9%98%B6%22%22%22%0A%20%20%20%20if%20n%20%3C%3D%201%3A%0A%20%20%20%20%20%20%20%20return%201%0A%20%20%20%20count%20%3D%20linear_log_recur%28n%20//%202%29%20%2B%20linear_log_recur%28n%20//%202%29%0A%20%20%20%20for%20_%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20linear_log_recur%28n%29%0A%20%20%20%20print%28%22%E7%BA%BF%E6%80%A7%E5%AF%B9%E6%95%B0%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
|
|
|
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20linear_log_recur%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%BA%BF%E6%80%A7%E5%AF%B9%E6%95%B0%E9%98%B6%22%22%22%0A%20%20%20%20if%20n%20%3C%3D%201%3A%0A%20%20%20%20%20%20%20%20return%201%0A%20%20%20%20count%20%3D%20linear_log_recur%28n%20//%202%29%20%2B%20linear_log_recur%28n%20//%202%29%0A%20%20%20%20for%20_%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20linear_log_recur%28n%29%0A%20%20%20%20print%28%22%E7%BA%BF%E6%80%A7%E5%AF%B9%E6%95%B0%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全螢幕觀看 ></a></div>
|
|
|
|
|
|
圖 2-13 展示了線性對數階的生成方式。二元樹的每一層的操作總數都為 $n$ ，樹共有 $\log_2 n + 1$ 層，因此時間複雜度為 $O(n \log n)$ 。
|
|
|
|
|
|
{ class="animation-figure" }
|
|
|
|
|
|
<p align="center"> 圖 2-13 線性對數階的時間複雜度 </p>
|
|
|
|
|
|
主流排序演算法的時間複雜度通常為 $O(n \log n)$ ，例如快速排序、合併排序、堆積排序等。
|
|
|
|
|
|
### 7. 階乘階 $O(n!)$ {data-toc-label="7. 階乘階"}
|
|
|
|
|
|
階乘階對應數學上的“全排列”問題。給定 $n$ 個互不重複的元素，求其所有可能的排列方案，方案數量為：
|
|
|
|
|
|
$$
|
|
|
n! = n \times (n - 1) \times (n - 2) \times \dots \times 2 \times 1
|
|
|
$$
|
|
|
|
|
|
階乘通常使用遞迴實現。如圖 2-14 和以下程式碼所示，第一層分裂出 $n$ 個，第二層分裂出 $n - 1$ 個，以此類推，直至第 $n$ 層時停止分裂：
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
```python title="time_complexity.py"
|
|
|
def factorial_recur(n: int) -> int:
|
|
|
"""階乘階（遞迴實現）"""
|
|
|
if n == 0:
|
|
|
return 1
|
|
|
count = 0
|
|
|
# 從 1 個分裂出 n 個
|
|
|
for _ in range(n):
|
|
|
count += factorial_recur(n - 1)
|
|
|
return count
|
|
|
```
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
```cpp title="time_complexity.cpp"
|
|
|
/* 階乘階（遞迴實現） */
|
|
|
int factorialRecur(int n) {
|
|
|
if (n == 0)
|
|
|
return 1;
|
|
|
int count = 0;
|
|
|
// 從 1 個分裂出 n 個
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
count += factorialRecur(n - 1);
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
```java title="time_complexity.java"
|
|
|
/* 階乘階（遞迴實現） */
|
|
|
int factorialRecur(int n) {
|
|
|
if (n == 0)
|
|
|
return 1;
|
|
|
int count = 0;
|
|
|
// 從 1 個分裂出 n 個
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
count += factorialRecur(n - 1);
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
```csharp title="time_complexity.cs"
|
|
|
/* 階乘階（遞迴實現） */
|
|
|
int FactorialRecur(int n) {
|
|
|
if (n == 0) return 1;
|
|
|
int count = 0;
|
|
|
// 從 1 個分裂出 n 個
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
count += FactorialRecur(n - 1);
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
```go title="time_complexity.go"
|
|
|
/* 階乘階（遞迴實現） */
|
|
|
