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---
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comments: true
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---
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# 时间复杂度
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## 统计算法运行时间
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运行时间能够直观且准确地体现出算法的效率水平。如果我们想要 **准确预估一段代码的运行时间** ,该如何做呢?
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1. 首先需要 **确定运行平台** ,包括硬件配置、编程语言、系统环境等,这些都会影响到代码的运行效率。
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2. 评估 **各种计算操作的所需运行时间** ,例如加法操作 `+` 需要 1 ns ,乘法操作 `*` 需要 10 ns ,打印操作需要 5 ns 等。
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3. 根据代码 **统计所有计算操作的数量** ,并将所有操作的执行时间求和,即可得到运行时间。
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例如以下代码,输入数据大小为 $n$ ,根据以上方法,可以得到算法运行时间为 $6n + 12$ 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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=== "Java"
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```java title=""
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// 在某运行平台下
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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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// 循环 n 次
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for (int i = 0; i < n; i++) { // 1 ns ,每轮都要执行 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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```cpp title=""
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// 在某运行平台下
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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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// 循环 n 次
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for (int i = 0; i < n; i++) { // 1 ns ,每轮都要执行 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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=== "Python"
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```python title=""
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# 在某运行平台下
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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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# 循环 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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=== "Go"
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```go title=""
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// 在某运行平台下
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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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// 循环 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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=== "JavaScript"
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```js title=""
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// 在某运行平台下
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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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// 循环 n 次
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for(let i = 0; i < n; i++) { // 1 ns ,每轮都要执行 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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=== "TypeScript"
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```typescript title=""
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// 在某运行平台下
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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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// 循环 n 次
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for(let i = 0; i < n; i++) { // 1 ns ,每轮都要执行 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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=== "C"
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```c title=""
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// 在某运行平台下
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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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// 循环 n 次
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for (int i = 0; i < n; i++) { // 1 ns ,每轮都要执行 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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=== "C#"
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```csharp title=""
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// 在某运行平台下
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void algorithm(int n)
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{
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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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// 循环 n 次
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for (int i = 0; i < n; i++)
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{ // 1 ns ,每轮都要执行 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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=== "Swift"
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```swift title=""
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// 在某运行平台下
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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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// 循环 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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但实际上, **统计算法的运行时间既不合理也不现实**。首先,我们不希望预估时间和运行平台绑定,毕竟算法需要跑在各式各样的平台之上。其次,我们很难获知每一种操作的运行时间,这为预估过程带来了极大的难度。
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## 统计时间增长趋势
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「时间复杂度分析」采取了不同的做法,其统计的不是算法运行时间,而是 **算法运行时间随着数据量变大时的增长趋势** 。
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“时间增长趋势”这个概念比较抽象,我们借助一个例子来理解。设输入数据大小为 $n$ ,给定三个算法 `A` , `B` , `C` 。
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- 算法 `A` 只有 $1$ 个打印操作,算法运行时间不随着 $n$ 增大而增长。我们称此算法的时间复杂度为「常数阶」。
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- 算法 `B` 中的打印操作需要循环 $n$ 次,算法运行时间随着 $n$ 增大成线性增长。此算法的时间复杂度被称为「线性阶」。
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- 算法 `C` 中的打印操作需要循环 $1000000$ 次,但运行时间仍与输入数据大小 $n$ 无关。因此 `C` 的时间复杂度和 `A` 相同,仍为「常数阶」。
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=== "Java"
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```java title=""
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// 算法 A 时间复杂度:常数阶
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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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// 算法 B 时间复杂度:线性阶
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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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// 算法 C 时间复杂度:常数阶
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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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```cpp title=""
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// 算法 A 时间复杂度:常数阶
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void algorithm_A(int n) {
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cout << 0 << endl;
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}
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// 算法 B 时间复杂度:线性阶
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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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// 算法 C 时间复杂度:常数阶
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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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=== "Python"
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```python title=""
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# 算法 A 时间复杂度:常数阶
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def algorithm_A(n):
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print(0)
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# 算法 B 时间复杂度:线性阶
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def algorithm_B(n):
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for _ in range(n):
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print(0)
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# 算法 C 时间复杂度:常数阶
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def algorithm_C(n):
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for _ in range(1000000):
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print(0)
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```
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=== "Go"
