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hello-algo/docs/chapter_computational_compl.../time_complexity.md

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build
1 year ago
---
comments: true
---
# 2.3   时间复杂度
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12 months ago
运行时间可以直观且准确地反映算法的效率。如果我们想准确预估一段代码的运行时间，应该如何操作呢？
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1 year ago
1. **确定运行平台**，包括硬件配置、编程语言、系统环境等，这些因素都会影响代码的运行效率。
2. **评估各种计算操作所需的运行时间**，例如加法操作 `+` 需要 1 ns ，乘法操作 `*` 需要 10 ns ，打印操作 `print()` 需要 5 ns 等。
3. **统计代码中所有的计算操作**，并将所有操作的执行时间求和，从而得到运行时间。
例如在以下代码中，输入数据大小为 $n$ ：
=== "Python"
```python title=""
# 在某运行平台下
def algorithm(n: int):
a = 2 # 1 ns
a = a + 1 # 1 ns
a = a * 2 # 10 ns
# 循环 n 次
for _ in range(n): # 1 ns
print(0) # 5 ns
```
=== "C++"
```cpp title=""
// 在某运行平台下
void algorithm(int n) {
int a = 2; // 1 ns
a = a + 1; // 1 ns
a = a * 2; // 10 ns
// 循环 n 次
for (int i = 0; i < n; i++) { // 1 ns ，每轮都要执行 i++
cout << 0 << endl; // 5 ns
}
}
```
=== "Java"
```java title=""
// 在某运行平台下
void algorithm(int n) {
int a = 2; // 1 ns
a = a + 1; // 1 ns
a = a * 2; // 10 ns
// 循环 n 次
for (int i = 0; i < n; i++) { // 1 ns ，每轮都要执行 i++
System.out.println(0); // 5 ns
}
}
```
=== "C#"
```csharp title=""
// 在某运行平台下
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void Algorithm(int n) {
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int a = 2; // 1 ns
a = a + 1; // 1 ns
a = a * 2; // 10 ns
// 循环 n 次
for (int i = 0; i < n; i++) { // 1 ns ，每轮都要执行 i++
Console.WriteLine(0); // 5 ns
}
}
```
=== "Go"
```go title=""
// 在某运行平台下
func algorithm(n int) {
a := 2 // 1 ns
a = a + 1 // 1 ns
a = a * 2 // 10 ns
// 循环 n 次
for i := 0; i < n; i++ { // 1 ns
fmt.Println(a) // 5 ns
}
}
```
=== "Swift"
```swift title=""
// 在某运行平台下
func algorithm(n: Int) {
var a = 2 // 1 ns
a = a + 1 // 1 ns
a = a * 2 // 10 ns
// 循环 n 次
for _ in 0 ..< n { // 1 ns
print(0) // 5 ns
}
}
```
=== "JS"
```javascript title=""
// 在某运行平台下
function algorithm(n) {
var a = 2; // 1 ns
a = a + 1; // 1 ns
a = a * 2; // 10 ns
// 循环 n 次
for(let i = 0; i < n; i++) { // 1 ns ，每轮都要执行 i++
console.log(0); // 5 ns
}
}
```
=== "TS"
```typescript title=""
// 在某运行平台下
function algorithm(n: number): void {
var a: number = 2; // 1 ns
a = a + 1; // 1 ns
a = a * 2; // 10 ns
// 循环 n 次
for(let i = 0; i < n; i++) { // 1 ns ，每轮都要执行 i++
console.log(0); // 5 ns
}
}
```
=== "Dart"
```dart title=""
// 在某运行平台下
void algorithm(int n) {
int a = 2; // 1 ns
a = a + 1; // 1 ns
a = a * 2; // 10 ns
// 循环 n 次
for (int i = 0; i < n; i++) { // 1 ns ，每轮都要执行 i++
print(0); // 5 ns
}
}
```
=== "Rust"
```rust title=""
// 在某运行平台下
fn algorithm(n: i32) {
let mut a = 2; // 1 ns
a = a + 1; // 1 ns
a = a * 2; // 10 ns
// 循环 n 次
for _ in 0..n { // 1 ns ，每轮都要执行 i++
println!("{}", 0); // 5 ns
}
}
```
=== "C"
```c title=""
// 在某运行平台下
void algorithm(int n) {
int a = 2; // 1 ns
a = a + 1; // 1 ns
a = a * 2; // 10 ns
// 循环 n 次
for (int i = 0; i < n; i++) { // 1 ns ，每轮都要执行 i++
printf("%d", 0); // 5 ns
}
}
```
=== "Zig"
```zig title=""
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1 year ago
// 在某运行平台下
fn algorithm(n: usize) void {
var a: i32 = 2; // 1 ns
a += 1; // 1 ns
a *= 2; // 10 ns
// 循环 n 次
for (0..n) |_| { // 1 ns
std.debug.print("{}\n", .{0}); // 5 ns
}
}
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```
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根据以上方法，可以得到算法的运行时间为 $(6n + 12)$ ns ：
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$$
1 + 1 + 10 + (1 + 5) \times n = 6n + 12
$$
但实际上，**统计算法的运行时间既不合理也不现实**。首先，我们不希望将预估时间和运行平台绑定，因为算法需要在各种不同的平台上运行。其次，我们很难获知每种操作的运行时间，这给预估过程带来了极大的难度。
## 2.3.1 &nbsp; 统计时间增长趋势
时间复杂度分析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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12 months ago
“时间增长趋势”这个概念比较抽象，我们通过一个例子来加以理解。假设输入数据大小为 $n$ ，给定三个算法 `A`、`B` 和 `C` ：
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=== "Python"
```python title=""
# 算法 A 的时间复杂度：常数阶
def algorithm_A(n: int):
print(0)
# 算法 B 的时间复杂度：线性阶
def algorithm_B(n: int):
for _ in range(n):
print(0)
# 算法 C 的时间复杂度：常数阶
def algorithm_C(n: int):
for _ in range(1000000):
print(0)
```
=== "C++"
```cpp title=""
// 算法 A 的时间复杂度：常数阶
void algorithm_A(int n) {
cout << 0 << endl;
}
// 算法 B 的时间复杂度：线性阶
void algorithm_B(int n) {
for (int i = 0; i < n; i++) {
cout << 0 << endl;
}
}
// 算法 C 的时间复杂度：常数阶
void algorithm_C(int n) {
for (int i = 0; i < 1000000; i++) {
cout << 0 << endl;
}
}
```
=== "Java"
```java title=""
// 算法 A 的时间复杂度：常数阶
void algorithm_A(int n) {
System.out.println(0);
}
// 算法 B 的时间复杂度：线性阶
void algorithm_B(int n) {
for (int i = 0; i < n; i++) {
System.out.println(0);
}
}
// 算法 C 的时间复杂度：常数阶
void algorithm_C(int n) {
for (int i = 0; i < 1000000; i++) {
System.out.println(0);
}
}
```
=== "C#"
```csharp title=""
// 算法 A 的时间复杂度：常数阶
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void AlgorithmA(int n) {
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Console.WriteLine(0);
}
// 算法 B 的时间复杂度：线性阶
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void AlgorithmB(int n) {
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for (int i = 0; i < n; i++) {
Console.WriteLine(0);
}
}
// 算法 C 的时间复杂度：常数阶
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void AlgorithmC(int n) {
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for (int i = 0; i < 1000000; i++) {
Console.WriteLine(0);
}
}
```
=== "Go"
```go title=""
// 算法 A 的时间复杂度：常数阶
func algorithm_A(n int) {
fmt.Println(0)
}
// 算法 B 的时间复杂度：线性阶
func algorithm_B(n int) {
for i := 0; i < n; i++ {
fmt.Println(0)
}
}
// 算法 C 的时间复杂度：常数阶
func algorithm_C(n int) {
for i := 0; i < 1000000; i++ {
fmt.Println(0)
}
}
```
=== "Swift"
```swift title=""
// 算法 A 的时间复杂度：常数阶
func algorithmA(n: Int) {
print(0)
}
// 算法 B 的时间复杂度：线性阶
func algorithmB(n: Int) {
for _ in 0 ..< n {
print(0)
}
}
// 算法 C 的时间复杂度：常数阶
func algorithmC(n: Int) {
for _ in 0 ..< 1000000 {
print(0)
}
}
```
=== "JS"
```javascript title=""
// 算法 A 的时间复杂度：常数阶
function algorithm_A(n) {
console.log(0);
}
// 算法 B 的时间复杂度：线性阶
function algorithm_B(n) {
for (let i = 0; i < n; i++) {
console.log(0);
}
}
// 算法 C 的时间复杂度：常数阶
function algorithm_C(n) {
for (let i = 0; i < 1000000; i++) {
console.log(0);
}
}
```
=== "TS"
```typescript title=""
// 算法 A 的时间复杂度：常数阶
function algorithm_A(n: number): void {
console.log(0);
}
// 算法 B 的时间复杂度：线性阶
function algorithm_B(n: number): void {
for (let i = 0; i < n; i++) {
console.log(0);
}
}
// 算法 C 的时间复杂度：常数阶
function algorithm_C(n: number): void {
for (let i = 0; i < 1000000; i++) {
console.log(0);
}
}
```
=== "Dart"
```dart title=""
// 算法 A 的时间复杂度：常数阶
void algorithmA(int n) {
print(0);
}
// 算法 B 的时间复杂度：线性阶
void algorithmB(int n) {
for (int i = 0; i < n; i++) {
print(0);
}
}
// 算法 C 的时间复杂度：常数阶
void algorithmC(int n) {
for (int i = 0; i < 1000000; i++) {
print(0);
}
}
```
=== "Rust"
```rust title=""
// 算法 A 的时间复杂度：常数阶
fn algorithm_A(n: i32) {
println!("{}", 0);
}
// 算法 B 的时间复杂度：线性阶
fn algorithm_B(n: i32) {
for _ in 0..n {
println!("{}", 0);
}
}
// 算法 C 的时间复杂度：常数阶
fn algorithm_C(n: i32) {
for _ in 0..1000000 {
println!("{}", 0);
}
}
```
=== "C"
```c title=""
// 算法 A 的时间复杂度：常数阶
void algorithm_A(int n) {
printf("%d", 0);
}
// 算法 B 的时间复杂度：线性阶
void algorithm_B(int n) {
for (int i = 0; i < n; i++) {
printf("%d", 0);
}
}
// 算法 C 的时间复杂度：常数阶
void algorithm_C(int n) {
for (int i = 0; i < 1000000; i++) {
printf("%d", 0);
}
}
```
=== "Zig"
```zig title=""
build
1 year ago
// 算法 A 的时间复杂度：常数阶
fn algorithm_A(n: usize) void {
_ = n;
std.debug.print("{}\n", .{0});
}
// 算法 B 的时间复杂度：线性阶
fn algorithm_B(n: i32) void {
for (0..n) |_| {
std.debug.print("{}\n", .{0});
}
}
// 算法 C 的时间复杂度：常数阶
fn algorithm_C(n: i32) void {
_ = n;
for (0..1000000) |_| {
std.debug.print("{}\n", .{0});
}
}
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1 year ago
```
图 2-7 展示了以上三个算法函数的时间复杂度。
- 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”。
- 算法 `B` 中的打印操作需要循环 $n$ 次，算法运行时间随着 $n$ 增大呈线性增长。此算法的时间复杂度被称为“线性阶”。
- 算法 `C` 中的打印操作需要循环 $1000000$ 次，虽然运行时间很长，但它与输入数据大小 $n$ 无关。因此 `C` 的时间复杂度和 `A` 相同，仍为“常数阶”。
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1 year ago
![算法 A、B 和 C 的时间增长趋势](time_complexity.assets/time_complexity_simple_example.png){ class="animation-figure" }