func factorialRecur(n int) int {
|
|
|
if n == 0 {
|
|
|
return 1
|
|
|
}
|
|
|
count := 0
|
|
|
// 從 1 個分裂出 n 個
|
|
|
for i := 0; i < n; i++ {
|
|
|
count += factorialRecur(n - 1)
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
```swift title="time_complexity.swift"
|
|
|
/* 階乘階（遞迴實現） */
|
|
|
func factorialRecur(n: Int) -> Int {
|
|
|
if n == 0 {
|
|
|
return 1
|
|
|
}
|
|
|
var count = 0
|
|
|
// 從 1 個分裂出 n 個
|
|
|
for _ in 0 ..< n {
|
|
|
count += factorialRecur(n: n - 1)
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
```javascript title="time_complexity.js"
|
|
|
/* 階乘階（遞迴實現） */
|
|
|
function factorialRecur(n) {
|
|
|
if (n === 0) return 1;
|
|
|
let count = 0;
|
|
|
// 從 1 個分裂出 n 個
|
|
|
for (let i = 0; i < n; i++) {
|
|
|
count += factorialRecur(n - 1);
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
```typescript title="time_complexity.ts"
|
|
|
/* 階乘階（遞迴實現） */
|
|
|
function factorialRecur(n: number): number {
|
|
|
if (n === 0) return 1;
|
|
|
let count = 0;
|
|
|
// 從 1 個分裂出 n 個
|
|
|
for (let i = 0; i < n; i++) {
|
|
|
count += factorialRecur(n - 1);
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
```dart title="time_complexity.dart"
|
|
|
/* 階乘階（遞迴實現） */
|
|
|
int factorialRecur(int n) {
|
|
|
if (n == 0) return 1;
|
|
|
int count = 0;
|
|
|
// 從 1 個分裂出 n 個
|
|
|
for (var i = 0; i < n; i++) {
|
|
|
count += factorialRecur(n - 1);
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
```rust title="time_complexity.rs"
|
|
|
/* 階乘階（遞迴實現） */
|
|
|
fn factorial_recur(n: i32) -> i32 {
|
|
|
if n == 0 {
|
|
|
return 1;
|
|
|
}
|
|
|
let mut count = 0;
|
|
|
// 從 1 個分裂出 n 個
|
|
|
for _ in 0..n {
|
|
|
count += factorial_recur(n - 1);
|
|
|
}
|
|
|
count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
```c title="time_complexity.c"
|
|
|
/* 階乘階（遞迴實現） */
|
|
|
int factorialRecur(int n) {
|
|
|
if (n == 0)
|
|
|
return 1;
|
|
|
int count = 0;
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
count += factorialRecur(n - 1);
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Kotlin"
|
|
|
|
|
|
```kotlin title="time_complexity.kt"
|
|
|
/* 階乘階（遞迴實現） */
|
|
|
fun factorialRecur(n: Int): Int {
|
|
|
if (n == 0)
|
|
|
return 1
|
|
|
var count = 0
|
|
|
// 從 1 個分裂出 n 個
|
|
|
for (i in 0..<n) {
|
|
|
count += factorialRecur(n - 1)
|
|
|
}
|
|
|
return count
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Ruby"
|
|
|
|
|
|
```ruby title="time_complexity.rb"
|
|
|
### 階乘階（遞迴實現）###
|
|
|
def factorial_recur(n)
|
|
|
return 1 if n == 0
|
|
|
|
|
|
count = 0
|
|
|
# 從 1 個分裂出 n 個
|
|
|
(0...n).each { count += factorial_recur(n - 1) }
|
|
|
|
|
|
count
|
|
|
end
|
|
|
```
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
```zig title="time_complexity.zig"
|
|
|
// 階乘階（遞迴實現）
|
|
|
fn factorialRecur(n: i32) i32 {
|
|
|
if (n == 0) return 1;
|
|
|
var count: i32 = 0;
|
|
|
var i: i32 = 0;
|
|
|
// 從 1 個分裂出 n 個
|
|
|
while (i < n) : (i += 1) {
|
|
|