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```go title=""
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// 算法 A 时间复杂度:常数阶
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func algorithm_A(n int) {
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fmt.Println(0)
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}
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// 算法 B 时间复杂度:线性阶
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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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// 算法 C 时间复杂度:常数阶
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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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=== "JavaScript"
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```js title=""
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// 算法 A 时间复杂度:常数阶
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function algorithm_A(n) {
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console.log(0);
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}
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// 算法 B 时间复杂度:线性阶
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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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// 算法 C 时间复杂度:常数阶
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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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=== "TypeScript"
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```typescript title=""
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// 算法 A 时间复杂度:常数阶
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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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// 算法 B 时间复杂度:线性阶
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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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// 算法 C 时间复杂度:常数阶
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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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=== "C"
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```c title=""
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// 算法 A 时间复杂度:常数阶
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void algorithm_A(int n) {
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printf("%d", 0);
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}
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// 算法 B 时间复杂度:线性阶
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void algorithm_B(int n) {
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for (int i = 0; i < n; i++) {
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printf("%d", 0);
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}
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}
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// 算法 C 时间复杂度:常数阶
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void algorithm_C(int n) {
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for (int i = 0; i < 1000000; i++) {
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printf("%d", 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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// 算法 A 时间复杂度:常数阶
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void algorithm_A(int n)
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{
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Console.WriteLine(0);
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}
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// 算法 B 时间复杂度:线性阶
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void algorithm_B(int n)
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{
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for (int i = 0; i < n; i++)
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{
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|
|
Console.WriteLine(0);
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
// 算法 C 时间复杂度:常数阶
|
|
|
|
|
void algorithm_C(int n)
|
|
|
|
|
{
|
|
|
|
|
for (int i = 0; i < 1000000; i++)
|
|
|
|
|
{
|
|
|
|
|
Console.WriteLine(0);
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
|
|
|
|
```swift title=""
|
|
|
|
|
// 算法 A 时间复杂度:常数阶
|
|
|
|
|
func algorithmA(_ n: Int) {
|
|
|
|
|
print(0)
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
// 算法 B 时间复杂度:线性阶
|
|
|
|
|
func algorithmB(_ n: Int) {
|
|
|
|
|
for _ in 0 ..< n {
|
|
|
|
|
print(0)
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
// 算法 C 时间复杂度:常数阶
|
|
|
|
|
func algorithmC(_ n: Int) {
|
|
|
|
|
for _ in 0 ..< 1000000 {
|
|
|
|
|
print(0)
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
![time_complexity_first_example](time_complexity.assets/time_complexity_first_example.png)
|
|
|
|
|
|
|
|
|
|
<p align="center"> Fig. 算法 A, B, C 的时间增长趋势 </p>
|
|
|
|
|
|
|
|
|
|
相比直接统计算法运行时间,时间复杂度分析的做法有什么好处呢?以及有什么不足?
|
|
|
|
|
|
|
|
|
|
**时间复杂度可以有效评估算法效率**。算法 `B` 运行时间的增长是线性的,在 $n > 1$ 时慢于算法 `A` ,在 $n > 1000000$ 时慢于算法 `C` 。实质上,只要输入数据大小 $n$ 足够大,复杂度为「常数阶」的算法一定优于「线性阶」的算法,这也正是时间增长趋势的含义。
|
|
|
|
|
|
|
|
|
|
**时间复杂度的推算方法更加简便**。在时间复杂度分析中,我们可以将统计「计算操作的运行时间」简化为统计「计算操作的数量」,这是因为,无论是运行平台还是计算操作类型,都与算法运行时间的增长趋势无关。因而,我们可以简单地将所有计算操作的执行时间统一看作是相同的“单位时间”,这样的简化做法大大降低了估算难度。
|
|
|
|
|
|
|
|
|
|
**时间复杂度也存在一定的局限性**。比如,虽然算法 `A` 和 `C` 的时间复杂度相同,但是实际的运行时间有非常大的差别。再比如,虽然算法 `B` 比 `C` 的时间复杂度要更高,但在输入数据大小 $n$ 比较小时,算法 `B` 是要明显优于算法 `C` 的。对于以上情况,我们很难仅凭时间复杂度来判定算法效率高低。然而,即使存在这些问题,计算复杂度仍然是评判算法效率的最有效且常用的方法。
|
|
|
|
|
|
|
|
|
|
## 函数渐近上界
|
|
|
|
|
|
|
|
|
|
设算法「计算操作数量」为 $T(n)$ ,其是一个关于输入数据大小 $n$ 的函数。例如,以下算法的操作数量为
|
|
|
|
|
|
|
|
|
|
$$
|
|
|
|
|
T(n) = 3 + 2n
|
|
|
|
|
$$
|
|
|
|
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
|
|
|
|
```java title=""
|
|
|
|
|
void algorithm(int n) {
|
|
|
|
|
int a = 1; // +1
|
|
|
|
|
a = a + 1; // +1
|
|
|
|
|
a = a * 2; // +1
|
|
|
|
|
// 循环 n 次
|
|
|
|
|
for (int i = 0; i < n; i++) { // +1(每轮都执行 i ++)
|
|
|
|
|
System.out.println(0); // +1
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
|
|
|
|
```cpp title=""
|
|
|
|
|
void algorithm(int n) {
|
|
|
|
|
int a = 1; // +1
|
|
|
|
|
a = a + 1; // +1
|
|
|
|
|
a = a * 2; // +1
|
|
|
|
|
// 循环 n 次
|
|
|
|
|
for (int i = 0; i < n; i++) { // +1(每轮都执行 i ++)
|
|
|
|
|
cout << 0 << endl; // +1
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
|
|
|
|
```python title=""
|
|
|
|
|
def algorithm(n):
|
|
|
|
|
a = 1 # +1
|
|
|
|
|
a = a + 1 # +1
|
|
|
|
|
a = a * 2 # +1
|
|
|
|
|
# 循环 n 次
|
|
|
|
|
for i in range(n): # +1
|
|
|
|
|
print(0) # +1
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
|
|
|
|
```go title=""
|
|
|
|
|
func algorithm(n int) {
|
|
|
|
|
a := 1 // +1
|
|
|
|
|
a = a + 1 // +1
|
|
|
|
|
a = a * 2 // +1
|
|
|
|
|
// 循环 n 次
|
|
|
|
|
for i := 0; i < n; i++ { // +1
|
|
|
|
|
fmt.Println(a) // +1
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "JavaScript"
|
|
|
|
|
|
|
|
|
|
```js title=""
|
|
|
|
|
function algorithm(n){
|
|
|
|
|
var a = 1; // +1
|
|
|
|
|
a += 1; // +1
|
|
|
|
|
a *= 2; // +1
|
|
|
|
|
// 循环 n 次
|
|
|
|
|
for(let i = 0; i < n; i++){ // +1(每轮都执行 i ++)
|
|
|
|
|
console.log(0); // +1
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TypeScript"
|
|
|
|
|
|
|
|
|
|
```typescript title=""
|
|
|
|
|
function algorithm(n: number): void{
|
|
|
|
|
var a: number = 1; // +1
|
|
|
|
|
a += 1; // +1
|
|
|
|
|
a *= 2; // +1
|
|
|
|
|
// 循环 n 次
|
|
|
|
|
for(let i = 0; i < n; i++){ // +1(每轮都执行 i ++)
|
|
|
|
|
console.log(0); // +1
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
|
|
|
|
```c title=""
|
|
|
|
|
void algorithm(int n) {
|
|
|
|
|
int a = 1; // +1
|
|
|
|
|
a = a + 1; // +1
|
|
|
|
|
a = a * 2; // +1
|
|
|
|
|
// 循环 n 次
|
|
|
|
|
for (int i = 0; i < n; i++) { // +1(每轮都执行 i ++)
|
|
|
|
|
printf("%d", 0); // +1
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
|
|
|
|
```csharp title=""
|
|
|
|
|
void algorithm(int n) {
|
|
|
|
|
int a = 1; // +1
|
|
|
|
|
a = a + 1; // +1
|
|
|
|
|
a = a * 2; // +1
|
|
|
|
|
// 循环 n 次
|
|
|
|
|
for (int i = 0; i < n; i++) { // +1(每轮都执行 i ++)
|
|
|
|
|
Console.WriteLine(0); // +1
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
|
|
|
|
```swift title=""
|
|
|
|
|
func algorithm(n: Int) {
|
|
|
|
|
var a = 1 // +1
|
|
|
|
|
a = a + 1 // +1
|
|
|
|
|
a = a * 2 // +1
|
|
|
|
|
// 循环 n 次
|
|
|
|
|
for _ in 0 ..< n { // +1
|
|
|
|
|
print(0) // +1
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
$T(n)$ 是个一次函数,说明时间增长趋势是线性的,因此易得时间复杂度是线性阶。
|
|
|
|
|
|
|
|
|
|
我们将线性阶的时间复杂度记为 $O(n)$ ,这个数学符号被称为「大 $O$ 记号 Big-$O$ Notation」,代表函数 $T(n)$ 的「渐近上界 asymptotic upper bound」。
