42 - 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].

solution: 递归方法实现，从第0位开始，先交换当前位和后续1 到最后中的每个数，然后递归之。

class Solution {public:    void swap(vector<int> &num, int i, int j)    {        int temp = num[i];        num[i] = num[j];        num[j] = temp;    }        void per(vector< vector<int> >&res, vector<int> &num, int cur, int len)    {        if(cur == len)        {            res.push_back(num);            return;        }        for(int i = 0; i < len - cur; i++)        {            swap(num, cur, cur+i);            per(res, num, cur+1, len);            swap(num, cur, cur+i);        }    }        vector<vector<int> > permute(vector<int> &num) {        // Note: The Solution object is instantiated only once and is reused by each test case.        vector< vector<int> >res;        vector<int> result;        if(num.size() <= 0)            return res;                int len = num.size();                per(res, num, 0, len);        return res;    }};

在论坛上找到这样的代码，原理是相同的，由于引入了bool值的数组来标识已访问的数字，这样不需要改变原来数组，可靠性更强些，但由于每次递归都会做一次n的循环，所以效率会低。但该方法在实现permutations II 的时候显得很好用，所以sign下。

class Solution {public:vector<vector<int> > permute(vector<int> &num) {  // Start typing your C/C++ solution below  // DO NOT write int main() function  vector<vector<int> > ans;  if(num.size<()=0) return ans;  vector<int> s;  vector<bool> used(num.size(), false);  permuteRec(num, s, used, ans);  return ans;}void permuteRec(vector<int> &num, vector<int> &s, vector<bool> &used,    vector<vector<int> > &ans){  if(s.size()==num.size())  {    ans.push_back(s);    return;  }  for(int i=0; i<used.size(); i++)  {    if(!used[i])    {      s.push_back(num[i]);      used[i]=true;      permuteRec(num, s, used, ans);      used[i]=false;      s.pop_back();    }  }}};