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].

 

class Solution {
public:
    vector<vector<int> > re;
    vector<vector<int> > permuteUnique(vector<int> &num) {
        map<int , int> m;
        int size = num.size();
        vector <int> result;
        for(int i = 0; i < size ;i++)
        {
            m[num[i]]++;
        }
        per(result,0,0,m,size);
        return re;
        
        
    }
    void per(vector<int> now ,int flag,  int put,map<int , int> m,int size)
    {
        if(flag == 1)
        {
            now.push_back(put);
            m[put]--;
            if(now.size() == size)
            {
                re.push_back(now);
                return;
            }
        }
        else flag =1;
        for(map<int ,int>::iterator iter = m.begin();iter !=m.end() ;iter++)
        {
            if(iter->second > 0)
            {
                per(now ,1, iter->first,m,size);
            }
        }
        
    }
};

　　

用dfs应该会更快。但是用dfs需要先排一下序。 n log n，之后用n^2进行取元素。

我用map思路相似，但是每次取元素都需要加一个