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<p align="center"> 图 2-7 &nbsp; 算法 A、B 和 C 的时间增长趋势 </p>
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相较于直接统计算法的运行时间，时间复杂度分析有哪些特点呢？
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1 year ago
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- **时间复杂度能够有效评估算法效率**。例如，算法 `B` 的运行时间呈线性增长，在 $n > 1$ 时比算法 `A` 更慢，在 $n > 1000000$ 时比算法 `C` 更慢。事实上，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，这正是时间增长趋势的含义。
- **时间复杂度的推算方法更简便**。显然，运行平台和计算操作类型都与算法运行时间的增长趋势无关。因此在时间复杂度分析中，我们可以简单地将所有计算操作的执行时间视为相同的“单位时间”，从而将“计算操作运行时间统计”简化为“计算操作数量统计”，这样一来估算难度就大大降低了。
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1 year ago
- **时间复杂度也存在一定的局限性**。例如，尽管算法 `A` 和 `C` 的时间复杂度相同，但实际运行时间差别很大。同样，尽管算法 `B` 的时间复杂度比 `C` 高，但在输入数据大小 $n$ 较小时，算法 `B` 明显优于算法 `C` 。在这些情况下，我们很难仅凭时间复杂度判断算法效率的高低。当然，尽管存在上述问题，复杂度分析仍然是评判算法效率最有效且常用的方法。
## 2.3.2 &nbsp; 函数渐近上界
给定一个输入大小为 $n$ 的函数：
=== "Python"
```python title=""
def algorithm(n: int):
a = 1 # +1
a = a + 1 # +1
a = a * 2 # +1
# 循环 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
// 循环 n 次
for (int i = 0; i < n; i++) { // +1（每轮都执行 i ++）
cout << 0 << endl; // +1
}
}
```
=== "Java"
```java title=""
void algorithm(int n) {
int a = 1; // +1
a = a + 1; // +1
a = a * 2; // +1
// 循环 n 次
for (int i = 0; i < n; i++) { // +1（每轮都执行 i ++）
System.out.println(0); // +1
}
}
```
=== "C#"
```csharp title=""
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void Algorithm(int n) {
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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
}
}
```
=== "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
}
}
```
=== "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
}
}
```
=== "JS"
```javascript 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
}
}
```
=== "TS"
```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
}
}
```
=== "Dart"
```dart 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 ++）
print(0); // +1
}
}
```
=== "Rust"
```rust title=""
fn algorithm(n: i32) {
let mut a = 1; // +1
a = a + 1; // +1
a = a * 2; // +1
// 循环 n 次
for _ in 0..n { // +1（每轮都执行 i ++）
println!("{}", 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
}
}
```
=== "Zig"
```zig title=""
build
1 year ago
fn algorithm(n: usize) void {
var a: i32 = 1; // +1
a += 1; // +1
a *= 2; // +1
// 循环 n 次
for (0..n) |_| { // +1（每轮都执行 i ++）
std.debug.print("{}\n", .{0}); // +1
}
}
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1 year ago
```
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设算法的操作数量是一个关于输入数据大小 $n$ 的函数，记为 $T(n)$ ，则以上函数的操作数量为：
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1 year ago
$$
T(n) = 3 + 2n
$$
$T(n)$ 是一次函数，说明其运行时间的增长趋势是线性的，因此它的时间复杂度是线性阶。
我们将线性阶的时间复杂度记为 $O(n)$ ，这个数学符号称为「大 $O$ 记号 big-$O$ notation」，表示函数 $T(n)$ 的「渐近上界 asymptotic upper bound」。
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12 months ago
时间复杂度分析本质上是计算“操作数量 $T(n)$”的渐近上界，它具有明确的数学定义。
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!!! 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 $T(n) = O(f(n))$ 。
如图 2-8 所示，计算渐近上界就是寻找一个函数 $f(n)$ ，使得当 $n$ 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。
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1 year ago
![函数的渐近上界](time_complexity.assets/asymptotic_upper_bound.png){ class="animation-figure" }
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<p align="center"> 图 2-8 &nbsp; 函数的渐近上界 </p>
## 2.3.3 &nbsp; 推算方法
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12 months ago
渐近上界的数学味儿有点重，如果你感觉没有完全理解，也无须担心。我们可以先掌握推算方法，在不断的实践中，就可以逐渐领悟其数学意义。
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1 year ago
根据定义，确定 $f(n)$ 之后，我们便可得到时间复杂度 $O(f(n))$ 。那么如何确定渐近上界 $f(n)$ 呢？总体分为两步：首先统计操作数量，然后判断渐近上界。
### 1. &nbsp; 第一步：统计操作数量
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12 months ago
针对代码，逐行从上到下计算即可。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、常数项都可以忽略**。根据此原则，可以总结出以下计数简化技巧。
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1 year ago
1. **忽略 $T(n)$ 中的常数项**。因为它们都与 $n$ 无关，所以对时间复杂度不产生影响。
2. **省略所有系数**。例如，循环 $2n$ 次、$5n + 1$ 次等，都可以简化记为 $n$ 次，因为 $n$ 前面的系数对时间复杂度没有影响。
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第 `1.` 点和第 `2.` 点的技巧。
build
12 months ago
给定一个函数，我们可以用上述技巧来统计操作数量：
build
1 year ago
=== "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=""
build
1 year ago
void Algorithm(int n) {
build
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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);
}
}
}
```
=== "Zig"
```zig title=""
build
1 year ago
fn algorithm(n: usize) void {
var a: i32 = 1; // +0（技巧 1）
a = a + @as(i32, @intCast(n)); // +0（技巧 1）
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// +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});
}
}
}
build
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```
build
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以下公式展示了使用上述技巧前后的统计结果，两者推算出的时间复杂度都为 $O(n^2)$ 。
build
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$$
\begin{aligned}
T(n) & = 2n(n + 1) + (5n + 1) + 2 & \text{完整统计 (-.-|||)} \newline
& = 2n^2 + 7n + 3 \newline
T(n) & = n^2 + n & \text{偷懒统计 (o.O)}
\end{aligned}
$$
### 2. &nbsp; 第二步：判断渐近上界
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**时间复杂度由 $T(n)$ 中最高阶的项来决定**。这是因为在 $n$ 趋于无穷大时，最高阶的项将发挥主导作用，其他项的影响都可以忽略。
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表 2-2 展示了一些例子，其中一些夸张的值是为了强调“系数无法撼动阶数”这一结论。当 $n$ 趋于无穷大时，这些常数变得无足轻重。
<p align="center"> 表 2-2 &nbsp; 不同操作数量对应的时间复杂度 </p>
<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>
## 2.3.4 &nbsp; 常见类型
设输入数据大小为 $n$ ，常见的时间复杂度类型如图 2-9 所示（按照从低到高的顺序排列）。
$$
\begin{aligned}
O(1) < O(\log n) < O(n) < O(n \log n) < O(n^2) < O(2^n) < O(n!) \newline
\text{常数阶} < \text{对数阶} < \text{线性阶} < \text{线性对数阶} < \text{平方阶} < \text{指数阶} < \text{阶乘阶}
\end{aligned}
$$
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![常见的时间复杂度类型](time_complexity.assets/time_complexity_common_types.png){ class="animation-figure" }
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<p align="center"> 图 2-9 &nbsp; 常见的时间复杂度类型 </p>
### 1. &nbsp; 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。
在以下函数中，尽管操作数量 `size` 可能很大，但由于其与输入数据大小 $n$ 无关，因此时间复杂度仍为 $O(1)$ ：
=== "Python"
```python title="time_complexity.py"
build
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def constant(n: int) -> int:
"""常数阶"""
count = 0
size = 100000
for _ in range(size):
count += 1
return count
build
1 year ago
```
=== "C++"
```cpp title="time_complexity.cpp"
build
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/* 常数阶 */
int constant(int n) {
int count = 0;
int size = 100000;
for (int i = 0; i < size; i++)
count++;
return count;
}
build
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```
=== "Java"
```java title="time_complexity.java"
build
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/* 常数阶 */
int constant(int n) {
int count = 0;
int size = 100000;
for (int i = 0; i < size; i++)
count++;
return count;
}
build
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```
=== "C#"
```csharp title="time_complexity.cs"
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/* 常数阶 */
build
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int Constant(int n) {
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int count = 0;
int size = 100000;
for (int i = 0; i < size; i++)
count++;
return count;
}
build
1 year ago
```
=== "Go"
```go title="time_complexity.go"
build
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/* 常数阶 */
func constant(n int) int {
count := 0
size := 100000
for i := 0; i < size; i++ {
count++
}
return count
}
build
1 year ago
```
=== "Swift"
```swift title="time_complexity.swift"
build
1 year ago
/* 常数阶 */
func constant(n: Int) -> Int {
var count = 0
let size = 100_000
for _ in 0 ..< size {
count += 1
}
return count
}
build
1 year ago
```
=== "JS"
```javascript title="time_complexity.js"
build
1 year ago
/* 常数阶 */
function constant(n) {
let count = 0;
const size = 100000;
for (let i = 0; i < size; i++) count++;
return count;
}
build
1 year ago
```
=== "TS"
```typescript title="time_complexity.ts"
build
1 year ago
/* 常数阶 */
function constant(n: number): number {
let count = 0;
const size = 100000;
for (let i = 0; i < size; i++) count++;