count += factorialRecur(n - 1);
|
|
|
}
|
|
|
return count;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
??? pythontutor "視覺化執行"
|
|
|
|
|
|
<div style="height: 495px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20factorial_recur%28n%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E9%98%B6%E4%B9%98%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%22%22%22%0A%20%20%20%20if%20n%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20return%201%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20%23%20%E4%BB%8E%201%20%E4%B8%AA%E5%88%86%E8%A3%82%E5%87%BA%20n%20%E4%B8%AA%0A%20%20%20%20for%20_%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20count%20%2B%3D%20factorial_recur%28n%20-%201%29%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%204%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20factorial_recur%28n%29%0A%20%20%20%20print%28%22%E9%98%B6%E4%B9%98%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
|
|
|
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20factorial_recur%28n%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E9%98%B6%E4%B9%98%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%22%22%22%0A%20%20%20%20if%20n%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20return%201%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20%23%20%E4%BB%8E%201%20%E4%B8%AA%E5%88%86%E8%A3%82%E5%87%BA%20n%20%E4%B8%AA%0A%20%20%20%20for%20_%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20count%20%2B%3D%20factorial_recur%28n%20-%201%29%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%204%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20factorial_recur%28n%29%0A%20%20%20%20print%28%22%E9%98%B6%E4%B9%98%E9%98%B6%EF%BC%88%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%EF%BC%89%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全螢幕觀看 ></a></div>
|
|
|
|
|
|
{ class="animation-figure" }
|
|
|
|
|
|
<p align="center"> 圖 2-14 階乘階的時間複雜度 </p>
|
|
|
|
|
|
請注意，因為當 $n \geq 4$ 時恆有 $n! > 2^n$ ，所以階乘階比指數階增長得更快，在 $n$ 較大時也是不可接受的。
|
|
|
|
|
|
## 2.3.5 最差、最佳、平均時間複雜度
|
|
|
|
|
|
**演算法的時間效率往往不是固定的，而是與輸入資料的分佈有關**。假設輸入一個長度為 $n$ 的陣列 `nums` ，其中 `nums` 由從 $1$ 至 $n$ 的數字組成，每個數字只出現一次；但元素順序是隨機打亂的，任務目標是返回元素 $1$ 的索引。我們可以得出以下結論。
|
|
|
|
|
|
- 當 `nums = [?, ?, ..., 1]` ，即當末尾元素是 $1$ 時，需要完整走訪陣列，**達到最差時間複雜度 $O(n)$** 。
|
|
|
- 當 `nums = [1, ?, ?, ...]` ，即當首個元素為 $1$ 時，無論陣列多長都不需要繼續走訪，**達到最佳時間複雜度 $\Omega(1)$** 。
|
|
|
|
|
|
“最差時間複雜度”對應函式漸近上界，使用大 $O$ 記號表示。相應地，“最佳時間複雜度”對應函式漸近下界，用 $\Omega$ 記號表示：
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
```python title="worst_best_time_complexity.py"
|
|
|
def random_numbers(n: int) -> list[int]:
|
|
|
"""生成一個陣列，元素為: 1, 2, ..., n ，順序被打亂"""
|
|
|
# 生成陣列 nums =: 1, 2, 3, ..., n
|
|
|
nums = [i for i in range(1, n + 1)]
|
|
|
# 隨機打亂陣列元素
|
|
|
random.shuffle(nums)
|
|
|
return nums
|
|
|
|
|
|
def find_one(nums: list[int]) -> int:
|
|
|
"""查詢陣列 nums 中數字 1 所在索引"""
|
|
|
for i in range(len(nums)):
|
|
|
