Leetcode_permutations-ii

地址：http://oj.leetcode.com/problems/permutations-ii/

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

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

思路：跟上一题类似，http://blog.csdn.net/flyupliu/article/details/23609159，但是去重记得要对输入的vector排序。

对不改变的vector如num记得在参数前加上const，能用引用的参数尽量用引用，提高运算速度减少内存开销。

参考代码：

class Solution {public:    void dfs(vector<vector<int>>&ans, vector<bool>&visited, vector<int>&vec, const vector<int>&num)    {        if(vec.size()==num.size())        {            ans.push_back(vec);            return;        }        for(int i = 0; i<num.size(); ++i)        {            if(i && num[i]==num[i-1] && !visited[i-1])                continue;            if(!visited[i])            {                visited[i] = true;                vec.push_back(num[i]);                dfs(ans, visited, vec, num);                vec.pop_back();                visited[i] = false;            }        }    }    vector<vector<int> > permuteUnique(vector<int> &num) {        vector<vector<int> >ans;        if(num.empty())            return ans;        vector<bool>visited(num.size(), false);        vector<int>vec;        sort(num.begin(), num.end());        dfs(ans, visited, vec, num);        return ans;    }};