|
|
|
|
|
|
|
|
|
|
我们要推算时间复杂度,本质上是在计算「操作数量函数 $T(n)$ 」的渐近上界。下面我们先来看看函数渐近上界的数学定义。
|
|
|
|
|
|
|
|
|
|
!!! abstract "函数渐近上界"
|
|
|
|
|
|
|
|
|
|
若存在正实数 $c$ 和实数 $n_0$ ,使得对于所有的 $n > n_0$ ,均有
|
|
|
|
|
$$
|
|
|
|
|
T(n) \leq c \cdot f(n)
|
|
|
|
|
$$
|
|
|
|
|
则可认为 $f(n)$ 给出了 $T(n)$ 的一个渐近上界,记为
|
|
|
|
|
$$
|
|
|
|
|
T(n) = O(f(n))
|
|
|
|
|
$$
|
|
|
|
|
|
|
|
|
|
![asymptotic_upper_bound](time_complexity.assets/asymptotic_upper_bound.png)
|
|
|
|
|
|
|
|
|
|
<p align="center"> Fig. 函数的渐近上界 </p>
|
|
|
|
|
|
|
|
|
|
本质上看,计算渐近上界就是在找一个函数 $f(n)$ ,**使得在 $n$ 趋向于无穷大时,$T(n)$ 和 $f(n)$ 处于相同的增长级别(仅相差一个常数项 $c$ 的倍数)**。
|
|
|
|
|
|
|
|
|
|
!!! tip
|
|
|
|
|
|
|
|
|
|
渐近上界的数学味儿有点重,如果你感觉没有完全理解,无需担心,因为在实际使用中我们只需要会推算即可,数学意义可以慢慢领悟。
|
|
|
|
|
|
|
|
|
|
## 推算方法
|
|
|
|
|
|
|
|
|
|
推算出 $f(n)$ 后,我们就得到时间复杂度 $O(f(n))$ 。那么,如何来确定渐近上界 $f(n)$ 呢?总体分为两步,首先「统计操作数量」,然后「判断渐近上界」。
|
|
|
|
|
|
|
|
|
|
### 1. 统计操作数量
|
|
|
|
|
|
|
|
|
|
对着代码,从上到下一行一行地计数即可。然而,**由于上述 $c \cdot f(n)$ 中的常数项 $c$ 可以取任意大小,因此操作数量 $T(n)$ 中的各种系数、常数项都可以被忽略**。根据此原则,可以总结出以下计数偷懒技巧:
|
|
|
|
|
|
|
|
|
|
1. **跳过数量与 $n$ 无关的操作**。因为他们都是 $T(n)$ 中的常数项,对时间复杂度不产生影响。
|
|
|
|
|
2. **省略所有系数**。例如,循环 $2n$ 次、$5n + 1$ 次、……,都可以化简记为 $n$ 次,因为 $n$ 前面的系数对时间复杂度也不产生影响。
|
|
|
|
|
3. **循环嵌套时使用乘法**。总操作数量等于外层循环和内层循环操作数量之积,每一层循环依然可以分别套用上述 `1.` 和 `2.` 技巧。
|
|
|
|
|
|
|
|
|
|
根据以下示例,使用上述技巧前、后的统计结果分别为
|
|
|
|
|
|
|
|
|
|
$$
|
|
|
|
|
\begin{aligned}
|
|
|
|
|
T(n) & = 2n(n + 1) + (5n + 1) + 2 & \text{完整统计 (-.-|||)} \newline
|
|
|
|
|
& = 2n^2 + 7n + 3 \newline
|
|
|
|
|
T(n) & = n^2 + n & \text{偷懒统计 (o.O)}
|
|
|
|
|
\end{aligned}
|
|
|
|
|
$$
|
|
|
|
|
|
|
|
|
|
最终,两者都能推出相同的时间复杂度结果,即 $O(n^2)$ 。
|
|
|
|
|
|
|
|
|
|
=== "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++"
|
|
|
|
|
|
|
|
|
|
```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;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
|
|
|
|
```python title=""
|
|
|
|
|
def algorithm(n):
|
|
|
|
|
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)
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "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)
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "JavaScript"
|
|
|
|
|
|
|
|
|
|
```js 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);
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TypeScript"
|
|
|
|
|
|
|
|
|
|
```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);
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "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);
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "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);
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "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)
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
### 2. 判断渐近上界
|
|
|
|
|
|
|
|
|
|
**时间复杂度由多项式 $T(n)$ 中最高阶的项来决定**。这是因为在 $n$ 趋于无穷大时,最高阶的项将处于主导作用,其它项的影响都可以被忽略。
|
|
|
|
|
|
|
|
|
|
以下表格给出了一些例子,其中有一些夸张的值,是想要向大家强调 **系数无法撼动阶数** 这一结论。在 $n$ 趋于无穷大时,这些常数都是“浮云”。
|
|
|
|
|
|
|
|
|
|
<div class="center-table" markdown>
|
|
|
|
|
|
|
|
|
|
| 操作数量 $T(n)$ | 时间复杂度 $O(f(n))$ |
|
|
|
|
|
| ---------------------- | -------------------- |
|
|
|
|
|
| $100000$ | $O(1)$ |
|
|
|
|
|
| $3n + 2$ | $O(n)$ |
|
|
|
|
|
| $2n^2 + 3n + 2$ | $O(n^2)$ |
|
|
|
|
|
| $n^3 + 10000n^2$ | $O(n^3)$ |
|
|
|
|
|
| $2^n + 10000n^{10000}$ | $O(2^n)$ |
|
|
|
|
|
|
|
|
|
|
</div>
|
|
|
|
|
|
|
|
|
|
## 常见类型
|
|
|
|
|
|
|
|
|
|
设输入数据大小为 $n$ ,常见的时间复杂度类型有(从低到高排列)
|
|
|
|
|
|
|
|
|
|
$$
|
|
|
|
|
\begin{aligned}
|
|
|
|
|
O(1) < O(\log n) < O(n) < O(n \log n) < O(n^2) < O(2^n) < O(n!) \newline
|
|
|
|
|
\text{常数阶} < \text{对数阶} < \text{线性阶} < \text{线性对数阶} < \text{平方阶} < \text{指数阶} < \text{阶乘阶}
|
|
|
|
|
\end{aligned}
|
|
|
|
|
$$
|
|
|
|
|
|
|
|
|
|
![time_complexity_common_types](time_complexity.assets/time_complexity_common_types.png)
|
|
|
|
|
|
|
|
|
|
<p align="center"> Fig. 时间复杂度的常见类型 </p>
|
|
|
|
|
|
|
|
|
|
!!! tip
|
|
|
|
|
|
|
|
|
|
部分示例代码需要一些前置知识,包括数组、递归算法等。如果遇到看不懂的地方无需担心,可以在学习完后面章节后再来复习,现阶段先聚焦在理解时间复杂度含义和推算方法上。
|
|
|
|
|
|
|
|
|
|
### 常数阶 $O(1)$
|
|
|
|
|
|
|
|
|
|
常数阶的操作数量与输入数据大小 $n$ 无关,即不随着 $n$ 的变化而变化。
|
|
|
|
|
|
|
|
|
|
对于以下算法,无论操作数量 `size` 有多大,只要与数据大小 $n$ 无关,时间复杂度就仍为 $O(1)$ 。
|
|
|
|
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
|
|
|
|
```java title="time_complexity.java"
|
|
|
|
|
/* 常数阶 */
|
|
|
|
|
int constant(int n) {
|
|
|
|
|
int count = 0;
|
|
|
|
|
int size = 100000;
|
|
|
|
|
for (int i = 0; i < size; i++)
|
|
|
|
|
count++;
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
|
|
|
|
```cpp title="time_complexity.cpp"
|
|
|
|
|
/* 常数阶 */
|
|
|
|
|
int constant(int n) {
|
|
|
|
|
int count = 0;
|
|
|
|
|
int size = 100000;
|
|
|
|
|
for (int i = 0; i < size; i++)
|
|
|
|
|
count++;
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
|
|
|
|
```python title="time_complexity.py"
|
|
|
|
|
""" 常数阶 """
|
|
|
|
|
def constant(n):
|
|
|
|
|
count = 0
|
|
|
|
|
size = 100000
|
|
|
|
|
for _ in range(size):
|
|
|
|
|
count += 1
|
|
|
|
|
return count
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
|
|
|
|
```go title="time_complexity.go"
|
|
|
|
|
/* 常数阶 */
|
|
|
|
|
func constant(n int) int {
|
|
|
|
|
count := 0
|
|
|
|
|
size := 100000
|
|
|
|
|
for i := 0; i < size; i++ {
|
|
|
|
|
count ++
|
|
|
|
|
}
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "JavaScript"
|
|
|
|
|
|
|
|
|
|
```js title="time_complexity.js"
|
|
|
|
|
/* 常数阶 */
|
|
|
|
|
function constant(n) {
|
|
|
|
|
let count = 0;
|
|
|
|
|
const size = 100000;
|
|
|
|
|
for (let i = 0; i < size; i++) count++;
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TypeScript"
|
|
|
|
|
|
|
|
|
|
```typescript title="time_complexity.ts"
|
|
|
|
|
/* 常数阶 */
|
|
|
|
|
function constant(n: number): number {
|
|
|
|
|
let count = 0;
|
|
|
|
|
const size = 100000;
|
|
|
|
|
for (let i = 0; i < size; i++) count++;
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
|
|
|
|
```c title="time_complexity.c"
|
|
|
|
|
/* 常数阶 */
|
|
|
|
|
int constant(int n) {
|
|
|
|
|
int count = 0;
|
|
|
|
|
int size = 100000;
|
|
|
|
|
int i = 0;
|
|
|
|
|
for (int i = 0; i < size; i++) {
|
|
|
|
|
count ++;
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
|
|
|
|
```csharp title="time_complexity.cs"
|
|
|
|
|
/* 常数阶 */
|
|
|
|
|
int constant(int n)
|
|
|
|
|
{
|
|
|
|
|
int count = 0;
|
|
|
|
|
int size = 100000;
|
|
|
|
|
for (int i = 0; i < size; i++)
|
|
|
|
|
count++;
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
|
|
|
|
```swift title="time_complexity.swift"
|
|
|
|
|
// 常数阶
|
|
|
|
|
func constant(n: Int) -> Int {
|
|
|
|
|
var count = 0
|
|
|
|
|
let size = 100000
|
|
|
|
|
for _ in 0 ..< size {
|
|
|
|
|
count += 1
|
|
|
|
|
}
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
### 线性阶 $O(n)$
|
|
|
|
|
|
|
|
|
|
线性阶的操作数量相对输入数据大小成线性级别增长。线性阶常出现于单层循环。
|
|
|
|
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
|
|
|
|
```java title="time_complexity.java"
|
|
|
|
|
/* 线性阶 */
|
|
|
|
|
int linear(int n) {
|
|
|
|
|
int count = 0;
|
|
|
|
|
for (int i = 0; i < n; i++)
|
|
|
|
|
count++;
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
|
|
|
|
```cpp title="time_complexity.cpp"
|
|
|
|
|
/* 线性阶 */
|
|
|
|
|
int linear(int n) {
|
|
|
|
|
int count = 0;
|
|
|
|
|
for (int i = 0; i < n; i++)
|
|
|
|
|
count++;
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
|
|
|
|
```python title="time_complexity.py"
|
|
|
|
|
""" 线性阶 """
|
|
|
|
|
def linear(n):
|
|
|
|
|
count = 0
|
|
|
|
|
for _ in range(n):
|
|
|
|
|
count += 1
|
|
|
|
|
return count
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
|
|
|
|
```go title="time_complexity.go"
|
|
|
|
|
/* 线性阶 */
|
|
|
|
|
func linear(n int) int {
|
|
|
|
|
count := 0
|
|
|
|
|
for i := 0; i < n; i++ {
|
|
|
|
|
count++
|
|
|
|
|
}
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "JavaScript"
|
|
|
|
|
|
|
|
|
|
```js title="time_complexity.js"
|
|
|
|
|
/* 线性阶 */
|
|
|
|
|
function linear(n) {
|
|
|
|
|
let count = 0;
|
|
|
|
|
for (let i = 0; i < n; i++) count++;
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TypeScript"
|
|
|
|
|
|
|
|
|
|
```typescript title="time_complexity.ts"
|
|
|
|
|
/* 线性阶 */
|
|
|
|
|
function linear(n: number): number {
|
|
|
|
|
let count = 0;
|
|
|
|
|
for (let i = 0; i < n; i++) count++;
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
|
|
|
|
```c title="time_complexity.c"
|
|
|
|
|
/* 线性阶 */
|
|
|
|
|
int linear(int n) {
|
|
|
|
|
int count = 0;
|
|
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
|
|
count ++;
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
|
|
|
|
```csharp title="time_complexity.cs"
|
|
|
|
|
/* 线性阶 */
|
|
|
|
|
int linear(int n)
|
|
|
|
|
{
|
|