# 當元素 1 在陣列頭部時，達到最佳時間複雜度 O(1)
|
|
|
# 當元素 1 在陣列尾部時，達到最差時間複雜度 O(n)
|
|
|
if nums[i] == 1:
|
|
|
return i
|
|
|
return -1
|
|
|
```
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
```cpp title="worst_best_time_complexity.cpp"
|
|
|
/* 生成一個陣列，元素為 { 1, 2, ..., n }，順序被打亂 */
|
|
|
vector<int> randomNumbers(int n) {
|
|
|
vector<int> nums(n);
|
|
|
// 生成陣列 nums = { 1, 2, 3, ..., n }
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
nums[i] = i + 1;
|
|
|
}
|
|
|
// 使用系統時間生成隨機種子
|
|
|
unsigned seed = chrono::system_clock::now().time_since_epoch().count();
|
|
|
// 隨機打亂陣列元素
|
|
|
shuffle(nums.begin(), nums.end(), default_random_engine(seed));
|
|
|
return nums;
|
|
|
}
|
|
|
|
|
|
/* 查詢陣列 nums 中數字 1 所在索引 */
|
|
|
int findOne(vector<int> &nums) {
|
|
|
for (int i = 0; i < nums.size(); i++) {
|
|
|
// 當元素 1 在陣列頭部時，達到最佳時間複雜度 O(1)
|
|
|
// 當元素 1 在陣列尾部時，達到最差時間複雜度 O(n)
|
|
|
if (nums[i] == 1)
|
|
|
return i;
|
|
|
}
|
|
|
return -1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
```java title="worst_best_time_complexity.java"
|
|
|
/* 生成一個陣列，元素為 { 1, 2, ..., n }，順序被打亂 */
|
|
|
int[] randomNumbers(int n) {
|
|
|
Integer[] nums = new Integer[n];
|
|
|
// 生成陣列 nums = { 1, 2, 3, ..., n }
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
nums[i] = i + 1;
|
|
|
}
|
|
|
// 隨機打亂陣列元素
|
|
|
Collections.shuffle(Arrays.asList(nums));
|
|
|
// Integer[] -> int[]
|
|
|
int[] res = new int[n];
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
res[i] = nums[i];
|
|
|
}
|
|
|
return res;
|
|
|
}
|
|
|
|
|
|
/* 查詢陣列 nums 中數字 1 所在索引 */
|
|
|
int findOne(int[] nums) {
|
|
|
for (int i = 0; i < nums.length; i++) {
|
|
|
// 當元素 1 在陣列頭部時，達到最佳時間複雜度 O(1)
|
|
|
// 當元素 1 在陣列尾部時，達到最差時間複雜度 O(n)
|
|
|
if (nums[i] == 1)
|
|
|
return i;
|
|
|
}
|
|
|
return -1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
```csharp title="worst_best_time_complexity.cs"
|
|
|
/* 生成一個陣列，元素為 { 1, 2, ..., n }，順序被打亂 */
|
|
|
int[] RandomNumbers(int n) {
|
|
|
int[] nums = new int[n];
|
|
|
// 生成陣列 nums = { 1, 2, 3, ..., n }
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
nums[i] = i + 1;
|
|
|
}
|
|
|
|
|
|
// 隨機打亂陣列元素
|
|
|
for (int i = 0; i < nums.Length; i++) {
|
|
|
int index = new Random().Next(i, nums.Length);
|
|
|
(nums[i], nums[index]) = (nums[index], nums[i]);
|
|
|
}
|
|
|
return nums;
|
|
|
}
|
|
|
|
|
|
/* 查詢陣列 nums 中數字 1 所在索引 */
|
|
|
int FindOne(int[] nums) {
|
|
|
for (int i = 0; i < nums.Length; i++) {
|
|
|
// 當元素 1 在陣列頭部時，達到最佳時間複雜度 O(1)
|
|
|
// 當元素 1 在陣列尾部時，達到最差時間複雜度 O(n)
|
|
|
if (nums[i] == 1)
|
|
|
return i;
|
|
|
}
|
|
|
return -1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
```go title="worst_best_time_complexity.go"
|
|
|
/* 生成一個陣列，元素為 { 1, 2, ..., n }，順序被打亂 */
|
|
|
func randomNumbers(n int) []int {
|
|
|
nums := make([]int, n)
|
|
|
// 生成陣列 nums = { 1, 2, 3, ..., n }
|
|
|
for i := 0; i < n; i++ {
|
|
|
nums[i] = i + 1
|
|
|
}
|
|
|
// 隨機打亂陣列元素
|
|
|
rand.Shuffle(len(nums), func(i, j int) {
