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---
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comments: true
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---
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# 2.2 迭代与递归
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在算法中，重复执行某个任务是很常见的，它与复杂度分析息息相关。因此，在介绍时间复杂度和空间复杂度之前，我们先来了解如何在程序中实现重复执行任务，即两种基本的程序控制结构：迭代、递归。
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## 2.2.1 迭代
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「迭代 iteration」是一种重复执行某个任务的控制结构。在迭代中，程序会在满足一定的条件下重复执行某段代码，直到这个条件不再满足。
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### 1. for 循环
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`for` 循环是最常见的迭代形式之一，**适合在预先知道迭代次数时使用**。
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以下函数基于 `for` 循环实现了求和 $1 + 2 + \dots + n$ ，求和结果使用变量 `res` 记录。需要注意的是，Python 中 `range(a, b)` 对应的区间是“左闭右开”的，对应的遍历范围为 $a, a + 1, \dots, b-1$ ：
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=== "Python"
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```python title="iteration.py"
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def for_loop(n: int) -> int:
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"""for 循环"""
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res = 0
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# 循环求和 1, 2, ..., n-1, n
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for i in range(1, n + 1):
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res += i
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return res
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```
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=== "C++"
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```cpp title="iteration.cpp"
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/* for 循环 */
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int forLoop(int n) {
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int res = 0;
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// 循环求和 1, 2, ..., n-1, n
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for (int i = 1; i <= n; ++i) {
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res += i;
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}
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return res;
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}
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```
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=== "Java"
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```java title="iteration.java"
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/* for 循环 */
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int forLoop(int n) {
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int res = 0;
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// 循环求和 1, 2, ..., n-1, n
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for (int i = 1; i <= n; i++) {
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res += i;
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}
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return res;
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}
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```
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=== "C#"
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```csharp title="iteration.cs"
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/* for 循环 */
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int ForLoop(int n) {
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int res = 0;
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// 循环求和 1, 2, ..., n-1, n
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for (int i = 1; i <= n; i++) {
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res += i;
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}
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return res;
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}
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```
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=== "Go"
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```go title="iteration.go"
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/* for 循环 */
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func forLoop(n int) int {
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res := 0
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// 循环求和 1, 2, ..., n-1, n
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for i := 1; i <= n; i++ {
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res += i
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}
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return res
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}
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```
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=== "Swift"
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```swift title="iteration.swift"
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/* for 循环 */
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func forLoop(n: Int) -> Int {
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var res = 0
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// 循环求和 1, 2, ..., n-1, n
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for i in 1 ... n {
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res += i
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}
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return res
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}
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```
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=== "JS"
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```javascript title="iteration.js"
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/* for 循环 */
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function forLoop(n) {
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let res = 0;
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// 循环求和 1, 2, ..., n-1, n
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for (let i = 1; i <= n; i++) {
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res += i;
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}
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return res;
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}
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```
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=== "TS"
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```typescript title="iteration.ts"
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/* for 循环 */
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function forLoop(n: number): number {
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let res = 0;
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// 循环求和 1, 2, ..., n-1, n
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for (let i = 1; i <= n; i++) {
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res += i;
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}
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return res;
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}
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```
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=== "Dart"
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```dart title="iteration.dart"
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/* for 循环 */
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int forLoop(int n) {
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int res = 0;
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// 循环求和 1, 2, ..., n-1, n
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for (int i = 1; i <= n; i++) {
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res += i;
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}
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return res;
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}
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```
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=== "Rust"
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```rust title="iteration.rs"
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/* for 循环 */
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fn for_loop(n: i32) -> i32 {
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let mut res = 0;
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// 循环求和 1, 2, ..., n-1, n
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for i in 1..=n {
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res += i;
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}
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res
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}
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```
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=== "C"
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```c title="iteration.c"
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/* for 循环 */
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int forLoop(int n) {
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int res = 0;
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// 循环求和 1, 2, ..., n-1, n
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for (int i = 1; i <= n; i++) {
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res += i;
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}
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return res;
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}
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```
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=== "Zig"
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```zig title="iteration.zig"
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// for 循环
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fn forLoop(n: usize) i32 {
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var res: i32 = 0;
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// 循环求和 1, 2, ..., n-1, n
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for (1..n+1) |i| {
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res = res + @as(i32, @intCast(i));
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}
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return res;
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}
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```
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??? pythontutor "可视化运行"
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<iframe width="800" height="600" frameborder="0" src="https://pythontutor.com/iframe-embed.html#code=def%20for_loop%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22for%20%E5%BE%AA%E7%8E%AF%22%22%22%0A%20%20%20%20res%20%3D%200%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E6%B1%82%E5%92%8C%201,%202,%20...,%20n-1,%20n%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20res%20%2B%3D%20i%0A%20%20%20%20return%20res%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%205%0A%20%20%20%20res%20%3D%20for_loop%28n%29%0A%20%20%20%20print%28f%22%5Cnfor%20%E5%BE%AA%E7%8E%AF%E7%9A%84%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=370&codeDivWidth=300&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&"> </iframe>
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<a href=https://pythontutor.com/iframe-embed.html#code=def%20for_loop%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22for%20%E5%BE%AA%E7%8E%AF%22%22%22%0A%20%20%20%20res%20%3D%200%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E6%B1%82%E5%92%8C%201,%202,%20...,%20n-1,%20n%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20res%20%2B%3D%20i%0A%20%20%20%20return%20res%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%205%0A%20%20%20%20res%20%3D%20for_loop%28n%29%0A%20%20%20%20print%28f%22%5Cnfor%20%E5%BE%AA%E7%8E%AF%E7%9A%84%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=370&codeDivWidth=300&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D& target="_blank" rel="noopener noreferrer">全屏观看 ></a>
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图 2-1 是该求和函数的流程框图。
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{ class="animation-figure" }
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<p align="center"> 图 2-1 求和函数的流程框图 </p>
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此求和函数的操作数量与输入数据大小 $n$ 成正比，或者说成“线性关系”。实际上，**时间复杂度描述的就是这个“线性关系”**。相关内容将会在下一节中详细介绍。
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### 2. while 循环
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与 `for` 循环类似，`while` 循环也是一种实现迭代的方法。在 `while` 循环中，程序每轮都会先检查条件，如果条件为真，则继续执行，否则就结束循环。
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下面我们用 `while` 循环来实现求和 $1 + 2 + \dots + n$ ：
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=== "Python"
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```python title="iteration.py"
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def while_loop(n: int) -> int:
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"""while 循环"""
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res = 0
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i = 1 # 初始化条件变量
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# 循环求和 1, 2, ..., n-1, n
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while i <= n:
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res += i
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i += 1 # 更新条件变量
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return res
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```
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=== "C++"
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```cpp title="iteration.cpp"
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/* while 循环 */
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int whileLoop(int n) {
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int res = 0;
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int i = 1; // 初始化条件变量
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// 循环求和 1, 2, ..., n-1, n
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while (i <= n) {
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res += i;
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i++; // 更新条件变量
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}
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return res;
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}
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```
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=== "Java"
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```java title="iteration.java"
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/* while 循环 */
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int whileLoop(int n) {
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int res = 0;
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int i = 1; // 初始化条件变量
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// 循环求和 1, 2, ..., n-1, n
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while (i <= n) {
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res += i;
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i++; // 更新条件变量
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}
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return res;
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}
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```
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=== "C#"
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```csharp title="iteration.cs"
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/* while 循环 */
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int WhileLoop(int n) {
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int res = 0;
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int i = 1; // 初始化条件变量
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// 循环求和 1, 2, ..., n-1, n
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while (i <= n) {
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res += i;
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i += 1; // 更新条件变量
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}
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return res;
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}
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```
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=== "Go"
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```go title="iteration.go"
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/* while 循环 */
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func whileLoop(n int) int {
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res := 0
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// 初始化条件变量
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i := 1
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// 循环求和 1, 2, ..., n-1, n
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for i <= n {
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res += i
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// 更新条件变量
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i++
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}
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return res
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}
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```
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=== "Swift"
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```swift title="iteration.swift"
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/* while 循环 */
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func whileLoop(n: Int) -> Int {
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var res = 0
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var i = 1 // 初始化条件变量
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// 循环求和 1, 2, ..., n-1, n
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while i <= n {
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res += i
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i += 1 // 更新条件变量
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}
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return res
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}
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```
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=== "JS"
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```javascript title="iteration.js"
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/* while 循环 */
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function whileLoop(n) {
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let res = 0;
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let i = 1; // 初始化条件变量