|
|
|
int count = 0;
|
|
|
|
|
for (int i = 0; i < n; i++)
|
|
|
|
|
count++;
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
|
|
|
|
```swift title="time_complexity.swift"
|
|
|
|
|
// 线性阶
|
|
|
|
|
func linear(n: Int) -> Int {
|
|
|
|
|
var count = 0
|
|
|
|
|
for _ in 0 ..< n {
|
|
|
|
|
count += 1
|
|
|
|
|
}
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
「遍历数组」和「遍历链表」等操作,时间复杂度都为 $O(n)$ ,其中 $n$ 为数组或链表的长度。
|
|
|
|
|
|
|
|
|
|
!!! tip
|
|
|
|
|
|
|
|
|
|
**数据大小 $n$ 是根据输入数据的类型来确定的**。比如,在上述示例中,我们直接将 $n$ 看作输入数据大小;以下遍历数组示例中,数据大小 $n$ 为数组的长度。
|
|
|
|
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
|
|
|
|
```java title="time_complexity.java"
|
|
|
|
|
/* 线性阶(遍历数组) */
|
|
|
|
|
int arrayTraversal(int[] nums) {
|
|
|
|
|
int count = 0;
|
|
|
|
|
// 循环次数与数组长度成正比
|
|
|
|
|
for (int num : nums) {
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
|
|
|
|
```cpp title="time_complexity.cpp"
|
|
|
|
|
/* 线性阶(遍历数组) */
|
|
|
|
|
int arrayTraversal(vector<int>& nums) {
|
|
|
|
|
int count = 0;
|
|
|
|
|
// 循环次数与数组长度成正比
|
|
|
|
|
for (int num : nums) {
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
|
|
|
|
```python title="time_complexity.py"
|
|
|
|
|
""" 线性阶(遍历数组)"""
|
|
|
|
|
def array_traversal(nums):
|
|
|
|
|
count = 0
|
|
|
|
|
# 循环次数与数组长度成正比
|
|
|
|
|
for num in nums:
|
|
|
|
|
count += 1
|
|
|
|
|
return count
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
|
|
|
|
```go title="time_complexity.go"
|
|
|
|
|
/* 线性阶(遍历数组) */
|
|
|
|
|
func arrayTraversal(nums []int) int {
|
|
|
|
|
count := 0
|
|
|
|
|
// 循环次数与数组长度成正比
|
|
|
|
|
for range nums {
|
|
|
|
|
count++
|
|
|
|
|
}
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "JavaScript"
|
|
|
|
|
|
|
|
|
|
```js title="time_complexity.js"
|
|
|
|
|
/* 线性阶(遍历数组) */
|
|
|
|
|
function arrayTraversal(nums) {
|
|
|
|
|
let count = 0;
|
|
|
|
|
// 循环次数与数组长度成正比
|
|
|
|
|
for (let i = 0; i < nums.length; i++) {
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TypeScript"
|
|
|
|
|
|
|
|
|
|
```typescript title="time_complexity.ts"
|
|
|
|
|
/* 线性阶(遍历数组) */
|
|
|
|
|
function arrayTraversal(nums: number[]): number {
|
|
|
|
|
let count = 0;
|
|
|
|
|
// 循环次数与数组长度成正比
|
|
|
|
|
for (let i = 0; i < nums.length; i++) {
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
|
|
|
|
```c title="time_complexity.c"
|
|
|
|
|
/* 线性阶(遍历数组) */
|
|
|
|
|
int arrayTraversal(int *nums, int n) {
|
|
|
|
|
int count = 0;
|
|
|
|
|
// 循环次数与数组长度成正比
|
|
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
|
|
count ++;
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
|
|
|
|
```csharp title="time_complexity.cs"
|
|
|
|
|
/* 线性阶(遍历数组) */
|
|
|
|
|
int arrayTraversal(int[] nums)
|
|
|
|
|
{
|
|
|
|
|
int count = 0;
|
|
|
|
|
// 循环次数与数组长度成正比
|
|
|
|
|
foreach(int num in nums)
|
|
|
|
|
{
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
|
|
|
|
```swift title="time_complexity.swift"
|
|
|
|
|
// 线性阶(遍历数组)
|
|
|
|
|
func arrayTraversal(nums: [Int]) -> Int {
|
|
|
|
|
var count = 0
|
|
|
|
|
// 循环次数与数组长度成正比
|
|
|
|
|
for _ in nums {
|
|
|
|
|
count += 1
|
|
|
|
|
}
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
### 平方阶 $O(n^2)$
|
|
|
|
|
|
|
|
|
|
平方阶的操作数量相对输入数据大小成平方级别增长。平方阶常出现于嵌套循环,外层循环和内层循环都为 $O(n)$ ,总体为 $O(n^2)$ 。
|
|
|
|
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
|
|
|
|
```java title="time_complexity.java"
|
|
|
|
|
/* 平方阶 */
|
|
|
|
|
int quadratic(int n) {
|
|
|
|
|
int count = 0;
|
|
|
|
|
// 循环次数与数组长度成平方关系
|
|
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
|
|
for (int j = 0; j < n; j++) {
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
|
|
|
|
```cpp title="time_complexity.cpp"
|
|
|
|
|
/* 平方阶 */
|
|
|
|
|
int quadratic(int n) {
|
|
|
|
|
int count = 0;
|
|
|
|
|
// 循环次数与数组长度成平方关系
|
|
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
|
|
for (int j = 0; j < n; j++) {
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
|
|
|
|
```python title="time_complexity.py"
|
|
|
|
|
""" 平方阶 """
|
|
|
|
|
def quadratic(n):
|
|
|
|
|
count = 0
|
|
|
|
|
# 循环次数与数组长度成平方关系
|
|
|
|
|
for i in range(n):
|
|
|
|
|
for j in range(n):
|
|
|
|
|
count += 1
|
|
|
|
|
return count
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
|
|
|
|
```go title="time_complexity.go"
|
|
|
|
|
/* 平方阶 */
|
|
|
|
|
func quadratic(n int) int {
|
|
|
|
|
count := 0
|
|
|
|
|
// 循环次数与数组长度成平方关系
|
|
|
|
|
for i := 0; i < n; i++ {
|
|
|
|
|
for j := 0; j < n; j++ {
|
|
|
|
|
count++
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "JavaScript"
|
|
|
|
|
|
|
|
|
|
```js title="time_complexity.js"
|
|
|
|
|
/* 平方阶 */
|
|
|
|
|
function quadratic(n) {
|
|
|
|
|
let count = 0;
|
|
|
|
|
// 循环次数与数组长度成平方关系
|
|
|
|
|
for (let i = 0; i < n; i++) {
|
|
|
|
|
for (let j = 0; j < n; j++) {
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TypeScript"
|
|
|
|
|
|
|
|
|
|
```typescript title="time_complexity.ts"
|
|
|
|
|
/* 平方阶 */
|
|
|
|
|
function quadratic(n: number): number {
|
|
|
|
|
let count = 0;
|
|
|
|
|
// 循环次数与数组长度成平方关系
|
|
|
|
|
for (let i = 0; i < n; i++) {
|
|
|
|
|
for (let j = 0; j < n; j++) {
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
|
|
|
|
```c title="time_complexity.c"
|
|
|
|
|
/* 平方阶 */
|
|
|
|
|
int quadratic(int n) {
|
|
|
|
|
int count = 0;
|
|
|
|
|
// 循环次数与数组长度成平方关系
|
|
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
|
|
for (int j = 0; j < n; j++) {
|
|
|
|
|
count ++;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
|
|
|
|
```csharp title="time_complexity.cs"
|
|
|
|
|
/* 平方阶 */
|
|
|
|
|
int quadratic(int n)
|
|
|
|
|
{
|
|
|
|
|
int count = 0;
|
|
|
|
|
// 循环次数与数组长度成平方关系
|
|
|
|
|
for (int i = 0; i < n; i++)
|
|
|
|
|
{
|
|
|
|
|
for (int j = 0; j < n; j++)
|
|
|
|
|
{
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
|
|
|
|
```swift title="time_complexity.swift"
|
|
|
|
|
// 平方阶
|
|
|
|
|
func quadratic(n: Int) -> Int {
|
|
|
|
|
var count = 0
|
|
|
|
|
// 循环次数与数组长度成平方关系
|
|
|
|
|
for _ in 0 ..< n {
|
|
|
|
|
for _ in 0 ..< n {
|
|
|
|
|
count += 1
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
![time_complexity_constant_linear_quadratic](time_complexity.assets/time_complexity_constant_linear_quadratic.png)
|
|
|
|
|
|
|
|
|
|
<p align="center"> Fig. 常数阶、线性阶、平方阶的时间复杂度 </p>
|
|
|
|
|
|
|
|
|
|
以「冒泡排序」为例,外层循环 $n - 1$ 次,内层循环 $n-1, n-2, \cdots, 2, 1$ 次,平均为 $\frac{n}{2}$ 次,因此时间复杂度为 $O(n^2)$ 。
|
|
|
|
|
|
|
|
|
|
$$
|
|
|
|
|
O((n - 1) \frac{n}{2}) = O(n^2)
|
|
|
|
|
$$
|
|
|
|
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
|
|
|
|
```java title="time_complexity.java"
|
|
|
|
|
/* 平方阶(冒泡排序) */
|
|
|
|
|
int bubbleSort(int[] nums) {
|
|
|
|
|
int count = 0; // 计数器
|
|
|
|
|
// 外循环:待排序元素数量为 n-1, n-2, ..., 1
|
|
|
|
|
for (int i = nums.length - 1; i > 0; i--) {
|
|
|
|
|
// 内循环:冒泡操作
|
|
|
|
|
for (int j = 0; j < i; j++) {
|
|
|
|
|
if (nums[j] > nums[j + 1]) {
|
|
|
|
|
// 交换 nums[j] 与 nums[j + 1]
|
|
|
|
|
int tmp = nums[j];
|
|
|
|
|
nums[j] = nums[j + 1];
|
|
|
|
|
nums[j + 1] = tmp;
|
|
|
|
|
count += 3; // 元素交换包含 3 个单元操作
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
|
|
|
|
```cpp title="time_complexity.cpp"
|
|
|
|
|
/* 平方阶(冒泡排序) */
|
|
|
|
|
int bubbleSort(vector<int>& nums) {
|
|
|
|
|
int count = 0; // 计数器
|
|
|
|
|
// 外循环:待排序元素数量为 n-1, n-2, ..., 1
|
|
|
|
|
for (int i = nums.size() - 1; i > 0; i--) {
|
|
|
|
|
// 内循环:冒泡操作
|
|
|
|
|
for (int j = 0; j < i; j++) {
|
|
|
|
|
if (nums[j] > nums[j + 1]) {
|
|
|
|
|
// 交换 nums[j] 与 nums[j + 1]
|
|
|
|
|
int tmp = nums[j];
|
|
|
|
|
nums[j] = nums[j + 1];
|
|
|
|
|
nums[j + 1] = tmp;
|
|
|
|
|
count += 3; // 元素交换包含 3 个单元操作
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
|
|
|
|
```python title="time_complexity.py"
|
|
|
|
|
""" 平方阶(冒泡排序)"""
|
|
|
|
|
def bubble_sort(nums):
|
|
|
|
|
count = 0 # 计数器
|
|
|
|
|
# 外循环:待排序元素数量为 n-1, n-2, ..., 1
|
|
|
|
|
for i in range(len(nums) - 1, 0, -1):
|
|
|
|
|
# 内循环:冒泡操作
|
|
|
|
|
for j in range(i):
|
|
|
|
|
if nums[j] > nums[j + 1]:
|
|
|
|
|
# 交换 nums[j] 与 nums[j + 1]
|
|
|
|
|
tmp = nums[j]
|
|
|
|
|
nums[j] = nums[j + 1]
|
|
|
|
|
nums[j + 1] = tmp
|
|
|
|
|
count += 3 # 元素交换包含 3 个单元操作
|
|
|
|
|
return count
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
|
|
|
|
```go title="time_complexity.go"
|
|
|
|
|
/* 平方阶(冒泡排序) */
|
|
|
|
|
func bubbleSort(nums []int) int {
|
|
|
|
|
count := 0 // 计数器
|
|
|
|
|
// 外循环:待排序元素数量为 n-1, n-2, ..., 1
|
|
|
|
|
for i := len(nums) - 1; i > 0; i-- {
|
|
|
|
|
// 内循环:冒泡操作
|
|
|
|
|
for j := 0; j < i; j++ {
|
|
|
|
|
if nums[j] > nums[j+1] {
|
|
|
|
|
// 交换 nums[j] 与 nums[j + 1]
|
|
|
|
|
tmp := nums[j]
|
|
|
|
|
nums[j] = nums[j+1]
|
|
|
|
|
nums[j+1] = tmp
|
|
|
|
|
count += 3 // 元素交换包含 3 个单元操作