return count;
}
build
1 year ago
```
=== "Dart"
```dart title="time_complexity.dart"
build
1 year ago
/* 常数阶 */
int constant(int n) {
int count = 0;
int size = 100000;
for (var i = 0; i < size; i++) {
count++;
}
return count;
}
build
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```
=== "Rust"
```rust title="time_complexity.rs"
build
1 year ago
/* 常数阶 */
fn constant(n: i32) -> i32 {
_ = n;
let mut count = 0;
let size = 100_000;
for _ in 0..size {
count += 1;
}
count
}
build
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```
=== "C"
```c title="time_complexity.c"
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/* 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;
}
build
1 year ago
```
=== "Zig"
```zig title="time_complexity.zig"
build
1 year ago
// 常数阶
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;
}
build
1 year ago
```
build
11 months ago
??? pythontutor "可视化运行"
build
10 months ago
<div style="height: 560px; 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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全屏观看 ></a>
build
11 months ago
build
1 year ago
### 2. &nbsp; 线性阶 $O(n)$
线性阶的操作数量相对于输入数据大小 $n$ 以线性级别增长。线性阶通常出现在单层循环中：
=== "Python"
```python title="time_complexity.py"
build
1 year ago
def linear(n: int) -> int:
"""线性阶"""
count = 0
for _ in range(n):
count += 1
return count
build
1 year ago
```
=== "C++"
```cpp title="time_complexity.cpp"
build
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/* 线性阶 */
int linear(int n) {
int count = 0;
for (int i = 0; i < n; i++)
count++;
return count;
}
build
1 year ago
```
=== "Java"
```java title="time_complexity.java"
build
1 year ago
/* 线性阶 */
int linear(int n) {
int count = 0;
for (int i = 0; i < n; i++)
count++;
return count;
}
build
1 year ago
```
=== "C#"
```csharp title="time_complexity.cs"
build
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/* 线性阶 */
build
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int Linear(int n) {
build
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int count = 0;
for (int i = 0; i < n; i++)
count++;
return count;
}
build
1 year ago
```
=== "Go"
```go title="time_complexity.go"
build
1 year ago
/* 线性阶 */
func linear(n int) int {
count := 0
for i := 0; i < n; i++ {
count++
}
return count
}
build
1 year ago
```
=== "Swift"
```swift title="time_complexity.swift"
build
1 year ago
/* 线性阶 */
func linear(n: Int) -> Int {
var count = 0
for _ in 0 ..< n {
count += 1
}
return count
}
build
1 year ago
```
=== "JS"
```javascript title="time_complexity.js"
build
1 year ago
/* 线性阶 */
function linear(n) {
let count = 0;
for (let i = 0; i < n; i++) count++;
return count;
}
build
1 year ago
```
=== "TS"
```typescript title="time_complexity.ts"
build
1 year ago
/* 线性阶 */
function linear(n: number): number {
let count = 0;
for (let i = 0; i < n; i++) count++;
return count;
}
build
1 year ago
```
=== "Dart"
```dart title="time_complexity.dart"
build
1 year ago
/* 线性阶 */
int linear(int n) {
int count = 0;
for (var i = 0; i < n; i++) {
count++;
}
return count;
}
build
1 year ago
```
=== "Rust"
```rust title="time_complexity.rs"
build
1 year ago
/* 线性阶 */
fn linear(n: i32) -> i32 {
let mut count = 0;
for _ in 0..n {
count += 1;
}
count
}
build
1 year ago
```
=== "C"
```c title="time_complexity.c"
build
1 year ago
/* 线性阶 */
int linear(int n) {
int count = 0;
for (int i = 0; i < n; i++) {
count++;
}
return count;
}
build
1 year ago
```
=== "Zig"
```zig title="time_complexity.zig"
build
1 year ago
// 线性阶
fn linear(n: i32) i32 {
var count: i32 = 0;
var i: i32 = 0;
while (i < n) : (i += 1) {
count += 1;
}
return count;
}
build
1 year ago
```
build
11 months ago
??? pythontutor "可视化运行"
build
10 months ago
<div style="height: 560px; 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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全屏观看 ></a>
build
11 months ago
build
1 year ago
遍历数组和遍历链表等操作的时间复杂度均为 $O(n)$ ，其中 $n$ 为数组或链表的长度：
=== "Python"
```python title="time_complexity.py"
build
1 year ago
def array_traversal(nums: list[int]) -> int:
"""线性阶（遍历数组）"""
count = 0
# 循环次数与数组长度成正比
for num in nums:
count += 1
return count
build
1 year ago
```
=== "C++"
```cpp title="time_complexity.cpp"
build
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/* 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;
}
build
1 year ago
```
=== "Java"
```java title="time_complexity.java"
build
1 year ago
/* 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;
}
build
1 year ago
```
=== "C#"
```csharp title="time_complexity.cs"
build
1 year ago
/* 线性阶（遍历数组） */
build
1 year ago
int ArrayTraversal(int[] nums) {
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int count = 0;
// 循环次数与数组长度成正比
foreach (int num in nums) {
count++;
}
return count;
}
build
1 year ago
```
=== "Go"
```go title="time_complexity.go"
build
1 year ago
/* 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
}
build
1 year ago
```
=== "Swift"
```swift title="time_complexity.swift"
build
1 year ago
/* 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
}
build
1 year ago
```
=== "JS"
```javascript title="time_complexity.js"
build
1 year ago
/* 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;
}
build
1 year ago
```
=== "TS"
```typescript title="time_complexity.ts"
build
1 year ago
/* 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;
}
build
1 year ago
```
=== "Dart"
```dart title="time_complexity.dart"
build
1 year ago
/* 线性阶（遍历数组） */
int arrayTraversal(List<int> nums) {
int count = 0;
// 循环次数与数组长度成正比
build
1 year ago
for (var _num in nums) {
build
1 year ago
count++;
}
return count;
}
build
1 year ago
```
=== "Rust"
```rust title="time_complexity.rs"
build
1 year ago
/* 线性阶（遍历数组） */
fn array_traversal(nums: &[i32]) -> i32 {
let mut count = 0;
// 循环次数与数组长度成正比
for _ in nums {
count += 1;
}
count
}
build
1 year ago
```
=== "C"
```c title="time_complexity.c"
build
1 year ago
/* 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;
}
build
1 year ago
```
=== "Zig"
```zig title="time_complexity.zig"
build
1 year ago
// 线性阶（遍历数组）
fn arrayTraversal(nums: []i32) i32 {
var count: i32 = 0;
// 循环次数与数组长度成正比
for (nums) |_| {
count += 1;
}
return count;
}
build
1 year ago
```
build
11 months ago
??? pythontutor "可视化运行"
build
10 months ago
<div style="height: 560px; 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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全屏观看 ></a>
build
11 months ago
build
1 year ago
值得注意的是，**输入数据大小 $n$ 需根据输入数据的类型来具体确定**。比如在第一个示例中，变量 $n$ 为输入数据大小；在第二个示例中，数组长度 $n$ 为数据大小。
### 3. &nbsp; 平方阶 $O(n^2)$
build
12 months ago
平方阶的操作数量相对于输入数据大小 $n$ 以平方级别增长。平方阶通常出现在嵌套循环中，外层循环和内层循环的时间复杂度都为 $O(n)$ ，因此总体的时间复杂度为 $O(n^2)$ ：
build
1 year ago
=== "Python"
```python title="time_complexity.py"
build
1 year ago
def quadratic(n: int) -> int:
"""平方阶"""
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
build
1 year ago
```
=== "C++"
```cpp title="time_complexity.cpp"
build
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/* 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;
}
build
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```
=== "Java"
```java title="time_complexity.java"
build
1 year ago
/* 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;
}
build
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```
=== "C#"
```csharp title="time_complexity.cs"
build
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/* 平方阶 */
build
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int Quadratic(int n) {
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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;
}
build
1 year ago
```
=== "Go"
```go title="time_complexity.go"
build
1 year ago
/* 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
}
build
1 year ago
```
=== "Swift"
```swift title="time_complexity.swift"
build
1 year ago
/* 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
}
build
1 year ago
```
=== "JS"
```javascript title="time_complexity.js"
build
1 year ago
/* 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;
}
build
1 year ago
```
=== "TS"
```typescript title="time_complexity.ts"
build
1 year ago
/* 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;
}
build
1 year ago
```
=== "Dart"
