两个古老问题的递归算法。

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1. 全排列

思路1：递归分治（基于交换）

#include <stdio.h>void perm(int* a, int start, int end);void swap(int& m, int& n);void output(const int* a, int size);int main() {int a[5] = {1, 2, 3, 4, 5};perm(a, 0, 5);return 0;}void perm(int* a, int start, int end) {if(start == end - 1) {output(a, end);} else {for(int i = start; i < end; ++i) {swap(a[start], a[i]);perm(a, start + 1, end);swap(a[start], a[i]);}}}void swap(int& m, int& n) {int t = m;m = n;n = t;}void output(const int* a, int size) {static int count;printf("%3d: ", ++count);for(int i = 0; i < size; ++i) {printf("%d ", a[i]);}printf("\n");}

思路2：剪枝回溯

#include <stdio.h> void perm(int* data, int* buf, int start, int end);void output(const int* a, int size);int main() {int data[5] = {1, 2, 3, 4, 5};int buf[5] = {0};perm(data, buf, 0, 5);return 0;}void perm(int* data, int* buf, int start, int end) {if(start == end) {output(buf, end);} else {for(int i = 0; i < end; i++) {if(!data[i]) continue;buf[start] = data[i];data[i] = 0;perm(data, buf, start + 1, end);data[i] = buf[start];}}}void output(const int* a, int size) {static int count;printf("%3d: ", ++count);for(int i = 0; i < size; ++i) {printf("%d ", a[i]);}printf("\n");}

2. 汉诺塔

#include <stdio.h>void hanoi(int n, char a, char b, char c);void move(int n, char from, char to);int main() {hanoi(5, 'A', 'B', 'C');return 0;}void hanoi(int n, char a, char b, char c) {if(n > 0) {hanoi(n - 1, a, c, b);move(n, a, b);hanoi(n - 1, c, b, a);}}void move(int n, char from, char to) {static int step;printf("Step %3d: No.%d %c -> %c.\n", ++step, n, from, to);}