|
|
|
nums[i], nums[j] = nums[j], nums[i]
|
|
|
})
|
|
|
return nums
|
|
|
}
|
|
|
|
|
|
/* 查詢陣列 nums 中數字 1 所在索引 */
|
|
|
func findOne(nums []int) int {
|
|
|
for i := 0; i < len(nums); i++ {
|
|
|
// 當元素 1 在陣列頭部時，達到最佳時間複雜度 O(1)
|
|
|
// 當元素 1 在陣列尾部時，達到最差時間複雜度 O(n)
|
|
|
if nums[i] == 1 {
|
|
|
return i
|
|
|
}
|
|
|
}
|
|
|
return -1
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
```swift title="worst_best_time_complexity.swift"
|
|
|
/* 生成一個陣列，元素為 { 1, 2, ..., n }，順序被打亂 */
|
|
|
func randomNumbers(n: Int) -> [Int] {
|
|
|
// 生成陣列 nums = { 1, 2, 3, ..., n }
|
|
|
var nums = Array(1 ... n)
|
|
|
// 隨機打亂陣列元素
|
|
|
nums.shuffle()
|
|
|
return nums
|
|
|
}
|
|
|
|
|
|
/* 查詢陣列 nums 中數字 1 所在索引 */
|
|
|
func findOne(nums: [Int]) -> Int {
|
|
|
for i in nums.indices {
|
|
|
// 當元素 1 在陣列頭部時，達到最佳時間複雜度 O(1)
|
|
|
// 當元素 1 在陣列尾部時，達到最差時間複雜度 O(n)
|
|
|
if nums[i] == 1 {
|
|
|
return i
|
|
|
}
|
|
|
}
|
|
|
return -1
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
```javascript title="worst_best_time_complexity.js"
|
|
|
/* 生成一個陣列，元素為 { 1, 2, ..., n }，順序被打亂 */
|
|
|
function randomNumbers(n) {
|
|
|
const nums = Array(n);
|
|
|
// 生成陣列 nums = { 1, 2, 3, ..., n }
|
|
|
for (let i = 0; i < n; i++) {
|
|
|
nums[i] = i + 1;
|
|
|
}
|
|
|
// 隨機打亂陣列元素
|
|
|
for (let i = 0; i < n; i++) {
|
|
|
const r = Math.floor(Math.random() * (i + 1));
|
|
|
const temp = nums[i];
|
|
|
nums[i] = nums[r];
|
|
|
nums[r] = temp;
|
|
|
}
|
|
|
return nums;
|
|
|
}
|
|
|
|
|
|
/* 查詢陣列 nums 中數字 1 所在索引 */
|
|
|
function findOne(nums) {
|
|
|
for (let i = 0; i < nums.length; i++) {
|
|
|
// 當元素 1 在陣列頭部時，達到最佳時間複雜度 O(1)
|
|
|
// 當元素 1 在陣列尾部時，達到最差時間複雜度 O(n)
|
|
|
if (nums[i] === 1) {
|
|
|
return i;
|
|
|
}
|
|
|
}
|
|
|
return -1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
```typescript title="worst_best_time_complexity.ts"
|
|
|
/* 生成一個陣列，元素為 { 1, 2, ..., n }，順序被打亂 */
|
|
|
function randomNumbers(n: number): number[] {
|
|
|
const nums = Array(n);
|
|
|
// 生成陣列 nums = { 1, 2, 3, ..., n }
|
|
|
for (let i = 0; i < n; i++) {
|
|
|
nums[i] = i + 1;
|
|
|
}
|
|
|
// 隨機打亂陣列元素
|
|
|
for (let i = 0; i < n; i++) {
|
|
|
const r = Math.floor(Math.random() * (i + 1));
|
|
|
const temp = nums[i];
|
|
|
nums[i] = nums[r];
|
|
|
nums[r] = temp;
|
|
|
}
|
|
|
return nums;
|
|
|
}
|
|
|
|
|
|
/* 查詢陣列 nums 中數字 1 所在索引 */
|
|
|
function findOne(nums: number[]): number {
|
|
|
for (let i = 0; i < nums.length; i++) {
|
|
|
// 當元素 1 在陣列頭部時，達到最佳時間複雜度 O(1)
|
|
|
// 當元素 1 在陣列尾部時，達到最差時間複雜度 O(n)
|
|
|
if (nums[i] === 1) {
|
|
|
return i;
|
|
|
}
|
|
|
}
|
|
|
return -1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
```dart title="worst_best_time_complexity.dart"
|
|
|
/* 生成一個陣列，元素為 { 1, 2, ..., n }，順序被打亂 */
|
|
|
List<int> randomNumbers(int n) {
|
|
|
final nums = List.filled(n, 0);
|
|
|
// 生成陣列 nums = { 1, 2, 3, ..., n }
|
|
|
for (var i = 0; i < n; i++) {
|
|
|
nums[i] = i + 1;
|
|
|
}
|
|
|
// 隨機打亂陣列元素
|
|
|