```dart title="time_complexity.dart"
build
1 year ago
/* 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;
}
build
1 year ago
```
=== "Rust"
```rust title="time_complexity.rs"
build
1 year ago
/* 平方阶 */
fn quadratic(n: i32) -> i32 {
let mut 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;
}
}
count
}
build
1 year ago
```
=== "C"
```c title="time_complexity.c"
build
1 year ago
/* 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;
}
build
1 year ago
```
=== "Zig"
```zig title="time_complexity.zig"
build
1 year ago
// 平方阶
fn quadratic(n: i32) i32 {
var count: i32 = 0;
var i: i32 = 0;
// 循环次数与数组长度成平方关系
while (i < n) : (i += 1) {
var j: i32 = 0;
while (j < n) : (j += 1) {
count += 1;
}
}
return count;
}
build
1 year ago
```
build
11 months ago
??? pythontutor "可视化运行"
build
10 months ago
<div style="height: 560px; 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%E7%BB%84%E9%95%BF%E5%BA%A6%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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<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%E7%BB%84%E9%95%BF%E5%BA%A6%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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全屏观看 ></a>
build
11 months ago
build
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图 2-10 对比了常数阶、线性阶和平方阶三种时间复杂度。
build
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![常数阶、线性阶和平方阶的时间复杂度](time_complexity.assets/time_complexity_constant_linear_quadratic.png){ class="animation-figure" }
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<p align="center"> 图 2-10 &nbsp; 常数阶、线性阶和平方阶的时间复杂度 </p>
build
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以冒泡排序为例，外层循环执行 $n - 1$ 次，内层循环执行 $n-1$、$n-2$、$\dots$、$2$、$1$ 次，平均为 $n / 2$ 次，因此时间复杂度为 $O((n - 1) n / 2) = O(n^2)$ ：
build
1 year ago
=== "Python"
```python title="time_complexity.py"
build
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def bubble_sort(nums: list[int]) -> int:
"""平方阶（冒泡排序）"""
count = 0 # 计数器
# 外循环：未排序区间为 [0, i]
for i in range(len(nums) - 1, 0, -1):
# 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
for j in range(i):
if nums[j] > nums[j + 1]:
# 交换 nums[j] 与 nums[j + 1]
tmp: int = 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
build
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```
=== "C++"
```cpp title="time_complexity.cpp"
build
1 year ago
/* 平方阶（冒泡排序） */
int bubbleSort(vector<int> &nums) {
int count = 0; // 计数器
// 外循环：未排序区间为 [0, i]
for (int i = nums.size() - 1; i > 0; i--) {
// 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
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;
}
build
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```
=== "Java"
```java title="time_complexity.java"
build
1 year ago
/* 平方阶（冒泡排序） */
int bubbleSort(int[] nums) {
int count = 0; // 计数器
// 外循环：未排序区间为 [0, i]
for (int i = nums.length - 1; i > 0; i--) {
// 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
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;
}
build
1 year ago
```
=== "C#"
```csharp title="time_complexity.cs"
build
1 year ago
/* 平方阶（冒泡排序） */
build
1 year ago
int BubbleSort(int[] nums) {
build
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int count = 0; // 计数器
// 外循环：未排序区间为 [0, i]
for (int i = nums.Length - 1; i > 0; i--) {
// 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
for (int j = 0; j < i; j++) {
if (nums[j] > nums[j + 1]) {
// 交换 nums[j] 与 nums[j + 1]
(nums[j + 1], nums[j]) = (nums[j], nums[j + 1]);
count += 3; // 元素交换包含 3 个单元操作
}
}
}
return count;
}
build
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```
=== "Go"
```go title="time_complexity.go"
build
1 year ago
/* 平方阶（冒泡排序） */
func bubbleSort(nums []int) int {
count := 0 // 计数器
// 外循环：未排序区间为 [0, i]
for i := len(nums) - 1; i > 0; i-- {
// 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
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
}
build
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```
=== "Swift"
```swift title="time_complexity.swift"
build
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/* 平方阶（冒泡排序） */
func bubbleSort(nums: inout [Int]) -> Int {
var count = 0 // 计数器
// 外循环：未排序区间为 [0, i]
for i in stride(from: nums.count - 1, to: 0, by: -1) {
// 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
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
}
build
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```
=== "JS"
```javascript title="time_complexity.js"
build
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/* 平方阶（冒泡排序） */
function bubbleSort(nums) {
let count = 0; // 计数器
// 外循环：未排序区间为 [0, i]
for (let i = nums.length - 1; i > 0; i--) {
// 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
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;
}
build
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```
=== "TS"
```typescript title="time_complexity.ts"
build
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/* 平方阶（冒泡排序） */
function bubbleSort(nums: number[]): number {
let count = 0; // 计数器
// 外循环：未排序区间为 [0, i]
for (let i = nums.length - 1; i > 0; i--) {
// 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
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;
}
build
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```
=== "Dart"
```dart title="time_complexity.dart"
build
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/* 平方阶（冒泡排序） */
int bubbleSort(List<int> nums) {
int count = 0; // 计数器
// 外循环：未排序区间为 [0, i]
for (var i = nums.length - 1; i > 0; i--) {
// 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
for (var 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;
}
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```
=== "Rust"
```rust title="time_complexity.rs"
build
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/* 平方阶（冒泡排序） */
fn bubble_sort(nums: &mut [i32]) -> i32 {
let mut count = 0; // 计数器
// 外循环：未排序区间为 [0, i]
for i in (1..nums.len()).rev() {
// 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
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
}
}
}
count
}
build
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```
=== "C"
```c title="time_complexity.c"
build
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/* 平方阶（冒泡排序） */
int bubbleSort(int *nums, int n) {
int count = 0; // 计数器
// 外循环：未排序区间为 [0, i]
for (int i = n - 1; i > 0; i--) {
// 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
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;
}
build
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```
=== "Zig"
```zig title="time_complexity.zig"
build
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// 平方阶（冒泡排序）
fn bubbleSort(nums: []i32) i32 {
var count: i32 = 0; // 计数器
// 外循环：未排序区间为 [0, i]
var i: i32 = @as(i32, @intCast(nums.len)) - 1;
while (i > 0) : (i -= 1) {
var j: usize = 0;
// 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
while (j < i) : (j += 1) {
if (nums[j] > nums[j + 1]) {
// 交换 nums[j] 与 nums[j + 1]
var 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;
}
build
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```
build
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??? pythontutor "可视化运行"
build
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<div style="height: 560px; 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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全屏观看 ></a>
build
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### 4. &nbsp; 指数阶 $O(2^n)$
生物学的“细胞分裂”是指数阶增长的典型例子：初始状态为 $1$ 个细胞，分裂一轮后变为 $2$ 个，分裂两轮后变为 $4$ 个，以此类推，分裂 $n$ 轮后有 $2^n$ 个细胞。
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图 2-11 和以下代码模拟了细胞分裂的过程，时间复杂度为 $O(2^n)$ ：
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=== "Python"
```python title="time_complexity.py"
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def exponential(n: int) -> int:
"""指数阶（循环实现）"""
count = 0
base = 1
# 细胞每轮一分为二，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
build
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```
=== "C++"
```cpp title="time_complexity.cpp"
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/* 指数阶（循环实现） */
int exponential(int n) {
int count = 0, base = 1;
// 细胞每轮一分为二，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;