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// 循环求和 1, 2, ..., n-1, n
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while (i <= n) {
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res += i;
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i++; // 更新条件变量
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}
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return res;
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}
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```
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=== "TS"
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```typescript title="iteration.ts"
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/* while 循环 */
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function whileLoop(n: number): number {
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let res = 0;
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let i = 1; // 初始化条件变量
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// 循环求和 1, 2, ..., n-1, n
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while (i <= n) {
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res += i;
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i++; // 更新条件变量
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}
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return res;
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}
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```
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=== "Dart"
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```dart title="iteration.dart"
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/* while 循环 */
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int whileLoop(int n) {
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int res = 0;
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int i = 1; // 初始化条件变量
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// 循环求和 1, 2, ..., n-1, n
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while (i <= n) {
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res += i;
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i++; // 更新条件变量
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}
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return res;
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}
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```
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=== "Rust"
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```rust title="iteration.rs"
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/* while 循环 */
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fn while_loop(n: i32) -> i32 {
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let mut res = 0;
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let mut i = 1; // 初始化条件变量
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// 循环求和 1, 2, ..., n-1, n
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while i <= n {
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res += i;
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i += 1; // 更新条件变量
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}
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res
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}
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```
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=== "C"
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```c title="iteration.c"
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/* while 循环 */
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int whileLoop(int n) {
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int res = 0;
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int i = 1; // 初始化条件变量
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// 循环求和 1, 2, ..., n-1, n
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while (i <= n) {
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res += i;
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i++; // 更新条件变量
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}
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return res;
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}
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```
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=== "Zig"
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```zig title="iteration.zig"
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// while 循环
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fn whileLoop(n: i32) i32 {
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var res: i32 = 0;
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var i: i32 = 1; // 初始化条件变量
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// 循环求和 1, 2, ..., n-1, n
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while (i <= n) {
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res += @intCast(i);
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i += 1;
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}
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return res;
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}
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```
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??? pythontutor "可视化运行"
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<iframe width="800" height="600" frameborder="0" src="https://pythontutor.com/iframe-embed.html#code=def%20while_loop%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22while%20%E5%BE%AA%E7%8E%AF%22%22%22%0A%20%20%20%20res%20%3D%200%0A%20%20%20%20i%20%3D%201%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E6%9D%A1%E4%BB%B6%E5%8F%98%E9%87%8F%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E6%B1%82%E5%92%8C%201,%202,%20...,%20n-1,%20n%0A%20%20%20%20while%20i%20%3C%3D%20n%3A%0A%20%20%20%20%20%20%20%20res%20%2B%3D%20i%0A%20%20%20%20%20%20%20%20i%20%2B%3D%201%20%20%23%20%E6%9B%B4%E6%96%B0%E6%9D%A1%E4%BB%B6%E5%8F%98%E9%87%8F%0A%20%20%20%20return%20res%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%205%0A%20%20%20%20res%20%3D%20while_loop%28n%29%0A%20%20%20%20print%28f%22%5Cnwhile%20%E5%BE%AA%E7%8E%AF%E7%9A%84%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=370&codeDivWidth=300&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe>
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<a href=https://pythontutor.com/iframe-embed.html#code=def%20while_loop%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22while%20%E5%BE%AA%E7%8E%AF%22%22%22%0A%20%20%20%20res%20%3D%200%0A%20%20%20%20i%20%3D%201%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E6%9D%A1%E4%BB%B6%E5%8F%98%E9%87%8F%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E6%B1%82%E5%92%8C%201,%202,%20...,%20n-1,%20n%0A%20%20%20%20while%20i%20%3C%3D%20n%3A%0A%20%20%20%20%20%20%20%20res%20%2B%3D%20i%0A%20%20%20%20%20%20%20%20i%20%2B%3D%201%20%20%23%20%E6%9B%B4%E6%96%B0%E6%9D%A1%E4%BB%B6%E5%8F%98%E9%87%8F%0A%20%20%20%20return%20res%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%205%0A%20%20%20%20res%20%3D%20while_loop%28n%29%0A%20%20%20%20print%28f%22%5Cnwhile%20%E5%BE%AA%E7%8E%AF%E7%9A%84%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=370&codeDivWidth=300&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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**`while` 循环比 `for` 循环的自由度更高**。在 `while` 循环中，我们可以自由地设计条件变量的初始化和更新步骤。
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例如在以下代码中，条件变量 $i$ 每轮进行两次更新，这种情况就不太方便用 `for` 循环实现：
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=== "Python"
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```python title="iteration.py"
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def while_loop_ii(n: int) -> int:
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"""while 循环（两次更新）"""
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res = 0
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i = 1 # 初始化条件变量
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# 循环求和 1, 4, 10, ...
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while i <= n:
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res += i
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# 更新条件变量
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i += 1
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i *= 2
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return res
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```
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=== "C++"
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```cpp title="iteration.cpp"
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/* while 循环（两次更新） */
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int whileLoopII(int n) {
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int res = 0;
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int i = 1; // 初始化条件变量
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// 循环求和 1, 4, 10, ...
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while (i <= n) {
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res += i;
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// 更新条件变量
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i++;
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i *= 2;
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}
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return res;
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}
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```
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=== "Java"
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```java title="iteration.java"
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/* while 循环（两次更新） */
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int whileLoopII(int n) {
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int res = 0;
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int i = 1; // 初始化条件变量
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// 循环求和 1, 4, 10, ...
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while (i <= n) {
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res += i;
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// 更新条件变量
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i++;
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i *= 2;
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}
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return res;
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}
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```
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=== "C#"
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```csharp title="iteration.cs"
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/* while 循环（两次更新） */
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int WhileLoopII(int n) {
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int res = 0;
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int i = 1; // 初始化条件变量
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// 循环求和 1, 2, 4, 5...
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while (i <= n) {
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res += i;
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// 更新条件变量
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i += 1;
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i *= 2;
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}
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return res;
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}
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```
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=== "Go"
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```go title="iteration.go"
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/* while 循环（两次更新） */
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func whileLoopII(n int) int {
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res := 0
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// 初始化条件变量
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i := 1
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// 循环求和 1, 4, 10, ...
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for i <= n {
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res += i
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// 更新条件变量
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i++
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i *= 2
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}
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return res
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}
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```
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|
=== "Swift"
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```swift title="iteration.swift"
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/* while 循环（两次更新） */
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func whileLoopII(n: Int) -> Int {
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var res = 0
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var i = 1 // 初始化条件变量
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|
// 循环求和 1, 4, 10, ...
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|
while i <= n {
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res += i
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// 更新条件变量
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i += 1
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|
i *= 2
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}
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return res
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}
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```
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|
=== "JS"
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```javascript title="iteration.js"
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/* while 循环（两次更新） */
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function whileLoopII(n) {
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let res = 0;
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let i = 1; // 初始化条件变量
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|
// 循环求和 1, 4, 10, ...
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while (i <= n) {
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res += i;
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// 更新条件变量
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|
i++;
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i *= 2;
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|
}
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|
return res;
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}
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|
```
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|
=== "TS"
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|
|
```typescript title="iteration.ts"
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/* while 循环（两次更新） */
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function whileLoopII(n: number): number {
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let res = 0;
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let i = 1; // 初始化条件变量
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|
// 循环求和 1, 4, 10, ...
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|
while (i <= n) {
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res += i;
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|
// 更新条件变量
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|
i++;
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|
i *= 2;
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|
}
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|
|
return res;
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|
}
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|
```
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|
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|
|
=== "Dart"
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|
|
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|
|
```dart title="iteration.dart"
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|
|
/* while 循环（两次更新） */
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|
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int whileLoopII(int n) {
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int res = 0;
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|
int i = 1; // 初始化条件变量
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|
// 循环求和 1, 4, 10, ...
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|
|
while (i <= n) {
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|
res += i;
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|
// 更新条件变量
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|
i++;
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|
i *= 2;
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|
}
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return res;
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|
}
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|
```
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|
|
=== "Rust"
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|
|
```rust title="iteration.rs"
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/* while 循环（两次更新） */
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fn while_loop_ii(n: i32) -> i32 {
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|
let mut res = 0;
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|
let mut i = 1; // 初始化条件变量
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|
|
// 循环求和 1, 4, 10, ...
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|
while i <= n {
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|
res += i;
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|
// 更新条件变量
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|
|
i += 1;
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|
|
i *= 2;
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|
}
|
|
|
res
|
|
|
}
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|
|
```
|
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|
|
=== "C"
|
|
|
|
|
|
```c title="iteration.c"
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|
|
/* while 循环（两次更新） */
|
|
|
int whileLoopII(int n) {
|
|
|
int res = 0;
|
|
|
int i = 1; // 初始化条件变量
|
|
|
// 循环求和 1, 4, 10, ...
|
|
|
while (i <= n) {
|
|
|
res += i;
|
|
|
// 更新条件变量
|
|
|
i++;
|
|
|
i *= 2;
|
|
|
}
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
```zig title="iteration.zig"
|
|
|
// while 循环（两次更新）
|
|
|
fn whileLoopII(n: i32) i32 {
|
|
|
var res: i32 = 0;
|
|
|
var i: i32 = 1; // 初始化条件变量
|
|
|
// 循环求和 1, 4, 10, ...