nums.shuffle();
|
|
|
|
|
|
return nums;
|
|
|
}
|
|
|
|
|
|
/* 查詢陣列 nums 中數字 1 所在索引 */
|
|
|
int findOne(List<int> nums) {
|
|
|
for (var i = 0; i < nums.length; i++) {
|
|
|
// 當元素 1 在陣列頭部時，達到最佳時間複雜度 O(1)
|
|
|
// 當元素 1 在陣列尾部時，達到最差時間複雜度 O(n)
|
|
|
if (nums[i] == 1) return i;
|
|
|
}
|
|
|
|
|
|
return -1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
```rust title="worst_best_time_complexity.rs"
|
|
|
/* 生成一個陣列，元素為 { 1, 2, ..., n }，順序被打亂 */
|
|
|
fn random_numbers(n: i32) -> Vec<i32> {
|
|
|
// 生成陣列 nums = { 1, 2, 3, ..., n }
|
|
|
let mut nums = (1..=n).collect::<Vec<i32>>();
|
|
|
// 隨機打亂陣列元素
|
|
|
nums.shuffle(&mut thread_rng());
|
|
|
nums
|
|
|
}
|
|
|
|
|
|
/* 查詢陣列 nums 中數字 1 所在索引 */
|
|
|
fn find_one(nums: &[i32]) -> Option<usize> {
|
|
|
for i in 0..nums.len() {
|
|
|
// 當元素 1 在陣列頭部時，達到最佳時間複雜度 O(1)
|
|
|
// 當元素 1 在陣列尾部時，達到最差時間複雜度 O(n)
|
|
|
if nums[i] == 1 {
|
|
|
return Some(i);
|
|
|
}
|
|
|
}
|
|
|
None
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
```c title="worst_best_time_complexity.c"
|
|
|
/* 生成一個陣列，元素為 { 1, 2, ..., n }，順序被打亂 */
|
|
|
int *randomNumbers(int n) {
|
|
|
// 分配堆積區記憶體（建立一維可變長陣列：陣列中元素數量為 n ，元素型別為 int ）
|
|
|
int *nums = (int *)malloc(n * sizeof(int));
|
|
|
// 生成陣列 nums = { 1, 2, 3, ..., n }
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
nums[i] = i + 1;
|
|
|
}
|
|
|
// 隨機打亂陣列元素
|
|
|
for (int i = n - 1; i > 0; i--) {
|
|
|
int j = rand() % (i + 1);
|
|
|
int temp = nums[i];
|
|
|
nums[i] = nums[j];
|
|
|
nums[j] = temp;
|
|
|
}
|
|
|
return nums;
|
|
|
}
|
|
|
|
|
|
/* 查詢陣列 nums 中數字 1 所在索引 */
|
|
|
int findOne(int *nums, int n) {
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
// 當元素 1 在陣列頭部時，達到最佳時間複雜度 O(1)
|
|
|
// 當元素 1 在陣列尾部時，達到最差時間複雜度 O(n)
|
|
|
if (nums[i] == 1)
|
|
|
return i;
|
|
|
}
|
|
|
return -1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Kotlin"
|
|
|
|
|
|
```kotlin title="worst_best_time_complexity.kt"
|
|
|
/* 生成一個陣列，元素為 { 1, 2, ..., n }，順序被打亂 */
|
|
|
fun randomNumbers(n: Int): Array<Int?> {
|
|
|
val nums = IntArray(n)
|
|
|
// 生成陣列 nums = { 1, 2, 3, ..., n }
|
|
|
for (i in 0..<n) {
|
|
|
nums[i] = i + 1
|
|
|
}
|
|
|
// 隨機打亂陣列元素
|
|
|
val mutableList = nums.toMutableList()
|
|
|
mutableList.shuffle()
|
|
|
// Integer[] -> int[]
|
|
|
val res = arrayOfNulls<Int>(n)
|
|
|
for (i in 0..<n) {
|
|
|
res[i] = mutableList[i]
|
|
|
}
|
|
|
return res
|
|
|
}
|
|
|
|
|
|
/* 查詢陣列 nums 中數字 1 所在索引 */
|
|
|
fun findOne(nums: Array<Int?>): Int {
|
|
|
for (i in nums.indices) {
|
|
|
// 當元素 1 在陣列頭部時，達到最佳時間複雜度 O(1)
|
|
|
// 當元素 1 在陣列尾部時，達到最差時間複雜度 O(n)
|
|
|
if (nums[i] == 1)
|
|
|
return i
|
|
|
}
|
|
|
return -1
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Ruby"
|
|
|
|
|
|
```ruby title="worst_best_time_complexity.rb"
|
|
|
### 生成一個陣列，元素為: 1, 2, ..., n ，順序被打亂 ###
|
|
|
def random_numbers(n)
|
|
|
# 生成陣列 nums =: 1, 2, 3, ..., n
|
|
|
nums = Array.new(n) { |i| i + 1 }
|
|
|
# 隨機打亂陣列元素
|
|
|
nums.shuffle!
