按照字典序產生1–n的排列

遞迴用得很巧妙。

//按照字典序產生1--n的排列#include <iostream>#include <fstream>using namespace std;void print_permutation(int n, int* A, int cur){int i, j;if(cur == n)//遞迴邊界{for(i = 0; i < n; i++){cout<<A[i];}cout<<endl;}else {for(i = 1; i <= n; i++)//嘗試在A[cur]中填各種整數i{int ok = 1;for(j = 0; j < cur; j++){if(A[j] == i)//如果i已經在A[0] -- A[cur - 1]中出現過，則不能再選{ok = 0;break;}}if(ok){A[cur] = i;print_permutation(n, A, cur + 1);}}}}int main(){int A[5] = {0};int cur = 0;print_permutation(3, A, cur);}