[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;

    }

};