|
|
|
end
|
|
|
|
|
|
### 查詢陣列 nums 中數字 1 所在索引 ###
|
|
|
def find_one(nums)
|
|
|
for i in 0...nums.length
|
|
|
# 當元素 1 在陣列頭部時，達到最佳時間複雜度 O(1)
|
|
|
# 當元素 1 在陣列尾部時，達到最差時間複雜度 O(n)
|
|
|
return i if nums[i] == 1
|
|
|
end
|
|
|
|
|
|
-1
|
|
|
end
|
|
|
```
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
```zig title="worst_best_time_complexity.zig"
|
|
|
// 生成一個陣列，元素為 { 1, 2, ..., n }，順序被打亂
|
|
|
fn randomNumbers(comptime n: usize) [n]i32 {
|
|
|
var nums: [n]i32 = undefined;
|
|
|
// 生成陣列 nums = { 1, 2, 3, ..., n }
|
|
|
for (&nums, 0..) |*num, i| {
|
|
|
num.* = @as(i32, @intCast(i)) + 1;
|
|
|
}
|
|
|
// 隨機打亂陣列元素
|
|
|
const rand = std.crypto.random;
|
|
|
rand.shuffle(i32, &nums);
|
|
|
return nums;
|
|
|
}
|
|
|
|
|
|
// 查詢陣列 nums 中數字 1 所在索引
|
|
|
fn findOne(nums: []i32) i32 {
|
|
|
for (nums, 0..) |num, i| {
|
|
|
// 當元素 1 在陣列頭部時，達到最佳時間複雜度 O(1)
|
|
|
// 當元素 1 在陣列尾部時，達到最差時間複雜度 O(n)
|
|
|
if (num == 1) return @intCast(i);
|
|
|
}
|
|
|
return -1;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
??? pythontutor "視覺化執行"
|
|
|
|
|
|
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=import%20random%0A%0Adef%20random_numbers%28n%3A%20int%29%20-%3E%20list%5Bint%5D%3A%0A%20%20%20%20%22%22%22%E7%94%9F%E6%88%90%E4%B8%80%E4%B8%AA%E6%95%B0%E7%BB%84%EF%BC%8C%E5%85%83%E7%B4%A0%E4%B8%BA%3A%201,%202,%20...,%20n%20%EF%BC%8C%E9%A1%BA%E5%BA%8F%E8%A2%AB%E6%89%93%E4%B9%B1%22%22%22%0A%20%20%20%20%23%20%E7%94%9F%E6%88%90%E6%95%B0%E7%BB%84%20nums%20%3D%3A%201,%202,%203,%20...,%20n%0A%20%20%20%20nums%20%3D%20%5Bi%20for%20i%20in%20range%281,%20n%20%2B%201%29%5D%0A%20%20%20%20%23%20%E9%9A%8F%E6%9C%BA%E6%89%93%E4%B9%B1%E6%95%B0%E7%BB%84%E5%85%83%E7%B4%A0%0A%20%20%20%20random.shuffle%28nums%29%0A%20%20%20%20return%20nums%0A%0Adef%20find_one%28nums%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9F%A5%E6%89%BE%E6%95%B0%E7%BB%84%20nums%20%E4%B8%AD%E6%95%B0%E5%AD%97%201%20%E6%89%80%E5%9C%A8%E7%B4%A2%E5%BC%95%22%22%22%0A%20%20%20%20for%20i%20in%20range%28len%28nums%29%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%BD%93%E5%85%83%E7%B4%A0%201%20%E5%9C%A8%E6%95%B0%E7%BB%84%E5%A4%B4%E9%83%A8%E6%97%B6%EF%BC%8C%E8%BE%BE%E5%88%B0%E6%9C%80%E4%BD%B3%E6%97%B6%E9%97%B4%E5%A4%8D%E6%9D%82%E5%BA%A6%20O%281%29%0A%20%20%20%20%20%20%20%20%23%20%E5%BD%93%E5%85%83%E7%B4%A0%201%20%E5%9C%A8%E6%95%B0%E7%BB%84%E5%B0%BE%E9%83%A8%E6%97%B6%EF%BC%8C%E8%BE%BE%E5%88%B0%E6%9C%80%E5%B7%AE%E6%97%B6%E9%97%B4%E5%A4%8D%E6%9D%82%E5%BA%A6%20O%28n%29%0A%20%20%20%20%20%20%20%20if%20nums%5Bi%5D%20%3D%3D%201%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20return%20i%0A%20%20%20%20return%20-1%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%2010%0A%20%20%20%20nums%20%3D%20random_numbers%28n%29%0A%20%20%20%20index%20%3D%20find_one%28nums%29%0A%20%20%20%20print%28%22%5Cn%E6%95%B0%E7%BB%84%20%5B%201,%202,%20...,%20n%20%5D%20%E8%A2%AB%E6%89%93%E4%B9%B1%E5%90%8E%20%3D%22,%20nums%29%0A%20%20%20%20print%28%22%E6%95%B0%E5%AD%97%201%20%E7%9A%84%E7%B4%A2%E5%BC%95%E4%B8%BA%22,%20index%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=25&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