|
|
|
while (i <= n) {
|
|
|
res += @intCast(i);
|
|
|
// 更新条件变量
|
|
|
i += 1;
|
|
|
i *= 2;
|
|
|
}
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
??? pythontutor "可视化运行"
|
|
|
|
|
|
<iframe width="800" height="600" frameborder="0" src="https://pythontutor.com/iframe-embed.html#code=def%20while_loop_ii%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22while%20%E5%BE%AA%E7%8E%AF%EF%BC%88%E4%B8%A4%E6%AC%A1%E6%9B%B4%E6%96%B0%EF%BC%89%22%22%22%0A%20%20%20%20res%20%3D%200%0A%20%20%20%20i%20%3D%201%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E6%9D%A1%E4%BB%B6%E5%8F%98%E9%87%8F%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E6%B1%82%E5%92%8C%201,%204,%2010,%20...%0A%20%20%20%20while%20i%20%3C%3D%20n%3A%0A%20%20%20%20%20%20%20%20res%20%2B%3D%20i%0A%20%20%20%20%20%20%20%20%23%20%E6%9B%B4%E6%96%B0%E6%9D%A1%E4%BB%B6%E5%8F%98%E9%87%8F%0A%20%20%20%20%20%20%20%20i%20%2B%3D%201%0A%20%20%20%20%20%20%20%20i%20*%3D%202%0A%20%20%20%20return%20res%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%205%0A%20%20%20%20res%20%3D%20while_loop_ii%28n%29%0A%20%20%20%20print%28f%22%5Cnwhile%20%E5%BE%AA%E7%8E%AF%EF%BC%88%E4%B8%A4%E6%AC%A1%E6%9B%B4%E6%96%B0%EF%BC%89%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=370&codeDivWidth=300&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe>
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|
<a href=https://pythontutor.com/iframe-embed.html#code=def%20while_loop_ii%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22while%20%E5%BE%AA%E7%8E%AF%EF%BC%88%E4%B8%A4%E6%AC%A1%E6%9B%B4%E6%96%B0%EF%BC%89%22%22%22%0A%20%20%20%20res%20%3D%200%0A%20%20%20%20i%20%3D%201%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E6%9D%A1%E4%BB%B6%E5%8F%98%E9%87%8F%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E6%B1%82%E5%92%8C%201,%204,%2010,%20...%0A%20%20%20%20while%20i%20%3C%3D%20n%3A%0A%20%20%20%20%20%20%20%20res%20%2B%3D%20i%0A%20%20%20%20%20%20%20%20%23%20%E6%9B%B4%E6%96%B0%E6%9D%A1%E4%BB%B6%E5%8F%98%E9%87%8F%0A%20%20%20%20%20%20%20%20i%20%2B%3D%201%0A%20%20%20%20%20%20%20%20i%20*%3D%202%0A%20%20%20%20return%20res%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%205%0A%20%20%20%20res%20%3D%20while_loop_ii%28n%29%0A%20%20%20%20print%28f%22%5Cnwhile%20%E5%BE%AA%E7%8E%AF%EF%BC%88%E4%B8%A4%E6%AC%A1%E6%9B%B4%E6%96%B0%EF%BC%89%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=370&codeDivWidth=300&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>
|
|
|
|
|
|
总的来说，**`for` 循环的代码更加紧凑，`while` 循环更加灵活**，两者都可以实现迭代结构。选择使用哪一个应该根据特定问题的需求来决定。
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|
|
|
### 3. 嵌套循环
|
|
|
|
|
|
我们可以在一个循环结构内嵌套另一个循环结构，下面以 `for` 循环为例：
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
```python title="iteration.py"
|
|
|
def nested_for_loop(n: int) -> str:
|
|
|
"""双层 for 循环"""
|
|
|
res = ""
|
|
|
# 循环 i = 1, 2, ..., n-1, n
|
|
|
for i in range(1, n + 1):
|
|
|
# 循环 j = 1, 2, ..., n-1, n
|
|
|
for j in range(1, n + 1):
|
|
|
res += f"({i}, {j}), "
|
|
|
return res
|
|
|
```
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
```cpp title="iteration.cpp"
|
|
|
/* 双层 for 循环 */
|
|
|
string nestedForLoop(int n) {
|
|
|
ostringstream res;
|
|
|
// 循环 i = 1, 2, ..., n-1, n
|
|
|
for (int i = 1; i <= n; ++i) {
|
|
|
// 循环 j = 1, 2, ..., n-1, n
|
|
|
for (int j = 1; j <= n; ++j) {
|
|
|
res << "(" << i << ", " << j << "), ";
|
|
|
}
|
|
|
}
|
|
|
return res.str();
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
```java title="iteration.java"
|
|
|
/* 双层 for 循环 */
|
|
|
String nestedForLoop(int n) {
|
|
|
StringBuilder res = new StringBuilder();
|
|
|
// 循环 i = 1, 2, ..., n-1, n
|
|
|
for (int i = 1; i <= n; i++) {
|
|
|
// 循环 j = 1, 2, ..., n-1, n
|
|
|
for (int j = 1; j <= n; j++) {
|
|
|
res.append("(" + i + ", " + j + "), ");
|
|
|
}
|
|
|
}
|
|
|
return res.toString();
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
```csharp title="iteration.cs"
|
|
|
/* 双层 for 循环 */
|
|
|
string NestedForLoop(int n) {
|
|
|
StringBuilder res = new();
|
|
|
// 循环 i = 1, 2, ..., n-1, n
|
|
|
for (int i = 1; i <= n; i++) {
|
|
|
// 循环 j = 1, 2, ..., n-1, n
|
|
|
for (int j = 1; j <= n; j++) {
|
|
|
res.Append($"({i}, {j}), ");
|
|
|
}
|
|
|
}
|
|
|
return res.ToString();
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
```go title="iteration.go"
|
|
|
/* 双层 for 循环 */
|
|
|
func nestedForLoop(n int) string {
|
|
|
res := ""
|
|
|
// 循环 i = 1, 2, ..., n-1, n
|
|
|
for i := 1; i <= n; i++ {
|
|
|
for j := 1; j <= n; j++ {
|
|
|
// 循环 j = 1, 2, ..., n-1, n
|
|
|
res += fmt.Sprintf("(%d, %d), ", i, j)
|
|
|
}
|
|
|
}
|
|
|
return res
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
```swift title="iteration.swift"
|
|
|
/* 双层 for 循环 */
|
|
|
func nestedForLoop(n: Int) -> String {
|
|
|
var res = ""
|
|
|
// 循环 i = 1, 2, ..., n-1, n
|
|
|
for i in 1 ... n {
|
|
|
// 循环 j = 1, 2, ..., n-1, n
|
|
|
for j in 1 ... n {
|
|
|
res.append("(\(i), \(j)), ")
|
|
|
}
|
|
|
}
|
|
|
return res
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
```javascript title="iteration.js"
|
|
|
/* 双层 for 循环 */
|
|
|
function nestedForLoop(n) {
|
|
|
let res = '';
|
|
|
// 循环 i = 1, 2, ..., n-1, n
|
|
|
for (let i = 1; i <= n; i++) {
|
|
|
// 循环 j = 1, 2, ..., n-1, n
|
|
|
for (let j = 1; j <= n; j++) {
|
|
|
res += `(${i}, ${j}), `;
|
|
|
}
|
|
|
}
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
```typescript title="iteration.ts"
|
|
|
/* 双层 for 循环 */
|
|
|
function nestedForLoop(n: number): string {
|
|
|
let res = '';
|
|
|
// 循环 i = 1, 2, ..., n-1, n
|
|
|
for (let i = 1; i <= n; i++) {
|
|
|
// 循环 j = 1, 2, ..., n-1, n
|
|
|
for (let j = 1; j <= n; j++) {
|
|
|
res += `(${i}, ${j}), `;
|
|
|
}
|
|
|
}
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
```dart title="iteration.dart"
|
|
|
/* 双层 for 循环 */
|
|
|
String nestedForLoop(int n) {
|
|
|
String res = "";
|
|
|
// 循环 i = 1, 2, ..., n-1, n
|
|
|
for (int i = 1; i <= n; i++) {
|
|
|
// 循环 j = 1, 2, ..., n-1, n
|
|
|
for (int j = 1; j <= n; j++) {
|
|
|
res += "($i, $j), ";
|
|
|
}
|
|
|
}
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
```rust title="iteration.rs"
|
|
|
/* 双层 for 循环 */
|
|
|
fn nested_for_loop(n: i32) -> String {
|
|
|
let mut res = vec![];
|
|
|
// 循环 i = 1, 2, ..., n-1, n
|
|
|
for i in 1..=n {
|
|
|
// 循环 j = 1, 2, ..., n-1, n
|
|
|
for j in 1..=n {
|
|
|
res.push(format!("({}, {}), ", i, j));
|
|
|
}
|
|
|
}
|
|
|
res.join("")
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
```c title="iteration.c"
|
|
|
/* 双层 for 循环 */
|
|
|
char *nestedForLoop(int n) {
|
|
|
// n * n 为对应点数量，"(i, j), " 对应字符串长最大为 6+10*2，加上最后一个空字符 \0 的额外空间
|
|
|
int size = n * n * 26 + 1;
|
|
|
char *res = malloc(size * sizeof(char));
|
|
|
// 循环 i = 1, 2, ..., n-1, n
|
|
|
for (int i = 1; i <= n; i++) {
|
|
|
// 循环 j = 1, 2, ..., n-1, n
|
|
|
for (int j = 1; j <= n; j++) {
|
|
|
char tmp[26];
|
|
|
snprintf(tmp, sizeof(tmp), "(%d, %d), ", i, j);
|
|
|
strncat(res, tmp, size - strlen(res) - 1);
|
|
|
}
|
|
|
}
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
```zig title="iteration.zig"
|
|
|
// 双层 for 循环
|
|
|
fn nestedForLoop(allocator: Allocator, n: usize) ![]const u8 {
|
|
|
var res = std.ArrayList(u8).init(allocator);
|
|
|
defer res.deinit();
|
|
|
var buffer: [20]u8 = undefined;
|
|
|
// 循环 i = 1, 2, ..., n-1, n
|
|
|
for (1..n+1) |i| {
