[LeetCode] Permutations

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].

Solution:

对一个没有重复元素的序列做全排序，可以考虑采用递归的思路。n个元素的全排列为n个元素分别作为第一个元素，并加上分别去除首元素剩下元素的全排列。当序列递归到只有一个元素时，

递归终止，一个可行的全排列产生。

class Solution {
public:
    vector<vector<int> > ans;
    vector<int> src;
    int len;
    
    void perm(int i)
    {
        if(i == len - 1)
        {
            //a new permutation comes
            vector<int> tmp;
            for(int i = 0;i < len;i++)
                tmp.push_back(src[i]);
            ans.push_back(tmp);
            return;
        }
        else
        {
            for(int k = i;k < len;k++)
            {
                int tmp = src[i];
                src[i] = src[k];
                src[k] = tmp;
                perm(i + 1);
                src[k] = src[i];
                src[i] = tmp;
            }
        }
    }
    
    vector<vector<int> > permute(vector<int> &num) {
        len = num.size();
        if(len == 0) return ans;
        src = num;
        perm(0);
        return ans;
    }
};