C ++ classical algorithm problem - permutations and combinations

27.Algorithm Gossip: permutations

Explanation

A set of numbers, letters or symbols arranged in order to obtain different combinations, for example of permutation and combination of these three digits 123 are:

1 2 3、1 3 2、2 1 3、2 3 1、3 1 2、3 2 1。

solution

Problems may be used to recursively cut into smaller units permutations and combinations, for example arrangement 1234 can be divided into

1 [2 3 4]、2 [1 3 4]、3 [1 2 4]、4 [1 2 3]

Arranges side by a spin method, a first rotation interval is set to 0, the rightmost digit is rotated to the left, and gradually increase the rotational intervals, for example:

1 2 3 4 -> 旋转1 -> 继续将右边2 3 4进行递回处理
2 1 3 4 -> 旋转1 2 变为 2 1-> 继续将右边1 3 4进行递回处理
3 1 2 4 -> 旋转1 2 3变为 3 1 2 -> 继续将右边1 2 4进行递回处理
4 1 2 3 -> 旋转1 2 3 4变为4 1 2 3 -> 继续将右边1 2 3进行递回处理

The sample code

#include <stdio.h>
#include <stdlib.h> 
#define N 4
    void perm(int*, int); int main(void) {
        int num[N+1], i;
        for(i = 1; i <= N; i++) num[i] = i;
        perm(num, 1);
        return 0;
    }

    void perm(int* num, int i) { int j, k, tmp;

        if(i < N) {
            for(j = i; j <= N; j++) { tmp = num[j];
                // 旋转该区段最右边数字至最左边
                for(k = j; k > i; k--)
                    num[k] = num[k-1]; num[i] = tmp; perm(num, i+1);
                // 还原
                for(k = i; k < j; k++) num[k] = num[k+1];
                num[j] = tmp;
            }
        }
        else {	// 显示此次排列
            for(j = 1; j <= N; j++) printf("%d ", num[j]);
            printf("\n");
        }
    }