|
|
|
// 循环 j = 1, 2, ..., n-1, n
|
|
|
for (1..n+1) |j| {
|
|
|
var _str = try std.fmt.bufPrint(&buffer, "({d}, {d}), ", .{i, j});
|
|
|
try res.appendSlice(_str);
|
|
|
}
|
|
|
}
|
|
|
return res.toOwnedSlice();
|
|
|
}
|
|
|
```
|
|
|
|
|
|
??? pythontutor "可视化运行"
|
|
|
|
|
|
<iframe width="800" height="600" frameborder="0" src="https://pythontutor.com/iframe-embed.html#code=def%20nested_for_loop%28n%3A%20int%29%20-%3E%20str%3A%0A%20%20%20%20%22%22%22%E5%8F%8C%E5%B1%82%20for%20%E5%BE%AA%E7%8E%AF%22%22%22%0A%20%20%20%20res%20%3D%20%22%22%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%20i%20%3D%201,%202,%20...,%20n-1,%20n%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%20j%20%3D%201,%202,%20...,%20n-1,%20n%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20res%20%2B%3D%20f%22%28%7Bi%7D,%20%7Bj%7D%29,%20%22%0A%20%20%20%20return%20res%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%205%0A%20%20%20%20res%20%3D%20nested_for_loop%28n%29%0A%20%20%20%20print%28f%22%5Cn%E5%8F%8C%E5%B1%82%20for%20%E5%BE%AA%E7%8E%AF%E7%9A%84%E9%81%8D%E5%8E%86%E7%BB%93%E6%9E%9C%20%7Bres%7D%22%29&codeDivHeight=370&codeDivWidth=300&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe>
|
|
|
<a href=https://pythontutor.com/iframe-embed.html#code=def%20nested_for_loop%28n%3A%20int%29%20-%3E%20str%3A%0A%20%20%20%20%22%22%22%E5%8F%8C%E5%B1%82%20for%20%E5%BE%AA%E7%8E%AF%22%22%22%0A%20%20%20%20res%20%3D%20%22%22%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%20i%20%3D%201,%202,%20...,%20n-1,%20n%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%20j%20%3D%201,%202,%20...,%20n-1,%20n%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20res%20%2B%3D%20f%22%28%7Bi%7D,%20%7Bj%7D%29,%20%22%0A%20%20%20%20return%20res%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%205%0A%20%20%20%20res%20%3D%20nested_for_loop%28n%29%0A%20%20%20%20print%28f%22%5Cn%E5%8F%8C%E5%B1%82%20for%20%E5%BE%AA%E7%8E%AF%E7%9A%84%E9%81%8D%E5%8E%86%E7%BB%93%E6%9E%9C%20%7Bres%7D%22%29&codeDivHeight=370&codeDivWidth=300&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>
|
|
|
|
|
|
图 2-2 是该嵌套循环的流程框图。
|
|
|
|
|
|
{ class="animation-figure" }
|
|
|
|
|
|
<p align="center"> 图 2-2 嵌套循环的流程框图 </p>
|
|
|
|
|
|
在这种情况下，函数的操作数量与 $n^2$ 成正比，或者说算法运行时间和输入数据大小 $n$ 成“平方关系”。
|
|
|
|
|
|
我们可以继续添加嵌套循环，每一次嵌套都是一次“升维”，将会使时间复杂度提高至“立方关系”“四次方关系”，以此类推。
|
|
|
|
|
|
## 2.2.2 递归
|
|
|
|
|
|
「递归 recursion」是一种算法策略，通过函数调用自身来解决问题。它主要包含两个阶段。
|
|
|
|
|
|
1. **递**：程序不断深入地调用自身，通常传入更小或更简化的参数，直到达到“终止条件”。
|
|
|
2. **归**：触发“终止条件”后，程序从最深层的递归函数开始逐层返回，汇聚每一层的结果。
|
|
|
|
|
|
而从实现的角度看，递归代码主要包含三个要素。
|
|
|
|
|
|
1. **终止条件**：用于决定什么时候由“递”转“归”。
|
|
|
2. **递归调用**：对应“递”，函数调用自身，通常输入更小或更简化的参数。
|
|
|
3. **返回结果**：对应“归”，将当前递归层级的结果返回至上一层。
|
|
|
|
|
|
观察以下代码，我们只需调用函数 `recur(n)` ，就可以完成 $1 + 2 + \dots + n$ 的计算：
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
```python title="recursion.py"
|
|
|
def recur(n: int) -> int:
|
|
|
"""递归"""
|
|
|
# 终止条件
|
|
|
if n == 1:
|
|
|
return 1
|
|
|
# 递：递归调用
|
|
|
res = recur(n - 1)
|
|
|
# 归：返回结果
|
|
|
return n + res
|
|
|
```
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
```cpp title="recursion.cpp"
|
|
|
/* 递归 */
|
|
|
int recur(int n) {
|
|
|
// 终止条件
|
|
|
if (n == 1)
|
|
|
return 1;
|
|
|
// 递：递归调用
|
|
|
int res = recur(n - 1);
|
|
|
// 归：返回结果
|
|
|
return n + res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
```java title="recursion.java"
|
|
|
/* 递归 */
|
|
|
int recur(int n) {
|
|
|
// 终止条件
|
|
|
if (n == 1)
|
|
|
return 1;
|
|
|
// 递：递归调用
|
|
|
int res = recur(n - 1);
|
|
|
// 归：返回结果
|
|
|
return n + res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
```csharp title="recursion.cs"
|
|
|
/* 递归 */
|
|
|
int Recur(int n) {
|
|
|
// 终止条件
|
|
|
if (n == 1)
|
|
|
return 1;
|
|
|
// 递：递归调用
|
|
|
int res = Recur(n - 1);
|
|
|
// 归：返回结果
|
|
|
return n + res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
```go title="recursion.go"
|
|
|
/* 递归 */
|
|
|
func recur(n int) int {
|
|
|
// 终止条件
|
|
|
if n == 1 {
|
|
|
return 1
|
|
|
}
|
|
|
// 递：递归调用
|
|
|
res := recur(n - 1)
|
|
|
// 归：返回结果
|
|
|
return n + res
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
```swift title="recursion.swift"
|
|
|
/* 递归 */
|
|
|
func recur(n: Int) -> Int {
|
|
|
// 终止条件
|
|
|
if n == 1 {
|
|
|
return 1
|
|
|
}
|
|
|
// 递：递归调用
|
|
|
let res = recur(n: n - 1)
|
|
|
// 归：返回结果
|
|
|
return n + res
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
```javascript title="recursion.js"
|
|
|
/* 递归 */
|
|
|
function recur(n) {
|
|
|
// 终止条件
|
|
|
if (n === 1) return 1;
|
|
|
// 递：递归调用
|
|
|
const res = recur(n - 1);
|
|
|
// 归：返回结果
|
|
|
return n + res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
```typescript title="recursion.ts"
|
|
|
/* 递归 */
|
|
|
function recur(n: number): number {
|
|
|
// 终止条件
|
|
|
if (n === 1) return 1;
|
|
|
// 递：递归调用
|
|
|
const res = recur(n - 1);
|
|
|
// 归：返回结果
|
|
|
return n + res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
```dart title="recursion.dart"
|
|
|
/* 递归 */
|
|
|
int recur(int n) {
|
|
|
// 终止条件
|
|
|
if (n == 1) return 1;
|
|
|
// 递：递归调用
|
|
|
int res = recur(n - 1);
|
|
|
// 归：返回结果
|
|
|
return n + res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
```rust title="recursion.rs"
|
|
|
/* 递归 */
|
|
|
fn recur(n: i32) -> i32 {
|
|
|
// 终止条件
|
|
|
if n == 1 {
|
|
|
return 1;
|
|
|
}
|
|
|
// 递：递归调用
|
|
|
let res = recur(n - 1);
|
|
|
// 归：返回结果
|
|
|
n + res
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
```c title="recursion.c"
|
|
|
/* 递归 */
|
|
|
int recur(int n) {
|
|
|
// 终止条件
|
|
|
if (n == 1)
|
|
|
return 1;
|
|
|
// 递：递归调用
|
|
|
int res = recur(n - 1);
|
|
|
// 归：返回结果
|
|
|
return n + res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
```zig title="recursion.zig"
|
|
|
// 递归函数
|
|
|
fn recur(n: i32) i32 {
|
|
|
// 终止条件
|
|
|
if (n == 1) {
|
|
|
return 1;
|
|
|
}
|
|
|
// 递：递归调用
|
|
|
var res: i32 = recur(n - 1);
|
|
|
// 归：返回结果
|
|
|
return n + res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
??? pythontutor "可视化运行"
|
|
|
|
|
|
<iframe width="800" height="600" frameborder="0" src="https://pythontutor.com/iframe-embed.html#code=def%20recur%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E9%80%92%E5%BD%92%22%22%22%0A%20%20%20%20%23%20%E7%BB%88%E6%AD%A2%E6%9D%A1%E4%BB%B6%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%20%23%20%E9%80%92%EF%BC%9A%E9%80%92%E5%BD%92%E8%B0%83%E7%94%A8%0A%20%20%20%20res%20%3D%20recur%28n%20-%201%29%0A%20%20%20%20%23%20%E5%BD%92%EF%BC%9A%E8%BF%94%E5%9B%9E%E7%BB%93%E6%9E%9C%0A%20%20%20%20return%20n%20%2B%20res%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%205%0A%20%20%20%20res%20%3D%20recur%28n%29%0A%20%20%20%20print%28f%22%5Cn%E9%80%92%E5%BD%92%E5%87%BD%E6%95%B0%E7%9A%84%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=370&codeDivWidth=300&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe>
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<a href=https://pythontutor.com/iframe-embed.html#code=def%20recur%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E9%80%92%E5%BD%92%22%22%22%0A%20%20%20%20%23%20%E7%BB%88%E6%AD%A2%E6%9D%A1%E4%BB%B6%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%20%23%20%E9%80%92%EF%BC%9A%E9%80%92%E5%BD%92%E8%B0%83%E7%94%A8%0A%20%20%20%20res%20%3D%20recur%28n%20-%201%29%0A%20%20%20%20%23%20%E5%BD%92%EF%BC%9A%E8%BF%94%E5%9B%9E%E7%BB%93%E6%9E%9C%0A%20%20%20%20return%20n%20%2B%20res%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%205%0A%20%20%20%20res%20%3D%20recur%28n%29%0A%20%20%20%20print%28f%22%5Cn%E9%80%92%E5%BD%92%E5%87%BD%E6%95%B0%E7%9A%84%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=370&codeDivWidth=300&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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图 2-3 展示了该函数的递归过程。