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "JavaScript"
|
|
|
|
|
|
|
|
|
|
```js title="time_complexity.js"
|
|
|
|
|
/* 平方阶(冒泡排序) */
|
|
|
|
|
function bubbleSort(nums) {
|
|
|
|
|
let count = 0; // 计数器
|
|
|
|
|
// 外循环:待排序元素数量为 n-1, n-2, ..., 1
|
|
|
|
|
for (let i = nums.length - 1; i > 0; i--) {
|
|
|
|
|
// 内循环:冒泡操作
|
|
|
|
|
for (let j = 0; j < i; j++) {
|
|
|
|
|
if (nums[j] > nums[j + 1]) {
|
|
|
|
|
// 交换 nums[j] 与 nums[j + 1]
|
|
|
|
|
let tmp = nums[j];
|
|
|
|
|
nums[j] = nums[j + 1];
|
|
|
|
|
nums[j + 1] = tmp;
|
|
|
|
|
count += 3; // 元素交换包含 3 个单元操作
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TypeScript"
|
|
|
|
|
|
|
|
|
|
```typescript title="time_complexity.ts"
|
|
|
|
|
/* 平方阶(冒泡排序) */
|
|
|
|
|
function bubbleSort(nums: number[]): number {
|
|
|
|
|
let count = 0; // 计数器
|
|
|
|
|
// 外循环:待排序元素数量为 n-1, n-2, ..., 1
|
|
|
|
|
for (let i = nums.length - 1; i > 0; i--) {
|
|
|
|
|
// 内循环:冒泡操作
|
|
|
|
|
for (let j = 0; j < i; j++) {
|
|
|
|
|
if (nums[j] > nums[j + 1]) {
|
|
|
|
|
// 交换 nums[j] 与 nums[j + 1]
|
|
|
|
|
let tmp = nums[j];
|
|
|
|
|
nums[j] = nums[j + 1];
|
|
|
|
|
nums[j + 1] = tmp;
|
|
|
|
|
count += 3; // 元素交换包含 3 个单元操作
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
|
|
|
|
```c title="time_complexity.c"
|
|
|
|
|
/* 平方阶(冒泡排序) */
|
|
|
|
|
int bubbleSort(int *nums, int n) {
|
|
|
|
|
int count = 0; // 计数器
|
|
|
|
|
// 外循环:待排序元素数量为 n-1, n-2, ..., 1
|
|
|
|
|
for (int i = n - 1; i > 0; i--) {
|
|
|
|
|
// 内循环:冒泡操作
|
|
|
|
|
for (int j = 0; j < i; j++) {
|
|
|
|
|
// 交换 nums[j] 与 nums[j + 1]
|
|
|
|
|
int tmp = nums[j];
|
|
|
|
|
nums[j] = nums[j + 1];
|
|
|
|
|
nums[j + 1] = tmp;
|
|
|
|
|
count += 3; // 元素交换包含 3 个单元操作
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
|
|
|
|
```csharp title="time_complexity.cs"
|
|
|
|
|
/* 平方阶(冒泡排序) */
|
|
|
|
|
int bubbleSort(int[] nums)
|
|
|
|
|
{
|
|
|
|
|
int count = 0; // 计数器
|
|
|
|
|
// 外循环:待排序元素数量为 n-1, n-2, ..., 1
|
|
|
|
|
for (int i = nums.Length - 1; i > 0; i--)
|
|
|
|
|
{
|
|
|
|
|
// 内循环:冒泡操作
|
|
|
|
|
for (int j = 0; j < i; j++)
|
|
|
|
|
{
|
|
|
|
|
if (nums[j] > nums[j + 1])
|
|
|
|
|
{
|
|
|
|
|
// 交换 nums[j] 与 nums[j + 1]
|
|
|
|
|
int tmp = nums[j];
|
|
|
|
|
nums[j] = nums[j + 1];
|
|
|
|
|
nums[j + 1] = tmp;
|
|
|
|
|
count += 3; // 元素交换包含 3 个单元操作
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
|
|
|
|
```swift title="time_complexity.swift"
|
|
|
|
|
// 平方阶(冒泡排序)
|
|
|
|
|
func bubbleSort(nums: inout [Int]) -> Int {
|
|
|
|
|
var count = 0 // 计数器
|
|
|
|
|
// 外循环:待排序元素数量为 n-1, n-2, ..., 1
|
|
|
|
|
for i in sequence(first: nums.count - 1, next: { $0 > 0 ? $0 - 1 : nil }) {
|
|
|
|
|
// 内循环:冒泡操作
|
|
|
|
|
for j in 0 ..< i {
|
|
|
|
|
if nums[j] > nums[j + 1] {
|
|
|
|
|
// 交换 nums[j] 与 nums[j + 1]
|
|
|
|
|
let tmp = nums[j]
|
|
|
|
|
nums[j] = nums[j + 1]
|
|
|
|
|
nums[j + 1] = tmp
|
|
|
|
|
count += 3 // 元素交换包含 3 个单元操作
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
### 指数阶 $O(2^n)$
|
|
|
|
|
|
|
|
|
|
!!! note
|
|
|
|
|
|
|
|
|
|
生物学科中的“细胞分裂”即是指数阶增长:初始状态为 $1$ 个细胞,分裂一轮后为 $2$ 个,分裂两轮后为 $4$ 个,……,分裂 $n$ 轮后有 $2^n$ 个细胞。
|
|
|
|
|
|
|
|
|
|
指数阶增长得非常快,在实际应用中一般是不能被接受的。若一个问题使用「暴力枚举」求解的时间复杂度是 $O(2^n)$ ,那么一般都需要使用「动态规划」或「贪心算法」等算法来求解。
|
|
|
|
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
|
|
|
|
```java title="time_complexity.java"
|
|
|
|
|
/* 指数阶(循环实现) */
|
|
|
|
|
int exponential(int n) {
|
|
|
|
|
int count = 0, base = 1;
|
|
|
|
|
// cell 每轮一分为二,形成数列 1, 2, 4, 8, ..., 2^(n-1)
|
|
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
|
|
for (int j = 0; j < base; j++) {
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
base *= 2;
|
|
|
|
|
}
|
|
|
|
|
// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
|
|
|
|
```cpp title="time_complexity.cpp"
|
|
|
|
|
/* 指数阶(循环实现) */
|
|
|
|
|
int exponential(int n) {
|
|
|
|
|
int count = 0, base = 1;
|
|
|
|
|
// cell 每轮一分为二,形成数列 1, 2, 4, 8, ..., 2^(n-1)
|
|
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
|
|
for (int j = 0; j < base; j++) {
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
base *= 2;
|
|
|
|
|
}
|
|
|
|
|
// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
|
|
|
|
```python title="time_complexity.py"
|
|
|
|
|
""" 指数阶(循环实现)"""
|
|
|
|
|
def exponential(n):
|
|
|
|
|
count, base = 0, 1
|
|
|
|
|
# cell 每轮一分为二,形成数列 1, 2, 4, 8, ..., 2^(n-1)
|
|
|
|
|
for _ in range(n):
|
|
|
|
|
for _ in range(base):
|
|
|
|
|
count += 1
|
|
|
|
|
base *= 2
|
|
|
|
|
# count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
|
|
|
|
|
return count
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
|
|
|
|
```go title="time_complexity.go"
|
|
|
|
|
/* 指数阶(循环实现)*/
|
|
|
|
|
func exponential(n int) int {
|
|
|
|
|
count, base := 0, 1
|
|
|
|
|
// cell 每轮一分为二,形成数列 1, 2, 4, 8, ..., 2^(n-1)
|
|
|
|
|
for i := 0; i < n; i++ {
|
|
|
|
|
for j := 0; j < base; j++ {
|
|
|
|
|
count++
|
|
|
|
|
}
|
|
|
|
|
base *= 2
|
|
|
|
|
}
|
|
|
|
|
// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "JavaScript"
|
|
|
|
|
|
|
|
|
|
```js title="time_complexity.js"
|
|
|
|
|
/* 指数阶(循环实现) */
|
|
|
|
|
function exponential(n) {
|
|
|
|
|
let count = 0,
|
|
|
|
|
base = 1;
|
|
|
|
|
// cell 每轮一分为二,形成数列 1, 2, 4, 8, ..., 2^(n-1)
|
|
|
|
|
for (let i = 0; i < n; i++) {
|
|
|
|
|
for (let j = 0; j < base; j++) {
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
base *= 2;
|
|
|
|
|
}
|
|
|
|
|
// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TypeScript"
|
|
|
|
|
|
|
|
|
|
```typescript title="time_complexity.ts"
|
|
|
|
|
/* 指数阶(循环实现) */
|
|
|
|
|
function exponential(n: number): number {
|
|
|
|
|
let count = 0,
|
|
|
|
|
base = 1;
|
|
|
|
|
// cell 每轮一分为二,形成数列 1, 2, 4, 8, ..., 2^(n-1)
|
|
|
|
|
for (let i = 0; i < n; i++) {
|
|
|
|
|
for (let j = 0; j < base; j++) {
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
base *= 2;
|
|
|
|
|
}
|
|
|
|
|
// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
|
|
|
|
```c title="time_complexity.c"
|
|
|
|
|
/* 指数阶(循环实现) */
|
|
|
|
|
int exponential(int n) {
|
|
|
|
|
int count = 0;
|
|
|
|
|
int bas = 1;
|
|
|
|
|
// cell 每轮一分为二,形成数列 1, 2, 4, 8, ..., 2^(n-1)
|
|
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
|
|
for (int j = 0; j < bas; j++) {
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
bas *= 2;
|
|
|
|
|
}
|
|
|
|
|
// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
|
|
|
|
```csharp title="time_complexity.cs"
|
|
|
|
|
/* 指数阶(循环实现) */
|
|
|
|
|
int exponential(int n)
|
|
|
|
|
{
|
|
|
|
|
int count = 0, bas = 1;
|
|
|
|
|
// cell 每轮一分为二,形成数列 1, 2, 4, 8, ..., 2^(n-1)
|
|
|
|
|
for (int i = 0; i < n; i++)
|
|
|
|
|
{
|
|
|
|
|
for (int j = 0; j < bas; j++)
|
|
|
|
|
{
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
bas *= 2;
|
|
|
|
|
}
|
|
|
|
|
// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
|
|
|
|
```swift title="time_complexity.swift"
|
|
|
|
|
// 指数阶(循环实现)
|
|
|
|
|
func exponential(n: Int) -> Int {
|
|
|
|
|
var count = 0
|
|
|
|
|
var base = 1
|
|
|
|
|
// cell 每轮一分为二,形成数列 1, 2, 4, 8, ..., 2^(n-1)
|
|
|
|
|
for _ in 0 ..< n {
|
|
|
|
|
for _ in 0 ..< base {
|
|
|
|
|
count += 1
|
|
|
|
|
}
|
|
|
|
|
base *= 2
|
|
|
|
|
}
|
|
|
|
|
// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
![time_complexity_exponential](time_complexity.assets/time_complexity_exponential.png)
|
|
|
|
|
|
|
|
|
|
<p align="center"> Fig. 指数阶的时间复杂度 </p>
|
|
|
|
|
|
|
|
|
|
在实际算法中,指数阶常出现于递归函数。例如以下代码,不断地一分为二,分裂 $n$ 次后停止。
|
|
|
|
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
|
|
|
|
```java title="time_complexity.java"
|
|
|
|
|
/* 指数阶(递归实现) */
|
|
|
|
|
int expRecur(int n) {
|
|
|
|
|
if (n == 1) return 1;
|
|
|
|
|
return expRecur(n - 1) + expRecur(n - 1) + 1;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
|
|
|
|
```cpp title="time_complexity.cpp"
|
|
|
|
|
/* 指数阶(递归实现) */
|
|
|
|
|
int expRecur(int n) {
|
|
|
|
|
if (n == 1) return 1;
|
|
|
|
|
return expRecur(n - 1) + expRecur(n - 1) + 1;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
|
|
|
|
```python title="time_complexity.py"
|
|
|
|
|
""" 指数阶(递归实现)"""
|
|
|
|
|
def exp_recur(n):
|
|
|
|
|
if n == 1: return 1
|
|
|
|
|
return exp_recur(n - 1) + exp_recur(n - 1) + 1
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
|
|
|
|
```go title="time_complexity.go"
|
|
|
|
|
/* 指数阶(递归实现)*/
|
|
|
|
|
func expRecur(n int) int {
|
|
|
|
|
if n == 1 {
|
|
|
|
|
return 1
|
|
|
|
|
}
|
|
|
|
|
return expRecur(n-1) + expRecur(n-1) + 1
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "JavaScript"
|
|
|
|
|
|
|
|
|
|
```js title="time_complexity.js"
|
|
|
|
|
/* 指数阶(递归实现) */
|
|
|
|
|
function expRecur(n) {
|
|
|
|
|
if (n == 1) return 1;
|
|
|
|
|