}
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```
=== "Java"
```java title="time_complexity.java"
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/* 指数阶（循环实现） */
int exponential(int n) {
int count = 0, base = 1;
// 细胞每轮一分为二，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;
}
build
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```
=== "C#"
```csharp title="time_complexity.cs"
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/* 指数阶（循环实现） */
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int Exponential(int n) {
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int count = 0, bas = 1;
// 细胞每轮一分为二，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;
}
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```
=== "Go"
```go title="time_complexity.go"
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/* 指数阶（循环实现）*/
func exponential(n int) int {
count, base := 0, 1
// 细胞每轮一分为二，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
}
build
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```
=== "Swift"
```swift title="time_complexity.swift"
build
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/* 指数阶（循环实现） */
func exponential(n: Int) -> Int {
var count = 0
var base = 1
// 细胞每轮一分为二，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
}
build
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```
=== "JS"
```javascript title="time_complexity.js"
build
1 year ago
/* 指数阶（循环实现） */
function exponential(n) {
let count = 0,
base = 1;
// 细胞每轮一分为二，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;
}
build
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```
=== "TS"
```typescript title="time_complexity.ts"
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/* 指数阶（循环实现） */
function exponential(n: number): number {
let count = 0,
base = 1;
// 细胞每轮一分为二，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;
}
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```
=== "Dart"
```dart title="time_complexity.dart"
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/* 指数阶（循环实现） */
int exponential(int n) {
int count = 0, base = 1;
// 细胞每轮一分为二，形成数列 1, 2, 4, 8, ..., 2^(n-1)
for (var i = 0; i < n; i++) {
for (var j = 0; j < base; j++) {
count++;
}
base *= 2;
}
// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
return count;
}
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```
=== "Rust"
```rust title="time_complexity.rs"
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/* 指数阶（循环实现） */
fn exponential(n: i32) -> i32 {
let mut count = 0;
let mut base = 1;
// 细胞每轮一分为二，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
count
}
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```
=== "C"
```c title="time_complexity.c"
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/* 指数阶（循环实现） */
int exponential(int n) {
int count = 0;
int bas = 1;
// 细胞每轮一分为二，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;
}
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```
=== "Zig"
```zig title="time_complexity.zig"
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// 指数阶（循环实现）
fn exponential(n: i32) i32 {
var count: i32 = 0;
var bas: i32 = 1;
var i: i32 = 0;
// 细胞每轮一分为二，形成数列 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;
}
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```
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??? pythontutor "可视化运行"
build
10 months ago
<div style="height: 560px; 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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全屏观看 ></a>
build
11 months ago
build
1 year ago
![指数阶的时间复杂度](time_complexity.assets/time_complexity_exponential.png){ class="animation-figure" }
build
1 year ago
<p align="center"> 图 2-11 &nbsp; 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中。例如在以下代码中，其递归地一分为二，经过 $n$ 次分裂后停止：
=== "Python"
```python title="time_complexity.py"
build
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def exp_recur(n: int) -> int:
"""指数阶（递归实现）"""
if n == 1:
return 1
return exp_recur(n - 1) + exp_recur(n - 1) + 1
build
1 year ago
```
=== "C++"
```cpp title="time_complexity.cpp"
build
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/* 指数阶（递归实现） */
int expRecur(int n) {
if (n == 1)
return 1;
return expRecur(n - 1) + expRecur(n - 1) + 1;
}
build
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```
=== "Java"
```java title="time_complexity.java"
build
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/* 指数阶（递归实现） */
int expRecur(int n) {
if (n == 1)
return 1;
return expRecur(n - 1) + expRecur(n - 1) + 1;
}
build
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```
=== "C#"
```csharp title="time_complexity.cs"
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/* 指数阶（递归实现） */
build
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int ExpRecur(int n) {
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if (n == 1) return 1;
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return ExpRecur(n - 1) + ExpRecur(n - 1) + 1;
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}
build
1 year ago
```
=== "Go"
```go title="time_complexity.go"
build
1 year ago
/* 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
}
build
1 year ago
```
=== "Swift"
```swift title="time_complexity.swift"
build
1 year ago
/* 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
}
build
1 year ago
```
=== "JS"
```javascript title="time_complexity.js"
build
1 year ago
/* 指数阶（递归实现） */
function expRecur(n) {
if (n === 1) return 1;
return expRecur(n - 1) + expRecur(n - 1) + 1;
}
build
1 year ago
```
=== "TS"
```typescript title="time_complexity.ts"
build
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/* 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;
}
build
1 year ago
```
=== "Dart"
```dart title="time_complexity.dart"
build
1 year ago
/* 指数阶（递归实现） */
int expRecur(int n) {
if (n == 1) return 1;
return expRecur(n - 1) + expRecur(n - 1) + 1;
}
build
1 year ago
```
=== "Rust"
```rust title="time_complexity.rs"
build
1 year ago
/* 指数阶（递归实现） */
fn exp_recur(n: i32) -> i32 {
if n == 1 {
return 1;
}
exp_recur(n - 1) + exp_recur(n - 1) + 1
}
build
1 year ago
```
=== "C"
```c title="time_complexity.c"
build
1 year ago
/* 指数阶（递归实现） */
int expRecur(int n) {
if (n == 1)
return 1;
return expRecur(n - 1) + expRecur(n - 1) + 1;
}
build
1 year ago
```
=== "Zig"
```zig title="time_complexity.zig"
build
1 year ago
// 指数阶（递归实现）
fn expRecur(n: i32) i32 {
if (n == 1) return 1;
return expRecur(n - 1) + expRecur(n - 1) + 1;
}
build
1 year ago
```
build
11 months ago
??? pythontutor "可视化运行"
build
10 months ago
<div style="height: 560px; 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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全屏观看 ></a>
build
11 months ago
build
12 months ago
指数阶增长非常迅速，在穷举法（暴力搜索、回溯等）中比较常见。对于数据规模较大的问题，指数阶是不可接受的，通常需要使用动态规划或贪心算法等来解决。
build
1 year ago
### 5. &nbsp; 对数阶 $O(\log n)$
与指数阶相反，对数阶反映了“每轮缩减到一半”的情况。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。
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图 2-12 和以下代码模拟了“每轮缩减到一半”的过程，时间复杂度为 $O(\log_2 n)$ ，简记为 $O(\log n)$ ：
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=== "Python"
```python title="time_complexity.py"
build
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def logarithmic(n: float) -> int:
"""对数阶（循环实现）"""
count = 0
while n > 1:
n = n / 2
count += 1
return count
build
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```
=== "C++"
```cpp title="time_complexity.cpp"
build
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/* 对数阶（循环实现） */
int logarithmic(float n) {
int count = 0;
while (n > 1) {
n = n / 2;
count++;
}
return count;
}
build
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```
=== "Java"
```java title="time_complexity.java"
build
1 year ago
/* 对数阶（循环实现） */
int logarithmic(float n) {
int count = 0;
while (n > 1) {
n = n / 2;
count++;
}
return count;
}
build
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```
=== "C#"
```csharp title="time_complexity.cs"
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/* 对数阶（循环实现） */
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int Logarithmic(float n) {