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{ class="animation-figure" }
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<p align="center"> 图 2-3 求和函数的递归过程 </p>
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虽然从计算角度看，迭代与递归可以得到相同的结果，**但它们代表了两种完全不同的思考和解决问题的范式**。
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- **迭代**：“自下而上”地解决问题。从最基础的步骤开始，然后不断重复或累加这些步骤，直到任务完成。
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- **递归**：“自上而下”地解决问题。将原问题分解为更小的子问题，这些子问题和原问题具有相同的形式。接下来将子问题继续分解为更小的子问题，直到基本情况时停止（基本情况的解是已知的）。
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以上述求和函数为例，设问题 $f(n) = 1 + 2 + \dots + n$ 。
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- **迭代**：在循环中模拟求和过程，从 $1$ 遍历到 $n$ ，每轮执行求和操作，即可求得 $f(n)$ 。
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- **递归**：将问题分解为子问题 $f(n) = n + f(n-1)$ ，不断（递归地）分解下去，直至基本情况 $f(1) = 1$ 时终止。
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### 1. 调用栈
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递归函数每次调用自身时，系统都会为新开启的函数分配内存，以存储局部变量、调用地址和其他信息等。这将导致两方面的结果。
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- 函数的上下文数据都存储在称为“栈帧空间”的内存区域中，直至函数返回后才会被释放。因此，**递归通常比迭代更加耗费内存空间**。
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- 递归调用函数会产生额外的开销。**因此递归通常比循环的时间效率更低**。
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如图 2-4 所示，在触发终止条件前，同时存在 $n$ 个未返回的递归函数，**递归深度为 $n$** 。
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{ class="animation-figure" }
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<p align="center"> 图 2-4 递归调用深度 </p>
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在实际中，编程语言允许的递归深度通常是有限的，过深的递归可能导致栈溢出错误。
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### 2. 尾递归
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有趣的是，**如果函数在返回前的最后一步才进行递归调用**，则该函数可以被编译器或解释器优化，使其在空间效率上与迭代相当。这种情况被称为「尾递归 tail recursion」。
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- **普通递归**：当函数返回到上一层级的函数后，需要继续执行代码，因此系统需要保存上一层调用的上下文。
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- **尾递归**：递归调用是函数返回前的最后一个操作，这意味着函数返回到上一层级后，无须继续执行其他操作，因此系统无须保存上一层函数的上下文。
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以计算 $1 + 2 + \dots + n$ 为例，我们可以将结果变量 `res` 设为函数参数，从而实现尾递归：
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=== "Python"
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```python title="recursion.py"
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def tail_recur(n, res):
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"""尾递归"""
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# 终止条件
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if n == 0:
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return res
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# 尾递归调用
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return tail_recur(n - 1, res + n)
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```
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|
=== "C++"
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|
```cpp title="recursion.cpp"
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/* 尾递归 */
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int tailRecur(int n, int res) {
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// 终止条件
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if (n == 0)
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return res;
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// 尾递归调用
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return tailRecur(n - 1, res + n);
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}
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```
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=== "Java"
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```java title="recursion.java"
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/* 尾递归 */
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int tailRecur(int n, int res) {
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// 终止条件
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if (n == 0)
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return res;
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// 尾递归调用
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return tailRecur(n - 1, res + n);
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}
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```
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=== "C#"
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|
```csharp title="recursion.cs"
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/* 尾递归 */
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int TailRecur(int n, int res) {
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// 终止条件
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if (n == 0)
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return res;
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// 尾递归调用
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return TailRecur(n - 1, res + n);
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}
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```
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=== "Go"
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```go title="recursion.go"
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/* 尾递归 */
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func tailRecur(n int, res int) int {
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// 终止条件
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if n == 0 {
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return res
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}
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// 尾递归调用
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return tailRecur(n-1, res+n)
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}
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```
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=== "Swift"
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```swift title="recursion.swift"
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/* 尾递归 */
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func tailRecur(n: Int, res: Int) -> Int {
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// 终止条件
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if n == 0 {
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return res
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}
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// 尾递归调用
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return tailRecur(n: n - 1, res: res + n)
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}
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```
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=== "JS"
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```javascript title="recursion.js"