return expRecur(n - 1) + expRecur(n - 1) + 1;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TypeScript"
|
|
|
|
|
|
|
|
|
|
```typescript title="time_complexity.ts"
|
|
|
|
|
/* 指数阶(递归实现) */
|
|
|
|
|
function expRecur(n: number): number {
|
|
|
|
|
if (n == 1) return 1;
|
|
|
|
|
return expRecur(n - 1) + expRecur(n - 1) + 1;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
|
|
|
|
```c title="time_complexity.c"
|
|
|
|
|
/* 指数阶(递归实现) */
|
|
|
|
|
int expRecur(int n) {
|
|
|
|
|
if (n == 1) return 1;
|
|
|
|
|
return expRecur(n - 1) + expRecur(n - 1) + 1;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
|
|
|
|
```csharp title="time_complexity.cs"
|
|
|
|
|
/* 指数阶(递归实现) */
|
|
|
|
|
int expRecur(int n)
|
|
|
|
|
{
|
|
|
|
|
if (n == 1) return 1;
|
|
|
|
|
return expRecur(n - 1) + expRecur(n - 1) + 1;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
|
|
|
|
```swift title="time_complexity.swift"
|
|
|
|
|
// 指数阶(递归实现)
|
|
|
|
|
func expRecur(n: Int) -> Int {
|
|
|
|
|
if n == 1 {
|
|
|
|
|
return 1
|
|
|
|
|
}
|
|
|
|
|
return expRecur(n: n - 1) + expRecur(n: n - 1) + 1
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
### 对数阶 $O(\log n)$
|
|
|
|
|
|
|
|
|
|
对数阶与指数阶正好相反,后者反映“每轮增加到两倍的情况”,而前者反映“每轮缩减到一半的情况”。对数阶仅次于常数阶,时间增长得很慢,是理想的时间复杂度。
|
|
|
|
|
|
|
|
|
|
对数阶常出现于「二分查找」和「分治算法」中,体现“一分为多”、“化繁为简”的算法思想。
|
|
|
|
|
|
|
|
|
|
设输入数据大小为 $n$ ,由于每轮缩减到一半,因此循环次数是 $\log_2 n$ ,即 $2^n$ 的反函数。
|
|
|
|
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
|
|
|
|
```java title="time_complexity.java"
|
|
|
|
|
/* 对数阶(循环实现) */
|
|
|
|
|
int logarithmic(float n) {
|
|
|
|
|
int count = 0;
|
|
|
|
|
while (n > 1) {
|
|
|
|
|
n = n / 2;
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
|
|
|
|
```cpp title="time_complexity.cpp"
|
|
|
|
|
/* 对数阶(循环实现) */
|
|
|
|
|
int logarithmic(float n) {
|
|
|
|
|
int count = 0;
|
|
|
|
|
while (n > 1) {
|
|
|
|
|
n = n / 2;
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
|
|
|
|
```python title="time_complexity.py"
|
|
|
|
|
""" 对数阶(循环实现)"""
|
|
|
|
|
def logarithmic(n):
|
|
|
|
|
count = 0
|
|
|
|
|
while n > 1:
|
|
|
|
|
n = n / 2
|
|
|
|
|
count += 1
|
|
|
|
|
return count
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
|
|
|
|
```go title="time_complexity.go"
|
|
|
|
|
/* 对数阶(循环实现)*/
|
|
|
|
|
func logarithmic(n float64) int {
|
|
|
|
|
count := 0
|
|
|
|
|
for n > 1 {
|
|
|
|
|
n = n / 2
|
|
|
|
|
count++
|
|
|
|
|
}
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "JavaScript"
|
|
|
|
|
|
|
|
|
|
```js title="time_complexity.js"
|
|
|
|
|
/* 对数阶(循环实现) */
|
|
|
|
|
function logarithmic(n) {
|
|
|
|
|
let count = 0;
|
|
|
|
|
while (n > 1) {
|
|
|
|
|
n = n / 2;
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TypeScript"
|
|
|
|
|
|
|
|
|
|
```typescript title="time_complexity.ts"
|
|
|
|
|
/* 对数阶(循环实现) */
|
|
|
|
|
function logarithmic(n: number): number {
|
|
|
|
|
let count = 0;
|
|
|
|
|
while (n > 1) {
|
|
|
|
|
n = n / 2;
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
|
|
|
|
```c title="time_complexity.c"
|
|
|
|
|
/* 对数阶(循环实现) */
|
|
|
|
|
int logarithmic(float n) {
|
|
|
|
|
int count = 0;
|
|
|
|
|
while (n > 1) {
|
|
|
|
|
n = n / 2;
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
|
|
|
|
```csharp title="time_complexity.cs"
|
|
|
|
|
/* 对数阶(循环实现) */
|
|
|
|
|
int logarithmic(float n)
|
|
|
|
|
{
|
|
|
|
|
int count = 0;
|
|
|
|
|
while (n > 1)
|
|
|
|
|
{
|
|
|
|
|
n = n / 2;
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
|
|
|
|
```swift title="time_complexity.swift"
|
|
|
|
|
// 对数阶(循环实现)
|
|
|
|
|
func logarithmic(n: Int) -> Int {
|
|
|
|
|
var count = 0
|
|
|
|
|
var n = n
|
|
|
|
|
while n > 1 {
|
|
|
|
|
n = n / 2
|
|
|
|
|
count += 1
|
|
|
|
|
}
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
![time_complexity_logarithmic](time_complexity.assets/time_complexity_logarithmic.png)
|
|
|
|
|
|
|
|
|
|
<p align="center"> Fig. 对数阶的时间复杂度 </p>
|
|
|
|
|
|
|
|
|
|
与指数阶类似,对数阶也常出现于递归函数。以下代码形成了一个高度为 $\log_2 n$ 的递归树。
|
|
|
|
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
|
|
|
|
```java title="time_complexity.java"
|
|
|
|
|
/* 对数阶(递归实现) */
|
|
|
|
|
int logRecur(float n) {
|
|
|
|
|
if (n <= 1) return 0;
|
|
|
|
|
return logRecur(n / 2) + 1;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
|
|
|
|
```cpp title="time_complexity.cpp"
|
|
|
|
|
/* 对数阶(递归实现) */
|
|
|
|
|
int logRecur(float n) {
|
|
|
|
|
if (n <= 1) return 0;
|
|
|
|
|
return logRecur(n / 2) + 1;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
|
|
|
|
```python title="time_complexity.py"
|
|
|
|
|
""" 对数阶(递归实现)"""
|
|
|
|
|
def log_recur(n):
|
|
|
|
|
if n <= 1: return 0
|
|
|
|
|
return log_recur(n / 2) + 1
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
|
|
|
|
```go title="time_complexity.go"
|
|
|
|
|
/* 对数阶(递归实现)*/
|
|
|
|
|
func logRecur(n float64) int {
|
|
|
|
|
if n <= 1 {
|
|
|
|
|
return 0
|
|
|
|
|
}
|
|
|
|
|
return logRecur(n/2) + 1
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "JavaScript"
|
|
|
|
|
|
|
|
|
|
```js title="time_complexity.js"
|
|
|
|
|
/* 对数阶(递归实现) */
|
|
|
|
|
function logRecur(n) {
|
|
|
|
|
if (n <= 1) return 0;
|
|
|
|
|
return logRecur(n / 2) + 1;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TypeScript"
|
|
|
|
|
|
|
|
|
|
```typescript title="time_complexity.ts"
|
|
|
|
|
/* 对数阶(递归实现) */
|
|
|
|
|
function logRecur(n: number): number {
|
|
|
|
|
if (n <= 1) return 0;
|
|
|
|
|
return logRecur(n / 2) + 1;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
|
|
|
|
```c title="time_complexity.c"
|
|
|
|
|
/* 对数阶(递归实现) */
|
|
|
|
|
int logRecur(float n) {
|
|
|
|
|
if (n <= 1) return 0;
|
|
|
|
|
return logRecur(n / 2) + 1;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
|
|
|
|
```csharp title="time_complexity.cs"
|
|
|
|
|
/* 对数阶(递归实现) */
|
|
|
|
|
int logRecur(float n)
|
|
|
|
|
{
|
|
|
|
|
if (n <= 1) return 0;
|
|
|
|
|
return logRecur(n / 2) + 1;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
|
|
|
|
```swift title="time_complexity.swift"
|
|
|
|
|
// 对数阶(递归实现)
|
|
|
|
|
func logRecur(n: Int) -> Int {
|
|
|
|
|
if n <= 1 {
|
|
|
|
|
return 0
|
|
|
|
|
}
|
|
|
|
|
return logRecur(n: n / 2) + 1
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
### 线性对数阶 $O(n \log n)$
|
|
|
|
|
|
|
|
|
|
线性对数阶常出现于嵌套循环中,两层循环的时间复杂度分别为 $O(\log n)$ 和 $O(n)$ 。
|
|
|
|
|
|
|
|
|
|
主流排序算法的时间复杂度都是 $O(n \log n )$ ,例如快速排序、归并排序、堆排序等。
|
|
|
|
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
|
|
|
|
```java title="time_complexity.java"
|
|
|
|
|
/* 线性对数阶 */
|
|
|
|
|
int linearLogRecur(float 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++"
|
|
|
|
|
|
|
|
|
|
```cpp title="time_complexity.cpp"
|
|
|
|
|
/* 线性对数阶 */
|
|
|
|
|
int linearLogRecur(float n) {
|
|
|
|
|
if (n <= 1) return 1;
|
|
|
|
|
int count = linearLogRecur(n / 2) +
|
|
|
|
|
linearLogRecur(n / 2);
|
|
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
|
|
|
|
```python title="time_complexity.py"
|
|
|
|
|
""" 线性对数阶 """
|
|
|
|
|
def linear_log_recur(n):
|
|
|
|
|
if n <= 1: return 1
|
|
|
|
|
count = linear_log_recur(n // 2) + \
|
|
|
|
|
linear_log_recur(n // 2)
|
|
|
|
|
for _ in range(n):
|
|
|
|
|
count += 1
|
|
|
|
|
return count
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
|
|
|
|
```go title="time_complexity.go"
|
|
|
|
|
/* 线性对数阶 */
|
|
|
|
|
func linearLogRecur(n float64) int {
|
|
|
|
|
if n <= 1 {
|
|
|
|
|
return 1
|
|
|
|
|
}
|
|
|
|
|
count := linearLogRecur(n/2) +
|
|
|
|
|
linearLogRecur(n/2)
|
|
|
|
|
for i := 0.0; i < n; i++ {
|
|
|
|
|
count++
|
|
|
|
|
}
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "JavaScript"
|
|
|
|
|
|
|
|
|
|
```js title="time_complexity.js"
|
|
|
|
|
/* 线性对数阶 */
|
|
|
|
|
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;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TypeScript"
|
|
|
|
|
|
|
|
|
|
```typescript title="time_complexity.ts"
|
|
|
|
|
/* 线性对数阶 */
|
|
|
|
|
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;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
|
|
|
|
```c title="time_complexity.c"
|
|
|
|
|
/* 线性对数阶 */
|
|
|
|
|
int linearLogRecur(float 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"
|
|
|
|
|
/* 线性对数阶 */
|
|
|
|
|
int linearLogRecur(float n)
|
|
|
|
|
{
|
|
|
|
|
if (n <= 1) return 1;
|
|
|
|
|
int count = linearLogRecur(n / 2) +
|
|
|
|
|
linearLogRecur(n / 2);
|
|
|
|
|
for (int i = 0; i < n; i++)
|
|
|
|
|
{
|
|
|
|
|
count++;
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
|
|
|
|
```swift title="time_complexity.swift"
|
|
|
|
|
// 线性对数阶
|
|
|
|
|
func linearLogRecur(n: Double) -> Int {
|
|
|
|
|
if n <= 1 {
|
|
|
|
|
return 1
|
|
|
|
|
}
|
|
|
|
|
var count = linearLogRecur(n: n / 2) + linearLogRecur(n: n / 2)