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int count = 0;
while (n > 1) {
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n /= 2;
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count++;
}
return count;
}
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```
=== "Go"
```go title="time_complexity.go"
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/* 对数阶（循环实现）*/
func logarithmic(n float64) int {
count := 0
for n > 1 {
n = n / 2
count++
}
return count
}
build
1 year ago
```
=== "Swift"
```swift title="time_complexity.swift"
build
1 year ago
/* 对数阶（循环实现） */
func logarithmic(n: Double) -> Int {
var count = 0
var n = n
while n > 1 {
n = n / 2
count += 1
}
return count
}
build
1 year ago
```
=== "JS"
```javascript title="time_complexity.js"
build
1 year ago
/* 对数阶（循环实现） */
function logarithmic(n) {
let count = 0;
while (n > 1) {
n = n / 2;
count++;
}
return count;
}
build
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```
=== "TS"
```typescript title="time_complexity.ts"
build
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/* 对数阶（循环实现） */
function logarithmic(n: number): number {
let count = 0;
while (n > 1) {
n = n / 2;
count++;
}
return count;
}
build
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```
=== "Dart"
```dart title="time_complexity.dart"
build
1 year ago
/* 对数阶（循环实现） */
int logarithmic(num n) {
int count = 0;
while (n > 1) {
n = n / 2;
count++;
}
return count;
}
build
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```
=== "Rust"
```rust title="time_complexity.rs"
build
1 year ago
/* 对数阶（循环实现） */
fn logarithmic(mut n: f32) -> i32 {
let mut count = 0;
while n > 1.0 {
n = n / 2.0;
count += 1;
}
count
}
build
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```
=== "C"
```c title="time_complexity.c"
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1 year ago
/* 对数阶（循环实现） */
int logarithmic(float n) {
int count = 0;
while (n > 1) {
n = n / 2;
count++;
}
return count;
}
build
1 year ago
```
=== "Zig"
```zig title="time_complexity.zig"
build
1 year ago
// 对数阶（循环实现）
fn logarithmic(n: f32) i32 {
var count: i32 = 0;
var n_var = n;
while (n_var > 1)
{
n_var = n_var / 2;
count +=1;
}
return count;
}
build
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```
build
11 months ago
??? pythontutor "可视化运行"
build
10 months ago
<div style="height: 560px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20logarithmic%28n%3A%20float%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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<a href="https://pythontutor.com/iframe-embed.html#code=def%20logarithmic%28n%3A%20float%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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全屏观看 ></a>
build
11 months ago
build
1 year ago
![对数阶的时间复杂度](time_complexity.assets/time_complexity_logarithmic.png){ class="animation-figure" }
build
1 year ago
<p align="center"> 图 2-12 &nbsp; 对数阶的时间复杂度 </p>
build
12 months ago
与指数阶类似，对数阶也常出现于递归函数中。以下代码形成了一棵高度为 $\log_2 n$ 的递归树：
build
1 year ago
=== "Python"
```python title="time_complexity.py"
build
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def log_recur(n: float) -> int:
"""对数阶（递归实现）"""
if n <= 1:
return 0
return log_recur(n / 2) + 1
build
1 year ago
```
=== "C++"
```cpp title="time_complexity.cpp"
build
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/* 对数阶（递归实现） */
int logRecur(float n) {
if (n <= 1)
return 0;
return logRecur(n / 2) + 1;
}
build
1 year ago
```
=== "Java"
```java title="time_complexity.java"
build
1 year ago
/* 对数阶（递归实现） */
int logRecur(float n) {
if (n <= 1)
return 0;
return logRecur(n / 2) + 1;
}
build
1 year ago
```
=== "C#"
```csharp title="time_complexity.cs"
build
1 year ago
/* 对数阶（递归实现） */
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1 year ago
int LogRecur(float n) {
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if (n <= 1) return 0;
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return LogRecur(n / 2) + 1;
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1 year ago
}
build
1 year ago
```
=== "Go"
```go title="time_complexity.go"
build
1 year ago
/* 对数阶（递归实现）*/
func logRecur(n float64) int {
if n <= 1 {
return 0
}
return logRecur(n/2) + 1
}
build
1 year ago
```
=== "Swift"
```swift title="time_complexity.swift"
build
1 year ago
/* 对数阶（递归实现） */
func logRecur(n: Double) -> Int {
if n <= 1 {
return 0
}
return logRecur(n: n / 2) + 1
}
build
1 year ago
```
=== "JS"
```javascript title="time_complexity.js"
build
1 year ago
/* 对数阶（递归实现） */
function logRecur(n) {
if (n <= 1) return 0;
return logRecur(n / 2) + 1;
}
build
1 year ago
```
=== "TS"
```typescript title="time_complexity.ts"
build
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/* 对数阶（递归实现） */
function logRecur(n: number): number {
if (n <= 1) return 0;
return logRecur(n / 2) + 1;
}
build
1 year ago
```
=== "Dart"
```dart title="time_complexity.dart"
build
1 year ago
/* 对数阶（递归实现） */
int logRecur(num n) {
if (n <= 1) return 0;
return logRecur(n / 2) + 1;
}
build
1 year ago
```
=== "Rust"
```rust title="time_complexity.rs"
build
1 year ago
/* 对数阶（递归实现） */
fn log_recur(n: f32) -> i32 {
if n <= 1.0 {
return 0;
}
log_recur(n / 2.0) + 1
}
build
1 year ago
```
=== "C"
```c title="time_complexity.c"
build
1 year ago
/* 对数阶（递归实现） */
int logRecur(float n) {
if (n <= 1)
return 0;
return logRecur(n / 2) + 1;
}
build
1 year ago
```
=== "Zig"
```zig title="time_complexity.zig"
build
1 year ago
// 对数阶（递归实现）
fn logRecur(n: f32) i32 {
if (n <= 1) return 0;
return logRecur(n / 2) + 1;
}
build
1 year ago
```
build
11 months ago
??? pythontutor "可视化运行"
build
10 months ago
<div style="height: 560px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20log_recur%28n%3A%20float%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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<a href="https://pythontutor.com/iframe-embed.html#code=def%20log_recur%28n%3A%20float%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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全屏观看 ></a>
build
11 months ago
build
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对数阶常出现于基于分治策略的算法中，体现了“一分为多”和“化繁为简”的算法思想。它增长缓慢，是仅次于常数阶的理想的时间复杂度。
!!! tip "$O(\log n)$ 的底数是多少？"
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准确来说，“一分为 $m$”对应的时间复杂度是 $O(\log_m n)$ 。而通过对数换底公式，我们可以得到具有不同底数、相等的时间复杂度：
build
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$$
O(\log_m n) = O(\log_k n / \log_k m) = O(\log_k n)
$$
也就是说，底数 $m$ 可以在不影响复杂度的前提下转换。因此我们通常会省略底数 $m$ ，将对数阶直接记为 $O(\log n)$ 。
### 6. &nbsp; 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 $O(n)$ 。相关代码如下：
=== "Python"
```python title="time_complexity.py"
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def linear_log_recur(n: float) -> int:
"""线性对数阶"""
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
build
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```
=== "C++"
```cpp title="time_complexity.cpp"
build
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/* 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;
}
build
1 year ago
```
=== "Java"
```java title="time_complexity.java"
build
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/* 线性对数阶 */
int linearLogRecur(float n) {
if (n <= 1)
return 1;
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int count = linearLogRecur(n / 2) + linearLogRecur(n / 2);
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for (int i = 0; i < n; i++) {
count++;
}
return count;
}
build
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```
=== "C#"
```csharp title="time_complexity.cs"
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/* 线性对数阶 */
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int LinearLogRecur(float n) {
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if (n <= 1) return 1;
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int count = LinearLogRecur(n / 2) + LinearLogRecur(n / 2);
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for (int i = 0; i < n; i++) {
count++;
}
return count;
}
build
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```
=== "Go"
```go title="time_complexity.go"
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/* 线性对数阶 */
func linearLogRecur(n float64) int {