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/* 尾递归 */
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function tailRecur(n, res) {
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// 终止条件
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if (n === 0) return res;
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// 尾递归调用
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return tailRecur(n - 1, res + n);
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}
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```
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=== "TS"
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```typescript title="recursion.ts"
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/* 尾递归 */
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function tailRecur(n: number, res: number): number {
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// 终止条件
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if (n === 0) return res;
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// 尾递归调用
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return tailRecur(n - 1, res + n);
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}
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|
```
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|
=== "Dart"
|
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|
|
```dart title="recursion.dart"
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|
/* 尾递归 */
|
|
|
int tailRecur(int n, int res) {
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|
// 终止条件
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if (n == 0) return res;
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|
// 尾递归调用
|
|
|
return tailRecur(n - 1, res + n);
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}
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|
```
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|
|
=== "Rust"
|
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|
|
|
|
```rust title="recursion.rs"
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|
|
/* 尾递归 */
|
|
|
fn tail_recur(n: i32, res: i32) -> i32 {
|
|
|
// 终止条件
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|
if n == 0 {
|
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|
return res;
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|
}
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|
// 尾递归调用
|
|
|
tail_recur(n - 1, res + n)
|
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|
}
|
|
|
```
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|
|
=== "C"
|
|
|
|
|
|
```c title="recursion.c"
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|
|
/* 尾递归 */
|
|
|
int tailRecur(int n, int res) {
|
|
|
// 终止条件
|
|
|
if (n == 0)
|
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|
return res;
|
|
|
// 尾递归调用
|
|
|
return tailRecur(n - 1, res + n);
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
```zig title="recursion.zig"
|
|
|
// 尾递归函数
|
|
|
fn tailRecur(n: i32, res: i32) i32 {
|
|
|
// 终止条件
|
|
|
if (n == 0) {
|
|
|
return res;
|
|
|
}
|
|
|
// 尾递归调用
|
|
|
return tailRecur(n - 1, res + n);
|
|
|
}
|
|
|
```
|
|
|
|
|
|
??? pythontutor "可视化运行"
|
|
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|
|
|
<iframe width="800" height="600" frameborder="0" src="https://pythontutor.com/iframe-embed.html#code=def%20tail_recur%28n,%20res%29%3A%0A%20%20%20%20%22%22%22%E5%B0%BE%E9%80%92%E5%BD%92%22%22%22%0A%20%20%20%20%23%20%E7%BB%88%E6%AD%A2%E6%9D%A1%E4%BB%B6%0A%20%20%20%20if%20n%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20return%20res%0A%20%20%20%20%23%20%E5%B0%BE%E9%80%92%E5%BD%92%E8%B0%83%E7%94%A8%0A%20%20%20%20return%20tail_recur%28n%20-%201,%20res%20%2B%20n%29%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%205%0A%20%20%20%20res%20%3D%20tail_recur%28n,%200%29%0A%20%20%20%20print%28f%22%5Cn%E5%B0%BE%E9%80%92%E5%BD%92%E5%87%BD%E6%95%B0%E7%9A%84%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=370&codeDivWidth=300&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe>
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|
<a href=https://pythontutor.com/iframe-embed.html#code=def%20tail_recur%28n,%20res%29%3A%0A%20%20%20%20%22%22%22%E5%B0%BE%E9%80%92%E5%BD%92%22%22%22%0A%20%20%20%20%23%20%E7%BB%88%E6%AD%A2%E6%9D%A1%E4%BB%B6%0A%20%20%20%20if%20n%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20return%20res%0A%20%20%20%20%23%20%E5%B0%BE%E9%80%92%E5%BD%92%E8%B0%83%E7%94%A8%0A%20%20%20%20return%20tail_recur%28n%20-%201,%20res%20%2B%20n%29%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%205%0A%20%20%20%20res%20%3D%20tail_recur%28n,%200%29%0A%20%20%20%20print%28f%22%5Cn%E5%B0%BE%E9%80%92%E5%BD%92%E5%87%BD%E6%95%B0%E7%9A%84%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=370&codeDivWidth=300&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>
|
|
|
|
|
|
尾递归的执行过程如图 2-5 所示。对比普通递归和尾递归，两者的求和操作的执行点是不同的。
|
|
|
|
|
|
- **普通递归**：求和操作是在“归”的过程中执行的，每层返回后都要再执行一次求和操作。
|
|
|
- **尾递归**：求和操作是在“递”的过程中执行的，“归”的过程只需层层返回。
|
|
|
|
|
|
{ class="animation-figure" }
|
|
|
|
|
|
<p align="center"> 图 2-5 尾递归过程 </p>
|
|
|
|
|
|
!!! tip
|
|
|
|
|
|
请注意，许多编译器或解释器并不支持尾递归优化。例如，Python 默认不支持尾递归优化，因此即使函数是尾递归形式，仍然可能会遇到栈溢出问题。
|
|
|
|
|
|
### 3. 递归树
|
|
|
|
|
|
当处理与“分治”相关的算法问题时，递归往往比迭代的思路更加直观、代码更加易读。以“斐波那契数列”为例。
|
|
|
|
|
|
!!! question
|
|
|
|
|
|
给定一个斐波那契数列 $0, 1, 1, 2, 3, 5, 8, 13, \dots$ ，求该数列的第 $n$ 个数字。
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|
|
|
|
设斐波那契数列的第 $n$ 个数字为 $f(n)$ ，易得两个结论。
|
|
|
|
|
|
- 数列的前两个数字为 $f(1) = 0$ 和 $f(2) = 1$ 。
|
|
|
- 数列中的每个数字是前两个数字的和，即 $f(n) = f(n - 1) + f(n - 2)$ 。
|
|
|
|
|
|
按照递推关系进行递归调用，将前两个数字作为终止条件，便可写出递归代码。调用 `fib(n)` 即可得到斐波那契数列的第 $n$ 个数字：
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
```python title="recursion.py"
|
|
|
def fib(n: int) -> int:
|
|
|
"""斐波那契数列：递归"""
|
|
|
# 终止条件 f(1) = 0, f(2) = 1
|
|
|
if n == 1 or n == 2:
|
|
|
return n - 1
|
|
|
# 递归调用 f(n) = f(n-1) + f(n-2)
|
|
|
res = fib(n - 1) + fib(n - 2)
|
|
|
# 返回结果 f(n)
|
|
|
return res
|
|
|
```
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
```cpp title="recursion.cpp"
|
|
|
/* 斐波那契数列：递归 */
|
|
|
int fib(int n) {
|
|
|
// 终止条件 f(1) = 0, f(2) = 1
|
|
|
if (n == 1 || n == 2)
|
|
|
return n - 1;
|
|
|
// 递归调用 f(n) = f(n-1) + f(n-2)
|
|
|
int res = fib(n - 1) + fib(n - 2);
|
|
|
// 返回结果 f(n)
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
```java title="recursion.java"
|
|
|
/* 斐波那契数列：递归 */
|
|
|
int fib(int n) {
|
|
|
// 终止条件 f(1) = 0, f(2) = 1
|
|
|
if (n == 1 || n == 2)
|
|
|
return n - 1;
|
|
|
// 递归调用 f(n) = f(n-1) + f(n-2)
|
|
|
int res = fib(n - 1) + fib(n - 2);
|
|
|
// 返回结果 f(n)
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
```csharp title="recursion.cs"
|
|
|
/* 斐波那契数列：递归 */
|
|
|
int Fib(int n) {
|
|
|
// 终止条件 f(1) = 0, f(2) = 1
|
|
|
if (n == 1 || n == 2)
|
|
|
return n - 1;
|
|
|
// 递归调用 f(n) = f(n-1) + f(n-2)
|
|
|
int res = Fib(n - 1) + Fib(n - 2);
|
|
|
// 返回结果 f(n)
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
```go title="recursion.go"
|
|
|
/* 斐波那契数列：递归 */
|
|
|
func fib(n int) int {
|
|
|
// 终止条件 f(1) = 0, f(2) = 1
|
|
|
if n == 1 || n == 2 {
|
|
|
return n - 1
|
|
|
}
|
|
|
// 递归调用 f(n) = f(n-1) + f(n-2)
|
|
|
res := fib(n-1) + fib(n-2)
|
|
|
// 返回结果 f(n)
|
|
|
return res
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