|
|
|
|
|
for _ in 0 ..< Int(n) {
|
|
|
|
|
count += 1
|
|
|
|
|
}
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
![time_complexity_logarithmic_linear](time_complexity.assets/time_complexity_logarithmic_linear.png)
|
|
|
|
|
|
|
|
|
|
<p align="center"> Fig. 线性对数阶的时间复杂度 </p>
|
|
|
|
|
|
|
|
|
|
### 阶乘阶 $O(n!)$
|
|
|
|
|
|
|
|
|
|
阶乘阶对应数学上的「全排列」。即给定 $n$ 个互不重复的元素,求其所有可能的排列方案,则方案数量为
|
|
|
|
|
|
|
|
|
|
$$
|
|
|
|
|
n! = n \times (n - 1) \times (n - 2) \times \cdots \times 2 \times 1
|
|
|
|
|
$$
|
|
|
|
|
|
|
|
|
|
阶乘常使用递归实现。例如以下代码,第一层分裂出 $n$ 个,第二层分裂出 $n - 1$ 个,…… ,直至到第 $n$ 层时终止分裂。
|
|
|
|
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
|
|
|
|
```java title="time_complexity.java"
|
|
|
|
|
/* 阶乘阶(递归实现) */
|
|
|
|
|
int factorialRecur(int n) {
|
|
|
|
|
if (n == 0) return 1;
|
|
|
|
|
int count = 0;
|
|
|
|
|
// 从 1 个分裂出 n 个
|
|
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
|
|
count += factorialRecur(n - 1);
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
|
|
|
|
```cpp title="time_complexity.cpp"
|
|
|
|
|
/* 阶乘阶(递归实现) */
|
|
|
|
|
int factorialRecur(int n) {
|
|
|
|
|
if (n == 0) return 1;
|
|
|
|
|
int count = 0;
|
|
|
|
|
// 从 1 个分裂出 n 个
|
|
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
|
|
count += factorialRecur(n - 1);
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
|
|
|
|
```python title="time_complexity.py"
|
|
|
|
|
""" 阶乘阶(递归实现)"""
|
|
|
|
|
def factorial_recur(n):
|
|
|
|
|
if n == 0: return 1
|
|
|
|
|
count = 0
|
|
|
|
|
# 从 1 个分裂出 n 个
|
|
|
|
|
for _ in range(n):
|
|
|
|
|
count += factorial_recur(n - 1)
|
|
|
|
|
return count
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
|
|
|
|
```go title="time_complexity.go"
|
|
|
|
|
/* 阶乘阶(递归实现) */
|
|
|
|
|
func factorialRecur(n int) int {
|
|
|
|
|
if n == 0 {
|
|
|
|
|
return 1
|
|
|
|
|
}
|
|
|
|
|
count := 0
|
|
|
|
|
// 从 1 个分裂出 n 个
|
|
|
|
|
for i := 0; i < n; i++ {
|
|
|
|
|
count += factorialRecur(n - 1)
|
|
|
|
|
}
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "JavaScript"
|
|
|
|
|
|
|
|
|
|
```js title="time_complexity.js"
|
|
|
|
|
/* 阶乘阶(递归实现) */
|
|
|
|
|
function factorialRecur(n) {
|
|
|
|
|
if (n == 0) return 1;
|
|
|
|
|
let count = 0;
|
|
|
|
|
// 从 1 个分裂出 n 个
|
|
|
|
|
for (let i = 0; i < n; i++) {
|
|
|
|
|
count += factorialRecur(n - 1);
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TypeScript"
|
|
|
|
|
|
|
|
|
|
```typescript title="time_complexity.ts"
|
|
|
|
|
/* 阶乘阶(递归实现) */
|
|
|
|
|
function factorialRecur(n: number): number {
|
|
|
|
|
if (n == 0) return 1;
|
|
|
|
|
let count = 0;
|
|
|
|
|
// 从 1 个分裂出 n 个
|
|
|
|
|
for (let i = 0; i < n; i++) {
|
|
|
|
|
count += factorialRecur(n - 1);
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
|
|
|
|
```c title="time_complexity.c"
|
|
|
|
|
/* 阶乘阶(递归实现) */
|
|
|
|
|
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;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
|
|
|
|
```csharp title="time_complexity.cs"
|
|
|
|
|
/* 阶乘阶(递归实现) */
|
|
|
|
|
int factorialRecur(int n)
|
|
|
|
|
{
|
|
|
|
|
if (n == 0) return 1;
|
|
|
|
|
int count = 0;
|
|
|
|
|
// 从 1 个分裂出 n 个
|
|
|
|
|
for (int i = 0; i < n; i++)
|
|
|
|
|
{
|
|
|
|
|
count += factorialRecur(n - 1);
|
|
|
|
|
}
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
|
|
|
|
```swift title="time_complexity.swift"
|
|
|
|
|
// 阶乘阶(递归实现)
|
|
|
|
|
func factorialRecur(n: Int) -> Int {
|
|
|
|
|
if n == 0 {
|
|
|
|
|
return 1
|
|
|
|
|
}
|
|
|
|
|
var count = 0
|
|
|
|
|
// 从 1 个分裂出 n 个
|
|
|
|
|
for _ in 0 ..< n {
|
|
|
|
|
count += factorialRecur(n: n - 1)
|
|
|
|
|
}
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
![time_complexity_factorial](time_complexity.assets/time_complexity_factorial.png)
|
|
|
|
|
|
|
|
|
|
<p align="center"> Fig. 阶乘阶的时间复杂度 </p>
|
|
|
|
|
|
|
|
|
|
## 最差、最佳、平均时间复杂度
|
|
|
|
|
|
|
|
|
|
**某些算法的时间复杂度不是恒定的,而是与输入数据的分布有关**。举一个例子,输入一个长度为 $n$ 数组 `nums` ,其中 `nums` 由从 $1$ 至 $n$ 的数字组成,但元素顺序是随机打乱的;算法的任务是返回元素 $1$ 的索引。我们可以得出以下结论:
|
|
|
|
|
|
|
|
|
|
- 当 `nums = [?, ?, ..., 1]`,即当末尾元素是 $1$ 时,则需完整遍历数组,此时达到 **最差时间复杂度 $O(n)$** ;
|
|
|
|
|
- 当 `nums = [1, ?, ?, ...]` ,即当首个数字为 $1$ 时,无论数组多长都不需要继续遍历,此时达到 **最佳时间复杂度 $\Omega(1)$** ;
|
|
|
|
|
|
|
|
|
|
「函数渐近上界」使用大 $O$ 记号表示,代表「最差时间复杂度」。与之对应,「函数渐近下界」用 $\Omega$ 记号(Omega Notation)来表示,代表「最佳时间复杂度」。
|
|
|
|
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
|
|
|
|
```java title="worst_best_time_complexity.java"
|
|
|
|
|
public class worst_best_time_complexity {
|
|
|
|
|
/* 生成一个数组,元素为 { 1, 2, ..., n },顺序被打乱 */
|
|
|
|
|
int[] randomNumbers(int n) {
|
|
|
|
|
Integer[] nums = new Integer[n];
|
|
|
|
|
// 生成数组 nums = { 1, 2, 3, ..., n }
|
|
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
|
|
nums[i] = i + 1;
|
|
|
|
|
}
|
|
|
|
|
// 随机打乱数组元素
|
|
|
|
|
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;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 查找数组 nums 中数字 1 所在索引 */
|
|
|
|
|
int findOne(int[] nums) {
|
|
|
|
|
for (int i = 0; i < nums.length; i++) {
|
|
|
|
|
if (nums[i] == 1)
|
|
|
|
|
return i;
|
|
|
|
|
}
|
|
|
|
|
return -1;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* Driver Code */
|
|
|
|
|
public void main(String[] args) {
|
|
|
|
|
for (int i = 0; i < 10; i++) {
|
|
|
|
|
int n = 100;
|
|
|
|
|
int[] nums = randomNumbers(n);
|
|
|
|
|
int index = findOne(nums);
|
|
|
|
|
System.out.println("打乱后的数组为 " + Arrays.toString(nums));
|
|
|
|
|
System.out.println("数字 1 的索引为 " + index);
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
|
|
|
|
```cpp title="worst_best_time_complexity.cpp"
|
|
|
|
|
/* 生成一个数组,元素为 { 1, 2, ..., n },顺序被打乱 */
|
|
|
|
|
vector<int> randomNumbers(int n) {
|
|
|
|
|
vector<int> nums(n);
|
|
|
|
|
// 生成数组 nums = { 1, 2, 3, ..., n }
|
|
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
|
|
nums[i] = i + 1;
|
|
|
|
|
}
|
|
|
|
|
// 使用系统时间生成随机种子
|
|
|
|
|
unsigned seed = chrono::system_clock::now().time_since_epoch().count();
|
|
|
|
|
// 随机打乱数组元素
|
|
|
|
|
shuffle(nums.begin(), nums.end(), default_random_engine(seed));
|
|
|
|
|
return nums;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 查找数组 nums 中数字 1 所在索引 */
|
|
|
|
|
int findOne(vector<int>& nums) {
|
|
|
|
|
for (int i = 0; i < nums.size(); i++) {
|
|
|
|
|
if (nums[i] == 1)
|
|
|
|
|
return i;
|
|
|
|
|
}
|
|
|
|
|
return -1;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
/* Driver Code */
|
|
|
|
|
int main() {
|
|
|
|
|
for (int i = 0; i < 1000; i++) {
|
|
|
|
|
int n = 100;
|
|
|
|
|
vector<int> nums = randomNumbers(n);
|
|
|
|
|
int index = findOne(nums);
|
|
|
|
|
cout << "\n数组 [ 1, 2, ..., n ] 被打乱后 = ";
|
|
|
|
|
PrintUtil::printVector(nums);
|
|
|
|
|
cout << "数字 1 的索引为 " << index << endl;
|
|
|
|
|
}
|
|
|
|
|
return 0;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
|
|
|
|
```python title="worst_best_time_complexity.py"
|
|
|
|
|
""" 生成一个数组,元素为: 1, 2, ..., n ,顺序被打乱 """
|
|
|
|
|
def random_numbers(n):
|
|
|
|
|
# 生成数组 nums =: 1, 2, 3, ..., n
|
|
|
|
|
nums = [i for i in range(1, n + 1)]
|
|
|
|
|
# 随机打乱数组元素
|
|
|
|
|
random.shuffle(nums)
|
|
|
|
|
return nums
|
|
|
|
|
|
|
|
|
|
""" 查找数组 nums 中数字 1 所在索引 """
|
|
|
|
|
def find_one(nums):
|
|
|
|
|
for i in range(len(nums)):
|
|
|
|
|
if nums[i] == 1:
|
|
|
|
|
return i
|
|
|
|
|
return -1
|
|
|
|
|
|
|
|
|
|
""" Driver Code """
|
|
|
|
|
if __name__ == "__main__":
|
|
|
|
|
for i in range(10):
|
|
|
|
|
n = 100
|
|
|
|
|
nums = random_numbers(n)
|
|
|
|
|
index = find_one(nums)
|
|
|
|
|
print("\n数组 [ 1, 2, ..., n ] 被打乱后 =", nums)
|
|
|
|
|
print("数字 1 的索引为", index)
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
|
|
|
|
```go title="worst_best_time_complexity.go"
|
|
|
|
|
/* 生成一个数组,元素为 { 1, 2, ..., n },顺序被打乱 */
|
|
|
|
|
func randomNumbers(n int) []int {
|
|
|
|
|
nums := make([]int, n)
|
|
|
|
|
// 生成数组 nums = { 1, 2, 3, ..., n }
|
|
|
|
|
for i := 0; i < n; i++ {
|
|
|
|
|
nums[i] = i + 1
|
|
|
|
|
}
|
|
|
|
|
// 随机打乱数组元素
|
|
|
|
|
rand.Shuffle(len(nums), func(i, j int) {
|
|
|
|
|
nums[i], nums[j] = nums[j], nums[i]
|
|
|
|
|
})
|
|
|
|
|
return nums
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 查找数组 nums 中数字 1 所在索引 */
|
|
|
|
|
func findOne(nums []int) int {
|
|
|
|
|
for i := 0; i < len(nums); i++ {
|
|
|
|
|
if nums[i] == 1 {
|
|
|
|
|
return i
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
return -1
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* Driver Code */
|
|
|
|
|
func main() {
|
|
|
|
|
for i := 0; i < 10; i++ {
|
|
|
|
|
n := 100
|
|
|
|
|
nums := randomNumbers(n)
|
|
|
|
|
index := findOne(nums)
|
|
|
|
|
fmt.Println("\n数组 [ 1, 2, ..., n ] 被打乱后 =", nums)
|
|
|
|
|
fmt.Println("数字 1 的索引为", index)
