Permutations

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Given a collection of numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].

C++:

class Solution {
public:
    vector<vector<int> > arr;

    void dfs(vector<int> &num, int i, int n) {
        int j, tmp;
        if(i == n) arr.push_back(num);
        for(j = i; j < n; j++) {
            tmp = num[i];
            num[i] = num[j];
            num[j] = tmp;
            //printf("pre:i:%d, j:%d\n", i, j);
            dfs(num, i + 1, n);
            //printf("after:i:%d, j:%d\n", i, j);
            tmp = num[i];
            num[i] = num[j];
            num[j] = tmp;
        }
    }

    vector<vector<int> > permute(vector<int> &num) {
        dfs(num, 0, num.size()); 
        return arr;
    }
};

#include<stdio.h>
#include<stdlib.h>
#define N 3


int arr[100][N];
static int index = 0;


void outprint(int A[], int n) {
    int i;
    for(i = 0; i < n; i++) {
        arr[index][i] = A[i];
        printf("%d ", arr[index][i]);
    }
    printf("\n");
}


void dfs(int A[], int i, int n) {
    int j, tmp;
    if(i == n) outprint(A, n);
    for(j = i; j < n; j++) {
        tmp = A[i];
        A[i] = A[j];
        A[j] = tmp;
        //printf("pre:i:%d, j:%d\n", i, j);
        dfs(A, i + 1, n);
        //printf("after:i:%d, j:%d\n", i, j);
        tmp = A[i];
        A[i] = A[j];
        A[j] = tmp;
    }
}


void permute(int A[], int n) {
    dfs(A, 0, n);    
}


void main() {
    int A[] = {1, 2, 3};
    permute(A, 3);
}

Permutations

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原文地址：http://blog.csdn.net/uj_mosquito/article/details/42915827