if n <= 1 {
return 1
}
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count := linearLogRecur(n/2) + linearLogRecur(n/2)
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for i := 0.0; i < n; i++ {
count++
}
return count
}
build
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```
=== "Swift"
```swift title="time_complexity.swift"
build
1 year ago
/* 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 stride(from: 0, to: n, by: 1) {
count += 1
}
return count
}
build
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```
=== "JS"
```javascript title="time_complexity.js"
build
1 year ago
/* 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;
}
build
1 year ago
```
=== "TS"
```typescript title="time_complexity.ts"
build
1 year ago
/* 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;
}
build
1 year ago
```
=== "Dart"
```dart title="time_complexity.dart"
build
1 year ago
/* 线性对数阶 */
int linearLogRecur(num n) {
if (n <= 1) return 1;
int count = linearLogRecur(n / 2) + linearLogRecur(n / 2);
for (var i = 0; i < n; i++) {
count++;
}
return count;
}
build
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```
=== "Rust"
```rust title="time_complexity.rs"
build
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/* 线性对数阶 */
fn linear_log_recur(n: f32) -> i32 {
if n <= 1.0 {
return 1;
}
build
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let mut count = linear_log_recur(n / 2.0) + linear_log_recur(n / 2.0);
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for _ in 0 ..n as i32 {
count += 1;
}
return count
}
build
1 year ago
```
=== "C"
```c title="time_complexity.c"
build
1 year ago
/* 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;
}
build
1 year ago
```
=== "Zig"
```zig title="time_complexity.zig"
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// 线性对数阶
fn linearLogRecur(n: f32) i32 {
if (n <= 1) return 1;
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var count: i32 = linearLogRecur(n / 2) + linearLogRecur(n / 2);
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var i: f32 = 0;
while (i < n) : (i += 1) {
count += 1;
}
return count;
}
build
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```
build
11 months ago
??? pythontutor "可视化运行"
build
10 months ago
<div style="height: 560px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20linear_log_recur%28n%3A%20float%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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<a href="https://pythontutor.com/iframe-embed.html#code=def%20linear_log_recur%28n%3A%20float%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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全屏观看 ></a>
build
11 months ago
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图 2-13 展示了线性对数阶的生成方式。二叉树的每一层的操作总数都为 $n$ ，树共有 $\log_2 n + 1$ 层，因此时间复杂度为 $O(n \log n)$ 。
build
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![线性对数阶的时间复杂度](time_complexity.assets/time_complexity_logarithmic_linear.png){ class="animation-figure" }
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<p align="center"> 图 2-13 &nbsp; 线性对数阶的时间复杂度 </p>
主流排序算法的时间复杂度通常为 $O(n \log n)$ ，例如快速排序、归并排序、堆排序等。
### 7. &nbsp; 阶乘阶 $O(n!)$
阶乘阶对应数学上的“全排列”问题。给定 $n$ 个互不重复的元素，求其所有可能的排列方案，方案数量为：
$$
n! = n \times (n - 1) \times (n - 2) \times \dots \times 2 \times 1
$$
阶乘通常使用递归实现。如图 2-14 和以下代码所示，第一层分裂出 $n$ 个，第二层分裂出 $n - 1$ 个，以此类推，直至第 $n$ 层时停止分裂：
=== "Python"
```python title="time_complexity.py"
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def factorial_recur(n: int) -> int:
"""阶乘阶（递归实现）"""
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
build
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```
=== "C++"
```cpp title="time_complexity.cpp"
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/* 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;
}
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1 year ago
```
=== "Java"
```java title="time_complexity.java"
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/* 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;
}
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```
=== "C#"
```csharp title="time_complexity.cs"
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/* 阶乘阶（递归实现） */
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int FactorialRecur(int n) {
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if (n == 0) return 1;
int count = 0;
// 从 1 个分裂出 n 个
for (int i = 0; i < n; i++) {
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count += FactorialRecur(n - 1);
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}
return count;
}
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```
=== "Go"
```go title="time_complexity.go"
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1 year ago
/* 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
}
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```
=== "Swift"
```swift title="time_complexity.swift"
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/* 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
}
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```
=== "JS"
```javascript title="time_complexity.js"
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1 year ago
/* 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;
}
build
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```
=== "TS"
```typescript title="time_complexity.ts"
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1 year ago
/* 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;
}
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1 year ago
```
=== "Dart"
```dart title="time_complexity.dart"
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/* 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 (var i = 0; i < n; i++) {
count += factorialRecur(n - 1);
}
return count;
}
build
1 year ago
```
=== "Rust"
```rust title="time_complexity.rs"
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1 year ago
/* 阶乘阶（递归实现） */
fn factorial_recur(n: i32) -> i32 {
if n == 0 {
return 1;
}
let mut count = 0;
// 从 1 个分裂出 n 个
for _ in 0..n {
count += factorial_recur(n - 1);
}
count
}
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1 year ago
```
=== "C"
```c title="time_complexity.c"
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/* 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;
}
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1 year ago
```
=== "Zig"
```zig title="time_complexity.zig"
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1 year ago
// 阶乘阶（递归实现）
fn factorialRecur(n: i32) i32 {
if (n == 0) return 1;
var count: i32 = 0;
var i: i32 = 0;
// 从 1 个分裂出 n 个
while (i < n) : (i += 1) {
count += factorialRecur(n - 1);
}
return count;
}
build
1 year ago
```
build
11 months ago
??? pythontutor "可视化运行"
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<div style="height: 560px; 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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<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=470&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全屏观看 ></a>
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![阶乘阶的时间复杂度](time_complexity.assets/time_complexity_factorial.png){ class="animation-figure" }
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<p align="center"> 图 2-14 &nbsp; 阶乘阶的时间复杂度 </p>
请注意，因为当 $n \geq 4$ 时恒有 $n! > 2^n$ ，所以阶乘阶比指数阶增长得更快，在 $n$ 较大时也是不可接受的。
## 2.3.5 &nbsp; 最差、最佳、平均时间复杂度
**算法的时间效率往往不是固定的，而是与输入数据的分布有关**。假设输入一个长度为 $n$ 的数组 `nums` ，其中 `nums` 由从 $1$ 至 $n$ 的数字组成，每个数字只出现一次；但元素顺序是随机打乱的，任务目标是返回元素 $1$ 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，需要完整遍历数组，**达到最差时间复杂度 $O(n)$** 。
- 当 `nums = [1, ?, ?, ...]` ，即当首个元素为 $1$ 时，无论数组多长都不需要继续遍历，**达到最佳时间复杂度 $\Omega(1)$** 。
“最差时间复杂度”对应函数渐近上界，使用大 $O$ 记号表示。相应地，“最佳时间复杂度”对应函数渐近下界，用 $\Omega$ 记号表示：
=== "Python"
```python title="worst_best_time_complexity.py"
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def random_numbers(n: int) -> list[int]:
"""生成一个数组，元素为: 1, 2, ..., n ，顺序被打乱"""
# 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
def find_one(nums: list[int]) -> int:
"""查找数组 nums 中数字 1 所在索引"""
for i in range(len(nums)):