```swift title="recursion.swift"
|
|
|
/* 斐波那契数列：递归 */
|
|
|
func fib(n: Int) -> Int {
|
|
|
// 终止条件 f(1) = 0, f(2) = 1
|
|
|
if n == 1 || n == 2 {
|
|
|
return n - 1
|
|
|
}
|
|
|
// 递归调用 f(n) = f(n-1) + f(n-2)
|
|
|
let res = fib(n: n - 1) + fib(n: n - 2)
|
|
|
// 返回结果 f(n)
|
|
|
return res
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
```javascript title="recursion.js"
|
|
|
/* 斐波那契数列：递归 */
|
|
|
function fib(n) {
|
|
|
// 终止条件 f(1) = 0, f(2) = 1
|
|
|
if (n === 1 || n === 2) return n - 1;
|
|
|
// 递归调用 f(n) = f(n-1) + f(n-2)
|
|
|
const res = fib(n - 1) + fib(n - 2);
|
|
|
// 返回结果 f(n)
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
```typescript title="recursion.ts"
|
|
|
/* 斐波那契数列：递归 */
|
|
|
function fib(n: number): number {
|
|
|
// 终止条件 f(1) = 0, f(2) = 1
|
|
|
if (n === 1 || n === 2) return n - 1;
|
|
|
// 递归调用 f(n) = f(n-1) + f(n-2)
|
|
|
const res = fib(n - 1) + fib(n - 2);
|
|
|
// 返回结果 f(n)
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
```dart title="recursion.dart"
|
|
|
/* 斐波那契数列：递归 */
|
|
|
int fib(int n) {
|
|
|
// 终止条件 f(1) = 0, f(2) = 1
|
|
|
if (n == 1 || n == 2) return n - 1;
|
|
|
// 递归调用 f(n) = f(n-1) + f(n-2)
|
|
|
int res = fib(n - 1) + fib(n - 2);
|
|
|
// 返回结果 f(n)
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
```rust title="recursion.rs"
|
|
|
/* 斐波那契数列：递归 */
|
|
|
fn fib(n: i32) -> i32 {
|
|
|
// 终止条件 f(1) = 0, f(2) = 1
|
|
|
if n == 1 || n == 2 {
|
|
|
return n - 1;
|
|
|
}
|
|
|
// 递归调用 f(n) = f(n-1) + f(n-2)
|
|
|
let res = fib(n - 1) + fib(n - 2);
|
|
|
// 返回结果
|
|
|
res
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
```c title="recursion.c"
|
|
|
/* 斐波那契数列：递归 */
|
|
|
int fib(int n) {
|
|
|
// 终止条件 f(1) = 0, f(2) = 1
|
|
|
if (n == 1 || n == 2)
|
|
|
return n - 1;
|
|
|
// 递归调用 f(n) = f(n-1) + f(n-2)
|
|
|
int res = fib(n - 1) + fib(n - 2);
|
|
|
// 返回结果 f(n)
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
```zig title="recursion.zig"
|
|
|
// 斐波那契数列
|
|
|
fn fib(n: i32) i32 {
|
|
|
// 终止条件 f(1) = 0, f(2) = 1
|
|
|
if (n == 1 or n == 2) {
|
|
|
return n - 1;
|
|
|
}
|
|
|
// 递归调用 f(n) = f(n-1) + f(n-2)
|
|
|
var res: i32 = fib(n - 1) + fib(n - 2);
|
|
|
// 返回结果 f(n)
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
??? pythontutor "可视化运行"
|
|
|
|
|
|
<iframe width="800" height="600" frameborder="0" src="https://pythontutor.com/iframe-embed.html#code=def%20fib%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%96%90%E6%B3%A2%E9%82%A3%E5%A5%91%E6%95%B0%E5%88%97%EF%BC%9A%E9%80%92%E5%BD%92%22%22%22%0A%20%20%20%20%23%20%E7%BB%88%E6%AD%A2%E6%9D%A1%E4%BB%B6%20f%281%29%20%3D%200,%20f%282%29%20%3D%201%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20n%20-%201%0A%20%20%20%20%23%20%E9%80%92%E5%BD%92%E8%B0%83%E7%94%A8%20f%28n%29%20%3D%20f%28n-1%29%20%2B%20f%28n-2%29%0A%20%20%20%20res%20%3D%20fib%28n%20-%201%29%20%2B%20fib%28n%20-%202%29%0A%20%20%20%20%23%20%E8%BF%94%E5%9B%9E%E7%BB%93%E6%9E%9C%20f%28n%29%0A%20%20%20%20return%20res%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%205%0A%20%20%20%20res%20%3D%20fib%28n%29%0A%20%20%20%20print%28f%22%5Cn%E6%96%90%E6%B3%A2%E9%82%A3%E5%A5%91%E6%95%B0%E5%88%97%E7%9A%84%E7%AC%AC%20%7Bn%7D%20%E9%A1%B9%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=370&codeDivWidth=300&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe>
|
|
|
<a href=https://pythontutor.com/iframe-embed.html#code=def%20fib%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%96%90%E6%B3%A2%E9%82%A3%E5%A5%91%E6%95%B0%E5%88%97%EF%BC%9A%E9%80%92%E5%BD%92%22%22%22%0A%20%20%20%20%23%20%E7%BB%88%E6%AD%A2%E6%9D%A1%E4%BB%B6%20f%281%29%20%3D%200,%20f%282%29%20%3D%201%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20n%20-%201%0A%20%20%20%20%23%20%E9%80%92%E5%BD%92%E8%B0%83%E7%94%A8%20f%28n%29%20%3D%20f%28n-1%29%20%2B%20f%28n-2%29%0A%20%20%20%20res%20%3D%20fib%28n%20-%201%29%20%2B%20fib%28n%20-%202%29%0A%20%20%20%20%23%20%E8%BF%94%E5%9B%9E%E7%BB%93%E6%9E%9C%20f%28n%29%0A%20%20%20%20return%20res%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%205%0A%20%20%20%20res%20%3D%20fib%28n%29%0A%20%20%20%20print%28f%22%5Cn%E6%96%90%E6%B3%A2%E9%82%A3%E5%A5%91%E6%95%B0%E5%88%97%E7%9A%84%E7%AC%AC%20%7Bn%7D%20%E9%A1%B9%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=370&codeDivWidth=300&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>
|
|
|
|
|
|
观察以上代码，我们在函数内递归调用了两个函数，**这意味着从一个调用产生了两个调用分支**。如图 2-6 所示，这样不断递归调用下去，最终将产生一棵层数为 $n$ 的「递归树 recursion tree」。
|
|
|
|
|
|
{ class="animation-figure" }
|
|
|
|
|
|
<p align="center"> 图 2-6 斐波那契数列的递归树 </p>
|
|
|
|
|
|
从本质上看，递归体现了“将问题分解为更小子问题”的思维范式，这种分治策略至关重要。
|
|
|
|
|
|
- 从算法角度看，搜索、排序、回溯、分治、动态规划等许多重要算法策略直接或间接地应用了这种思维方式。
|
|
|
- 从数据结构角度看，递归天然适合处理链表、树和图的相关问题，因为它们非常适合用分治思想进行分析。
|
|
|
|
|
|
## 2.2.3 两者对比
|
|
|
|
|
|
总结以上内容，如表 2-1 所示，迭代和递归在实现、性能和适用性上有所不同。
|
|
|
|
|
|
<p align="center"> 表 2-1 迭代与递归特点对比 </p>
|
|
|
|
|
|
<div class="center-table" markdown>
|
|
|
|
|
|
| | 迭代 | 递归 |
|
|
|
| -------- | -------------------------------------- | ------------------------------------------------------------ |
|
|
|
| 实现方式 | 循环结构 | 函数调用自身 |
|
|
|
| 时间效率 | 效率通常较高，无函数调用开销 | 每次函数调用都会产生开销 |
|
|
|
| 内存使用 | 通常使用固定大小的内存空间 | 累积函数调用可能使用大量的栈帧空间 |
|
|
|
| 适用问题 | 适用于简单循环任务，代码直观、可读性好 | 适用于子问题分解，如树、图、分治、回溯等，代码结构简洁、清晰 |
|
|
|
|
|
|
</div>
|
|
|
|
|
|
!!! tip
|
|
|
|
|
|
如果感觉以下内容理解困难，可以在读完“栈”章节后再来复习。
|
|
|
|
|
|
那么，迭代和递归具有什么内在联系呢？以上述递归函数为例，求和操作在递归的“归”阶段进行。这意味着最初被调用的函数实际上是最后完成其求和操作的，**这种工作机制与栈的“先入后出”原则异曲同工**。
|
|
|
|
|
|
事实上，“调用栈”和“栈帧空间”这类递归术语已经暗示了递归与栈之间的密切关系。
|
|
|
|
|
|
1. **递**：当函数被调用时，系统会在“调用栈”上为该函数分配新的栈帧，用于存储函数的局部变量、参数、返回地址等数据。
|
|
|
2. **归**：当函数完成执行并返回时，对应的栈帧会被从“调用栈”上移除，恢复之前函数的执行环境。
|
|
|
|
|
|
因此，**我们可以使用一个显式的栈来模拟调用栈的行为**，从而将递归转化为迭代形式：
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
```python title="recursion.py"
|
|
|
def for_loop_recur(n: int) -> int:
|
|
|
"""使用迭代模拟递归"""
|
|
|
# 使用一个显式的栈来模拟系统调用栈
|
|
|
stack = []
|
|
|
res = 0
|
|
|
# 递：递归调用
|
|
|
for i in range(n, 0, -1):
|
|
|
# 通过“入栈操作”模拟“递”
|
|
|
stack.append(i)
|
|
|
# 归：返回结果
|
|
|
while stack:
|
|
|
# 通过“出栈操作”模拟“归”
|
|
|
res += stack.pop()
|
|
|
# res = 1+2+3+...+n
|
|
|
return res
|
|
|
```
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
```cpp title="recursion.cpp"
|
|
|
/* 使用迭代模拟递归 */
|
|
|
int forLoopRecur(int n) {
|
|
|
// 使用一个显式的栈来模拟系统调用栈
|
|
|
stack<int> stack;
|
|
|
int res = 0;
|
|
|
// 递：递归调用
|
|
|
for (int i = n; i > 0; i--) {
|
|
|
// 通过“入栈操作”模拟“递”
|
|
|
stack.push(i);
|
|
|
}
|
|
|
// 归：返回结果
|
|
|
while (!stack.empty()) {
|
|
|
// 通过“出栈操作”模拟“归”
|
|
|
res += stack.top();
|
|
|
stack.pop();
|
|
|
}
|
|
|
// res = 1+2+3+...+n
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
```java title="recursion.java"
|
|
|
/* 使用迭代模拟递归 */
|
|
|
int forLoopRecur(int n) {
|
|
|
// 使用一个显式的栈来模拟系统调用栈
|
|
|
Stack<Integer> stack = new Stack<>();
|
|
|
int res = 0;
|
|
|
// 递：递归调用
|
|
|
for (int i = n; i > 0; i--) {
|
|
|
// 通过“入栈操作”模拟“递”
|
|
|
stack.push(i);
|
|
|
}
|
|
|
// 归：返回结果
|
|
|
while (!stack.isEmpty()) {
|
|
|
// 通过“出栈操作”模拟“归”
|
|
|
res += stack.pop();
|
|
|
}
|
|
|
// res = 1+2+3+...+n
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
```csharp title="recursion.cs"
|
|
|
/* 使用迭代模拟递归 */
|
|
|
int ForLoopRecur(int n) {
|
|
|
// 使用一个显式的栈来模拟系统调用栈
|
|
|
Stack<int> stack = new();
|
|
|
int res = 0;
|
|
|
// 递：递归调用
|
|
|
for (int i = n; i > 0; i--) {
|
|
|
// 通过“入栈操作”模拟“递”
|
|
|
stack.Push(i);
|
|