|
|
|
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=import%20random%0A%0Adef%20random_numbers%28n%3A%20int%29%20-%3E%20list%5Bint%5D%3A%0A%20%20%20%20%22%22%22%E7%94%9F%E6%88%90%E4%B8%80%E4%B8%AA%E6%95%B0%E7%BB%84%EF%BC%8C%E5%85%83%E7%B4%A0%E4%B8%BA%3A%201,%202,%20...,%20n%20%EF%BC%8C%E9%A1%BA%E5%BA%8F%E8%A2%AB%E6%89%93%E4%B9%B1%22%22%22%0A%20%20%20%20%23%20%E7%94%9F%E6%88%90%E6%95%B0%E7%BB%84%20nums%20%3D%3A%201,%202,%203,%20...,%20n%0A%20%20%20%20nums%20%3D%20%5Bi%20for%20i%20in%20range%281,%20n%20%2B%201%29%5D%0A%20%20%20%20%23%20%E9%9A%8F%E6%9C%BA%E6%89%93%E4%B9%B1%E6%95%B0%E7%BB%84%E5%85%83%E7%B4%A0%0A%20%20%20%20random.shuffle%28nums%29%0A%20%20%20%20return%20nums%0A%0Adef%20find_one%28nums%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9F%A5%E6%89%BE%E6%95%B0%E7%BB%84%20nums%20%E4%B8%AD%E6%95%B0%E5%AD%97%201%20%E6%89%80%E5%9C%A8%E7%B4%A2%E5%BC%95%22%22%22%0A%20%20%20%20for%20i%20in%20range%28len%28nums%29%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%BD%93%E5%85%83%E7%B4%A0%201%20%E5%9C%A8%E6%95%B0%E7%BB%84%E5%A4%B4%E9%83%A8%E6%97%B6%EF%BC%8C%E8%BE%BE%E5%88%B0%E6%9C%80%E4%BD%B3%E6%97%B6%E9%97%B4%E5%A4%8D%E6%9D%82%E5%BA%A6%20O%281%29%0A%20%20%20%20%20%20%20%20%23%20%E5%BD%93%E5%85%83%E7%B4%A0%201%20%E5%9C%A8%E6%95%B0%E7%BB%84%E5%B0%BE%E9%83%A8%E6%97%B6%EF%BC%8C%E8%BE%BE%E5%88%B0%E6%9C%80%E5%B7%AE%E6%97%B6%E9%97%B4%E5%A4%8D%E6%9D%82%E5%BA%A6%20O%28n%29%0A%20%20%20%20%20%20%20%20if%20nums%5Bi%5D%20%3D%3D%201%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20return%20i%0A%20%20%20%20return%20-1%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%2010%0A%20%20%20%20nums%20%3D%20random_numbers%28n%29%0A%20%20%20%20index%20%3D%20find_one%28nums%29%0A%20%20%20%20print%28%22%5Cn%E6%95%B0%E7%BB%84%20%5B%201,%202,%20...,%20n%20%5D%20%E8%A2%AB%E6%89%93%E4%B9%B1%E5%90%8E%20%3D%22,%20nums%29%0A%20%20%20%20print%28%22%E6%95%B0%E5%AD%97%201%20%E7%9A%84%E7%B4%A2%E5%BC%95%E4%B8%BA%22,%20index%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=25&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全螢幕觀看 ></a></div>
|
|
|
|
|
|
值得說明的是，我們在實際中很少使用最佳時間複雜度，因為通常只有在很小機率下才能達到，可能會帶來一定的誤導性。**而最差時間複雜度更為實用，因為它給出了一個效率安全值**，讓我們可以放心地使用演算法。
|
|
|
|
|
|
從上述示例可以看出，最差時間複雜度和最佳時間複雜度只出現於“特殊的資料分佈”，這些情況的出現機率可能很小，並不能真實地反映演算法執行效率。相比之下，**平均時間複雜度可以體現演算法在隨機輸入資料下的執行效率**，用 $\Theta$ 記號來表示。
|
|
|
|
|
|
對於部分演算法，我們可以簡單地推算出隨機資料分佈下的平均情況。比如上述示例，由於輸入陣列是被打亂的，因此元素 $1$ 出現在任意索引的機率都是相等的，那麼演算法的平均迴圈次數就是陣列長度的一半 $n / 2$ ，平均時間複雜度為 $\Theta(n / 2) = \Theta(n)$ 。
|
|
|
|
|
|
但對於較為複雜的演算法，計算平均時間複雜度往往比較困難，因為很難分析出在資料分佈下的整體數學期望。在這種情況下，我們通常使用最差時間複雜度作為演算法效率的評判標準。
|
|
|
|
|
|
!!! question "為什麼很少看到 $\Theta$ 符號？"
|
|
|
|
|
|
可能由於 $O$ 符號過於朗朗上口，因此我們常常使用它來表示平均時間複雜度。但從嚴格意義上講，這種做法並不規範。在本書和其他資料中，若遇到類似“平均時間複雜度 $O(n)$”的表述，請將其直接理解為 $\Theta(n)$ 。
|