# 当元素 1 在数组头部时，达到最佳时间复杂度 O(1)
# 当元素 1 在数组尾部时，达到最差时间复杂度 O(n)
if nums[i] == 1:
return i
return -1
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```
=== "C++"
```cpp title="worst_best_time_complexity.cpp"
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/* 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;
}
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/* 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++) {
// 当元素 1 在数组头部时，达到最佳时间复杂度 O(1)
// 当元素 1 在数组尾部时，达到最差时间复杂度 O(n)
if (nums[i] == 1)
return i;
}
return -1;
}
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```
=== "Java"
```java title="worst_best_time_complexity.java"
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/* 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;
}
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/* 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++) {
// 当元素 1 在数组头部时，达到最佳时间复杂度 O(1)
// 当元素 1 在数组尾部时，达到最差时间复杂度 O(n)
if (nums[i] == 1)
return i;
}
return -1;
}
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```
=== "C#"
```csharp title="worst_best_time_complexity.cs"
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/* 生成一个数组，元素为 { 1, 2, ..., n }，顺序被打乱 */
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int[] RandomNumbers(int n) {
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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;
}
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// 随机打乱数组元素
for (int i = 0; i < nums.Length; i++) {
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int index = new Random().Next(i, nums.Length);
(nums[i], nums[index]) = (nums[index], nums[i]);
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}
return nums;
}
/* 查找数组 nums 中数字 1 所在索引 */
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int FindOne(int[] nums) {
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for (int i = 0; i < nums.Length; i++) {
// 当元素 1 在数组头部时，达到最佳时间复杂度 O(1)
// 当元素 1 在数组尾部时，达到最差时间复杂度 O(n)
if (nums[i] == 1)
return i;
}
return -1;
}
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```
=== "Go"
```go title="worst_best_time_complexity.go"
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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 {
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
}
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/* 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++ {
// 当元素 1 在数组头部时，达到最佳时间复杂度 O(1)
// 当元素 1 在数组尾部时，达到最差时间复杂度 O(n)
if nums[i] == 1 {
return i
}
}
return -1
}
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```
=== "Swift"
```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
}
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/* 查找数组 nums 中数字 1 所在索引 */
func findOne(nums: [Int]) -> Int {
for i in nums.indices {
// 当元素 1 在数组头部时，达到最佳时间复杂度 O(1)
// 当元素 1 在数组尾部时，达到最差时间复杂度 O(n)
if nums[i] == 1 {
return i
}
}
return -1
}
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```
=== "JS"
```javascript title="worst_best_time_complexity.js"
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/* 生成一个数组，元素为 { 1, 2, ..., n }，顺序被打乱 */
function randomNumbers(n) {
const 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++) {
const r = Math.floor(Math.random() * (i + 1));
const temp = nums[i];
nums[i] = nums[r];
nums[r] = temp;
}
return nums;
}
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/* 查找数组 nums 中数字 1 所在索引 */
function findOne(nums) {
for (let i = 0; i < nums.length; i++) {
// 当元素 1 在数组头部时，达到最佳时间复杂度 O(1)
// 当元素 1 在数组尾部时，达到最差时间复杂度 O(n)
if (nums[i] === 1) {
return i;
}
}
return -1;
}
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```
=== "TS"
```typescript title="worst_best_time_complexity.ts"
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/* 生成一个数组，元素为 { 1, 2, ..., n }，顺序被打乱 */
function randomNumbers(n: number): number[] {
const 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++) {
const r = Math.floor(Math.random() * (i + 1));
const temp = nums[i];
nums[i] = nums[r];
nums[r] = temp;
}
return nums;
}
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/* 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++) {
// 当元素 1 在数组头部时，达到最佳时间复杂度 O(1)
// 当元素 1 在数组尾部时，达到最差时间复杂度 O(n)
if (nums[i] === 1) {
return i;
}
}
return -1;
}
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```
=== "Dart"
```dart title="worst_best_time_complexity.dart"
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/* 生成一个数组，元素为 { 1, 2, ..., n }，顺序被打乱 */
List<int> randomNumbers(int n) {
final nums = List.filled(n, 0);
// 生成数组 nums = { 1, 2, 3, ..., n }
for (var i = 0; i < n; i++) {
nums[i] = i + 1;
}
// 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 */
int findOne(List<int> nums) {
for (var i = 0; i < nums.length; i++) {
// 当元素 1 在数组头部时，达到最佳时间复杂度 O(1)
// 当元素 1 在数组尾部时，达到最差时间复杂度 O(n)
if (nums[i] == 1) return i;
}
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return -1;
}
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```
=== "Rust"
```rust title="worst_best_time_complexity.rs"
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/* 生成一个数组，元素为 { 1, 2, ..., n }，顺序被打乱 */
fn random_numbers(n: i32) -> Vec<i32> {
// 生成数组 nums = { 1, 2, 3, ..., n }
let mut nums = (1..=n).collect::<Vec<i32>>();
// 随机打乱数组元素
nums.shuffle(&mut thread_rng());
nums
}
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/* 查找数组 nums 中数字 1 所在索引 */
fn find_one(nums: &[i32]) -> Option<usize> {
for i in 0..nums.len() {
// 当元素 1 在数组头部时，达到最佳时间复杂度 O(1)
// 当元素 1 在数组尾部时，达到最差时间复杂度 O(n)
if nums[i] == 1 {
return Some(i);
}
}
None
}
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```
=== "C"
```c title="worst_best_time_complexity.c"
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/* 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;
}
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/* 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++) {
// 当元素 1 在数组头部时，达到最佳时间复杂度 O(1)
// 当元素 1 在数组尾部时，达到最差时间复杂度 O(n)
if (nums[i] == 1)
return i;
}
return -1;
}
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```
=== "Zig"
```zig title="worst_best_time_complexity.zig"
// 生成一个数组，元素为 { 1, 2, ..., n }，顺序被打乱
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fn randomNumbers(comptime n: usize) [n]i32 {
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var nums: [n]i32 = undefined;
// 生成数组 nums = { 1, 2, 3, ..., n }
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for (&nums, 0..) |*num, i| {
num.* = @as(i32, @intCast(i)) + 1;
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}
// 随机打乱数组元素
const rand = std.crypto.random;
rand.shuffle(i32, &nums);
return nums;
}
// 查找数组 nums 中数字 1 所在索引
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fn findOne(nums: []i32) i32 {
for (nums, 0..) |num, i| {
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// 当元素 1 在数组头部时，达到最佳时间复杂度 O(1)
// 当元素 1 在数组尾部时，达到最差时间复杂度 O(n)
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if (num == 1) return @intCast(i);
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}
return -1;
}
```
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??? pythontutor "可视化运行"
build
10 months ago
<div style="height: 560px; 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=470&codeDivWidth=350&cumulative=false&curInstr=25&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<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=470&codeDivWidth=350&cumulative=false&curInstr=25&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全屏观看 ></a>
build
11 months ago
build
1 year ago
值得说明的是，我们在实际中很少使用最佳时间复杂度，因为通常只有在很小概率下才能达到，可能会带来一定的误导性。**而最差时间复杂度更为实用，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。
build
12 months ago
从上述示例可以看出，最差时间复杂度和最佳时间复杂度只出现于“特殊的数据分布”，这些情况的出现概率可能很小，并不能真实地反映算法运行效率。相比之下，**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 $\Theta$ 记号来表示。
build
1 year ago
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就是数组长度的一半 $n / 2$ ，平均时间复杂度为 $\Theta(n / 2) = \Theta(n)$ 。
build
12 months ago
但对于较为复杂的算法，计算平均时间复杂度往往比较困难，因为很难分析出在数据分布下的整体数学期望。在这种情况下，我们通常使用最差时间复杂度作为算法效率的评判标准。
build
1 year ago
!!! question "为什么很少看到 $\Theta$ 符号？"
build
12 months ago
可能由于 $O$ 符号过于朗朗上口，因此我们常常使用它来表示平均时间复杂度。但从严格意义上讲，这种做法并不规范。在本书和其他资料中，若遇到类似“平均时间复杂度 $O(n)$”的表述，请将其直接理解为 $\Theta(n)$ 。