|
}
|
|
|
// 归：返回结果
|
|
|
while (stack.Count > 0) {
|
|
|
// 通过“出栈操作”模拟“归”
|
|
|
res += stack.Pop();
|
|
|
}
|
|
|
// res = 1+2+3+...+n
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
```go title="recursion.go"
|
|
|
/* 使用迭代模拟递归 */
|
|
|
func forLoopRecur(n int) int {
|
|
|
// 使用一个显式的栈来模拟系统调用栈
|
|
|
stack := list.New()
|
|
|
res := 0
|
|
|
// 递：递归调用
|
|
|
for i := n; i > 0; i-- {
|
|
|
// 通过“入栈操作”模拟“递”
|
|
|
stack.PushBack(i)
|
|
|
}
|
|
|
// 归：返回结果
|
|
|
for stack.Len() != 0 {
|
|
|
// 通过“出栈操作”模拟“归”
|
|
|
res += stack.Back().Value.(int)
|
|
|
stack.Remove(stack.Back())
|
|
|
}
|
|
|
// res = 1+2+3+...+n
|
|
|
return res
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
```swift title="recursion.swift"
|
|
|
/* 使用迭代模拟递归 */
|
|
|
func forLoopRecur(n: Int) -> Int {
|
|
|
// 使用一个显式的栈来模拟系统调用栈
|
|
|
var stack: [Int] = []
|
|
|
var res = 0
|
|
|
// 递：递归调用
|
|
|
for i in stride(from: n, to: 0, by: -1) {
|
|
|
// 通过“入栈操作”模拟“递”
|
|
|
stack.append(i)
|
|
|
}
|
|
|
// 归：返回结果
|
|
|
while !stack.isEmpty {
|
|
|
// 通过“出栈操作”模拟“归”
|
|
|
res += stack.removeLast()
|
|
|
}
|
|
|
// res = 1+2+3+...+n
|
|
|
return res
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
```javascript title="recursion.js"
|
|
|
/* 使用迭代模拟递归 */
|
|
|
function forLoopRecur(n) {
|
|
|
// 使用一个显式的栈来模拟系统调用栈
|
|
|
const stack = [];
|
|
|
let res = 0;
|
|
|
// 递：递归调用
|
|
|
for (let i = 1; i <= n; i++) {
|
|
|
// 通过“入栈操作”模拟“递”
|
|
|
stack.push(i);
|
|
|
}
|
|
|
// 归：返回结果
|
|
|
while (stack.length) {
|
|
|
// 通过“出栈操作”模拟“归”
|
|
|
res += stack.pop();
|
|
|
}
|
|
|
// res = 1+2+3+...+n
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
```typescript title="recursion.ts"
|
|
|
/* 使用迭代模拟递归 */
|
|
|
function forLoopRecur(n: number): number {
|
|
|
// 使用一个显式的栈来模拟系统调用栈
|
|
|
const stack: number[] = [];
|
|
|
let res: number = 0;
|
|
|
// 递：递归调用
|
|
|
for (let i = 1; i <= n; i++) {
|
|
|
// 通过“入栈操作”模拟“递”
|
|
|
stack.push(i);
|
|
|
}
|
|
|
// 归：返回结果
|
|
|
while (stack.length) {
|
|
|
// 通过“出栈操作”模拟“归”
|
|
|
res += stack.pop();
|
|
|
}
|
|
|
// res = 1+2+3+...+n
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
```dart title="recursion.dart"
|
|
|
/* 使用迭代模拟递归 */
|
|
|
int forLoopRecur(int n) {
|
|
|
// 使用一个显式的栈来模拟系统调用栈
|
|
|
List<int> stack = [];
|
|
|
int res = 0;
|
|
|
// 递：递归调用
|
|
|
for (int i = n; i > 0; i--) {
|
|
|
// 通过“入栈操作”模拟“递”
|
|
|
stack.add(i);
|
|
|
}
|
|
|
// 归：返回结果
|
|
|
while (!stack.isEmpty) {
|
|
|
// 通过“出栈操作”模拟“归”
|
|
|
res += stack.removeLast();
|
|
|
}
|
|
|
// res = 1+2+3+...+n
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
```rust title="recursion.rs"
|
|
|
/* 使用迭代模拟递归 */
|
|
|
fn for_loop_recur(n: i32) -> i32 {
|
|
|
// 使用一个显式的栈来模拟系统调用栈
|
|
|
let mut stack = Vec::new();
|
|
|
let mut res = 0;
|
|
|
// 递：递归调用
|
|
|
for i in (1..=n).rev() {
|
|
|
// 通过“入栈操作”模拟“递”
|
|
|
stack.push(i);
|
|
|
}
|
|
|
// 归：返回结果
|
|
|
while !stack.is_empty() {
|
|
|
// 通过“出栈操作”模拟“归”
|
|
|
res += stack.pop().unwrap();
|
|
|
}
|
|
|
// res = 1+2+3+...+n
|
|
|
res
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
```c title="recursion.c"
|
|
|
/* 使用迭代模拟递归 */
|
|
|
int forLoopRecur(int n) {
|
|
|
int stack[1000]; // 借助一个大数组来模拟栈
|
|
|
int top = -1; // 栈顶索引
|
|
|
int res = 0;
|
|
|
// 递：递归调用
|
|
|
for (int i = n; i > 0; i--) {
|
|
|
// 通过“入栈操作”模拟“递”
|
|
|
stack[1 + top++] = i;
|
|
|
}
|
|
|
// 归：返回结果
|
|
|
while (top >= 0) {
|
|
|
// 通过“出栈操作”模拟“归”
|
|
|
res += stack[top--];
|
|
|
}
|
|
|
// res = 1+2+3+...+n
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
```zig title="recursion.zig"
|
|
|
// 使用迭代模拟递归
|
|
|
fn forLoopRecur(comptime n: i32) i32 {
|
|
|
// 使用一个显式的栈来模拟系统调用栈
|
|
|
var stack: [n]i32 = undefined;
|
|
|
var res: i32 = 0;
|
|
|
// 递：递归调用
|
|
|
var i: usize = n;
|
|
|
while (i > 0) {
|
|
|
stack[i - 1] = @intCast(i);
|
|
|
i -= 1;
|
|
|
}
|
|
|
// 归：返回结果
|
|
|
var index: usize = n;
|
|
|
while (index > 0) {
|
|
|
index -= 1;
|
|
|
res += stack[index];
|
|
|
}
|
|
|
// res = 1+2+3+...+n
|
|
|
return res;
|
|
|
}
|
|
|
```
|
|
|
|
|
|
??? pythontutor "可视化运行"
|
|
|
|
|
|
<iframe width="800" height="600" frameborder="0" src="https://pythontutor.com/iframe-embed.html#code=def%20for_loop_recur%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%BD%BF%E7%94%A8%E8%BF%AD%E4%BB%A3%E6%A8%A1%E6%8B%9F%E9%80%92%E5%BD%92%22%22%22%0A%20%20%20%20%23%20%E4%BD%BF%E7%94%A8%E4%B8%80%E4%B8%AA%E6%98%BE%E5%BC%8F%E7%9A%84%E6%A0%88%E6%9D%A5%E6%A8%A1%E6%8B%9F%E7%B3%BB%E7%BB%9F%E8%B0%83%E7%94%A8%E6%A0%88%0A%20%20%20%20stack%20%3D%20%5B%5D%0A%20%20%20%20res%20%3D%200%0A%20%20%20%20%23%20%E9%80%92%EF%BC%9A%E9%80%92%E5%BD%92%E8%B0%83%E7%94%A8%0A%20%20%20%20for%20i%20in%20range%28n,%200,%20-1%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E9%80%9A%E8%BF%87%E2%80%9C%E5%85%A5%E6%A0%88%E6%93%8D%E4%BD%9C%E2%80%9D%E6%A8%A1%E6%8B%9F%E2%80%9C%E9%80%92%E2%80%9D%0A%20%20%20%20%20%20%20%20stack.append%28i%29%0A%20%20%20%20%23%20%E5%BD%92%EF%BC%9A%E8%BF%94%E5%9B%9E%E7%BB%93%E6%9E%9C%0A%20%20%20%20while%20stack%3A%0A%20%20%20%20%20%20%20%20%23%20%E9%80%9A%E8%BF%87%E2%80%9C%E5%87%BA%E6%A0%88%E6%93%8D%E4%BD%9C%E2%80%9D%E6%A8%A1%E6%8B%9F%E2%80%9C%E5%BD%92%E2%80%9D%0A%20%20%20%20%20%20%20%20res%20%2B%3D%20stack.pop%28%29%0A%20%20%20%20%23%20res%20%3D%201%2B2%2B3%2B...%2Bn%0A%20%20%20%20return%20res%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%205%0A%20%20%20%20res%20%3D%20for_loop_recur%28n%29%0A%20%20%20%20print%28f%22%5Cn%E4%BD%BF%E7%94%A8%E8%BF%AD%E4%BB%A3%E6%A8%A1%E6%8B%9F%E9%80%92%E5%BD%92%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=370&codeDivWidth=300&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe>
|
|
|
<a href=https://pythontutor.com/iframe-embed.html#code=def%20for_loop_recur%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%BD%BF%E7%94%A8%E8%BF%AD%E4%BB%A3%E6%A8%A1%E6%8B%9F%E9%80%92%E5%BD%92%22%22%22%0A%20%20%20%20%23%20%E4%BD%BF%E7%94%A8%E4%B8%80%E4%B8%AA%E6%98%BE%E5%BC%8F%E7%9A%84%E6%A0%88%E6%9D%A5%E6%A8%A1%E6%8B%9F%E7%B3%BB%E7%BB%9F%E8%B0%83%E7%94%A8%E6%A0%88%0A%20%20%20%20stack%20%3D%20%5B%5D%0A%20%20%20%20res%20%3D%200%0A%20%20%20%20%23%20%E9%80%92%EF%BC%9A%E9%80%92%E5%BD%92%E8%B0%83%E7%94%A8%0A%20%20%20%20for%20i%20in%20range%28n,%200,%20-1%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E9%80%9A%E8%BF%87%E2%80%9C%E5%85%A5%E6%A0%88%E6%93%8D%E4%BD%9C%E2%80%9D%E6%A8%A1%E6%8B%9F%E2%80%9C%E9%80%92%E2%80%9D%0A%20%20%20%20%20%20%20%20stack.append%28i%29%0A%20%20%20%20%23%20%E5%BD%92%EF%BC%9A%E8%BF%94%E5%9B%9E%E7%BB%93%E6%9E%9C%0A%20%20%20%20while%20stack%3A%0A%20%20%20%20%20%20%20%20%23%20%E9%80%9A%E8%BF%87%E2%80%9C%E5%87%BA%E6%A0%88%E6%93%8D%E4%BD%9C%E2%80%9D%E6%A8%A1%E6%8B%9F%E2%80%9C%E5%BD%92%E2%80%9D%0A%20%20%20%20%20%20%20%20res%20%2B%3D%20stack.pop%28%29%0A%20%20%20%20%23%20res%20%3D%201%2B2%2B3%2B...%2Bn%0A%20%20%20%20return%20res%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%205%0A%20%20%20%20res%20%3D%20for_loop_recur%28n%29%0A%20%20%20%20print%28f%22%5Cn%E4%BD%BF%E7%94%A8%E8%BF%AD%E4%BB%A3%E6%A8%A1%E6%8B%9F%E9%80%92%E5%BD%92%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=370&codeDivWidth=300&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>
|
|
|
|
|
|
观察以上代码，当递归转化为迭代后，代码变得更加复杂了。尽管迭代和递归在很多情况下可以互相转化，但不一定值得这样做，有以下两点原因。
|
|
|
|
|
|
- 转化后的代码可能更加难以理解，可读性更差。
|
|
|
- 对于某些复杂问题，模拟系统调用栈的行为可能非常困难。
|
|
|
|
|
|
总之，**选择迭代还是递归取决于特定问题的性质**。在编程实践中，权衡两者的优劣并根据情境选择合适的方法至关重要。
|
