leetcode Permutations

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废话少说，先上题：

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].

解这题的思路是从数组中找一个数作为第一位，然后依次找下一位作为permuta的第二位，我这里采用的是找到一位就把那位从数组中删除，判断数组为空的时候推、退出递归。上代码（accept）：

class Solution {public:void dfs(vector<vector<int> >& result, vector<int>& tmp, vector<int>& num){if (num.empty()){result.push_back(tmp);return;}for (int i = 0; i < num.size(); i++){tmp.push_back(num[i]);auto iter = num.begin();iter += i;vector<int> tmpA = num;num.erase(iter);dfs(result, tmp, num);num = tmpA;tmp.pop_back();}}vector<vector<int> > permute(vector<int> &num){vector<vector<int> > result;if (num.empty()){return result;}vector<int> tmp;dfs(result, tmp, num);return result;}};

以前我是用next_permutation的算法做的，就是循环n！次next_permutation，也能accept，同样上代码：

class Solution {public:     void nextPermutation(vector<int> &num) {     if (num.empty() || num.size() == 1)     {         return;     }     auto iter = num.end() - 1;     for (; iter != num.begin(); iter--)     {         if (*iter > *(iter - 1))         {             break;         }     }     if (iter == num.begin())     {         reverse(num.begin(), num.end());         return;     }     else     {         if (iter == num.end() - 1)         {             swap(num[iter - num.begin()], num[iter - num.begin() - 1]);             return;         }         else         {            auto it = iter;            while (*(it + 1) > *(iter - 1) && it != num.end() - 1)            {                it++;            }            swap(num[it - num.begin()], num[iter - num.begin() - 1]);            sort(iter, num.end());            return;         }     }    }    int factoral(int n)    {        int nRet = 1;        for (int i = 1; i <= n; i++)        {            nRet *= i;        }        return nRet;    }    vector<vector<int> > permute(vector<int> &num) {        vector<vector<int>> result;        vector<int> tmp = num;        result.push_back(tmp);        int n = num.size();        int size = factoral(n);        for (int i = 0; i < size - 1; i++)        {            nextPermutation(tmp);            result.push_back(tmp);        }        return result;    }};

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