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "JavaScript"
|
|
|
|
|
|
|
|
|
|
```js title="worst_best_time_complexity.js"
|
|
|
|
|
/* 生成一个数组,元素为 { 1, 2, ..., n },顺序被打乱 */
|
|
|
|
|
function randomNumbers(n) {
|
|
|
|
|
let nums = Array(n);
|
|
|
|
|
// 生成数组 nums = { 1, 2, 3, ..., n }
|
|
|
|
|
for (let i = 0; i < n; i++) {
|
|
|
|
|
nums[i] = i + 1;
|
|
|
|
|
}
|
|
|
|
|
// 随机打乱数组元素
|
|
|
|
|
for (let i = 0; i < n; i++) {
|
|
|
|
|
let r = Math.floor(Math.random() * (i + 1));
|
|
|
|
|
let temp = nums[i];
|
|
|
|
|
nums[i] = nums[r];
|
|
|
|
|
nums[r] = temp;
|
|
|
|
|
}
|
|
|
|
|
return nums;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 查找数组 nums 中数字 1 所在索引 */
|
|
|
|
|
function findOne(nums) {
|
|
|
|
|
for (let i = 0; i < nums.length; i++) {
|
|
|
|
|
if (nums[i] === 1) {
|
|
|
|
|
return i;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
return -1;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* Driver Code */
|
|
|
|
|
function main() {
|
|
|
|
|
for (let i = 0; i < 10; i++) {
|
|
|
|
|
let n = 100;
|
|
|
|
|
let nums = randomNumbers(n);
|
|
|
|
|
let index = findOne(nums);
|
|
|
|
|
console.log(
|
|
|
|
|
"\n数组 [ 1, 2, ..., n ] 被打乱后 = [" + nums.join(", ") + "]"
|
|
|
|
|
);
|
|
|
|
|
console.log("数字 1 的索引为 " + index);
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TypeScript"
|
|
|
|
|
|
|
|
|
|
```typescript title="worst_best_time_complexity.ts"
|
|
|
|
|
/* 生成一个数组,元素为 { 1, 2, ..., n },顺序被打乱 */
|
|
|
|
|
function randomNumbers(n: number): number[] {
|
|
|
|
|
let nums = Array(n);
|
|
|
|
|
// 生成数组 nums = { 1, 2, 3, ..., n }
|
|
|
|
|
for (let i = 0; i < n; i++) {
|
|
|
|
|
nums[i] = i + 1;
|
|
|
|
|
}
|
|
|
|
|
// 随机打乱数组元素
|
|
|
|
|
for (let i = 0; i < n; i++) {
|
|
|
|
|
let r = Math.floor(Math.random() * (i + 1));
|
|
|
|
|
let temp = nums[i];
|
|
|
|
|
nums[i] = nums[r];
|
|
|
|
|
nums[r] = temp;
|
|
|
|
|
}
|
|
|
|
|
return nums;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 查找数组 nums 中数字 1 所在索引 */
|
|
|
|
|
function findOne(nums: number[]): number {
|
|
|
|
|
for (let i = 0; i < nums.length; i++) {
|
|
|
|
|
if (nums[i] === 1) {
|
|
|
|
|
return i;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
return -1;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* Driver Code */
|
|
|
|
|
function main(): void {
|
|
|
|
|
for (let i = 0; i < 10; i++) {
|
|
|
|
|
let n = 100;
|
|
|
|
|
let nums = randomNumbers(n);
|
|
|
|
|
let index = findOne(nums);
|
|
|
|
|
console.log(
|
|
|
|
|
"\n数组 [ 1, 2, ..., n ] 被打乱后 = [" + nums.join(", ") + "]"
|
|
|
|
|
);
|
|
|
|
|
console.log("数字 1 的索引为 " + index);
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
|
|
|
|
```c title="worst_best_time_complexity.c"
|
|
|
|
|
/* 生成一个数组,元素为 { 1, 2, ..., n },顺序被打乱 */
|
|
|
|
|
int *randomNumbers(int n) {
|
|
|
|
|
// 分配堆区内存(创建一维可变长数组:数组中元素数量为n,元素类型为int)
|
|
|
|
|
int *nums = (int *)malloc(n * sizeof(int));
|
|
|
|
|
// 生成数组 nums = { 1, 2, 3, ..., n }
|
|
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
|
|
nums[i] = i + 1;
|
|
|
|
|
}
|
|
|
|
|
// 随机打乱数组元素
|
|
|
|
|
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;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 查找数组 nums 中数字 1 所在索引 */
|
|
|
|
|
int findOne(int *nums, int n) {
|
|
|
|
|
for (int i = 0; i < n; i++) {
|
|
|
|
|
if (nums[i] == 1) return i;
|
|
|
|
|
}
|
|
|
|
|
return -1;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* Driver Code */
|
|
|
|
|
int main(int argc, char *argv[]) {
|
|
|
|
|
// 初始化随机数种子
|
|
|
|
|
srand((unsigned int)time(NULL));
|
|
|
|
|
for (int i = 0; i < 10; i++) {
|
|
|
|
|
int n = 100;
|
|
|
|
|
int *nums = randomNumbers(n);
|
|
|
|
|
int index = findOne(nums, n);
|
|
|
|
|
printf("\n数组 [ 1, 2, ..., n ] 被打乱后 = ");
|
|
|
|
|
printArray(nums, n);
|
|
|
|
|
printf("数字 1 的索引为 %d\n", index);
|
|
|
|
|
// 释放堆区内存
|
|
|
|
|
if (nums != NULL) {
|
|
|
|
|
free(nums);
|
|
|
|
|
nums = NULL;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
getchar();
|
|
|
|
|
return 0;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
|
|
|
|
```csharp title="worst_best_time_complexity.cs"
|
|
|
|
|
/* 生成一个数组,元素为 { 1, 2, ..., n },顺序被打乱 */
|
|
|
|
|
int[] randomNumbers(int n)
|
|
|
|
|
{
|
|
|
|
|
int[] nums = new int[n];
|
|
|
|
|
// 生成数组 nums = { 1, 2, 3, ..., n }
|
|
|
|
|
for (int i = 0; i < n; i++)
|
|
|
|
|
{
|
|
|
|
|
nums[i] = i + 1;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
// 随机打乱数组元素
|
|
|
|
|
for (int i = 0; i < nums.Length; i++)
|
|
|
|
|
{
|
|
|
|
|
var index = new Random().Next(i, nums.Length);
|
|
|
|
|
var tmp = nums[i];
|
|
|
|
|
var ran = nums[index];
|
|
|
|
|
nums[i] = ran;
|
|
|
|
|
nums[index] = tmp;
|
|
|
|
|
}
|
|
|
|
|
return nums;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 查找数组 nums 中数字 1 所在索引 */
|
|
|
|
|
int findOne(int[] nums)
|
|
|
|
|
{
|
|
|
|
|
for (int i = 0; i < nums.Length; i++)
|
|
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{
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if (nums[i] == 1)
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return i;
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}
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return -1;
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}
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|
/* Driver Code */
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|
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|
public void main(String[] args)
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|
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|
{
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|
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for (int i = 0; i < 10; i++)
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|
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{
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|
|
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|
int n = 100;
|
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|
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int[] nums = randomNumbers(n);
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|
int index = findOne(nums);
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|
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|
Console.WriteLine("\n数组 [ 1, 2, ..., n ] 被打乱后 = " + string.Join(",", nums));
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Console.WriteLine("数字 1 的索引为 " + index);
|
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|
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|
}
|
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|
}
|
|
|
|
|
```
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|
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|
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|
=== "Swift"
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|
|
|
|
|
|
```swift title="worst_best_time_complexity.swift"
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|
|
|
// 生成一个数组,元素为 { 1, 2, ..., n },顺序被打乱
|
|
|
|
|
func randomNumbers(n: Int) -> [Int] {
|
|
|
|
|
// 生成数组 nums = { 1, 2, 3, ..., n }
|
|
|
|
|
var nums = Array(1 ... n)
|
|
|
|
|
// 随机打乱数组元素
|
|
|
|
|
nums.shuffle()
|
|
|
|
|
return nums
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
// 查找数组 nums 中数字 1 所在索引
|
|
|
|
|
func findOne(nums: [Int]) -> Int {
|
|
|
|
|
for i in nums.indices {
|
|
|
|
|
if nums[i] == 1 {
|
|
|
|
|
return i
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
return -1
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
// Driver Code
|
|
|
|
|
func main() {
|
|
|
|
|
for _ in 0 ..< 10 {
|
|
|
|
|
let n = 100
|
|
|
|
|
let nums = randomNumbers(n: n)
|
|
|
|
|
let index = findOne(nums: nums)
|
|
|
|
|
print("数组 [ 1, 2, ..., n ] 被打乱后 =", nums)
|
|
|
|
|
print("数字 1 的索引为", index)
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
!!! tip
|
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|
|
|
|
|
|
|
我们在实际应用中很少使用「最佳时间复杂度」,因为往往只有很小概率下才能达到,会带来一定的误导性。反之,「最差时间复杂度」最为实用,因为它给出了一个“效率安全值”,让我们可以放心地使用算法。
|
|
|
|
|
|
|
|
|
|
从上述示例可以看出,最差或最佳时间复杂度只出现在“特殊分布的数据”中,这些情况的出现概率往往很小,因此并不能最真实地反映算法运行效率。**相对地,「平均时间复杂度」可以体现算法在随机输入数据下的运行效率,用 $\Theta$ 记号(Theta Notation)来表示**。
|
|
|
|
|
|
|
|
|
|
对于部分算法,我们可以简单地推算出随机数据分布下的平均情况。比如上述示例,由于输入数组是被打乱的,因此元素 $1$ 出现在任意索引的概率都是相等的,那么算法的平均循环次数则是数组长度的一半 $\frac{n}{2}$ ,平均时间复杂度为 $\Theta(\frac{n}{2}) = \Theta(n)$ 。
|
|
|
|
|
|
|
|
|
|
但在实际应用中,尤其是较为复杂的算法,计算平均时间复杂度比较困难,因为很难简便地分析出在数据分布下的整体数学期望。这种情况下,我们一般使用最差时间复杂度来作为算法效率的评判标准。
|
|
|
|
|
|
|
|
|
|
!!! question "为什么很少看到 $\Theta$ 符号?"
|
|
|
|
|
|
|
|
|
|
实际中我们经常使用「大 $O$ 符号」来表示「平均复杂度」,这样严格意义上来说是不规范的。这可能是因为 $O$ 符号实在是太朗朗上口了。</br>如果在本书和其他资料中看到类似 **平均时间复杂度 $O(n)$** 的表述,请你直接理解为 $\Theta(n)$ 即可。
|