LeetCode: Permutations II

思路：和一般的排列一样，不过在存入结果时，先检查一下是否已经存在这个结果了。

code:

class Solution {
public:
    void solvePermute(vector<int> &num, int pos, vector<vector<int> > &ret){
        if(pos == num.size()-1){
            if(find(ret.begin(),ret.end(),num) == ret.end())
            ret.push_back(num);
            return;
        }
        //solvePermute(num,pos+1,ret);
        for(int i = pos;i< num.size();i++){
            if(i != pos && num[i] == num[i-1])
                continue;
            vector<int> temp = num;
            int s = temp[pos];
            temp[pos] = temp[i];
            temp[i] = s;
            solvePermute(temp,pos+1,ret);
        }
    }
    vector<vector<int> > permuteUnique(vector<int> &num) {
         sort(num.begin(),num.end());
         vector<vector<int> > ret;
         vector<int> curRet;
         int n = num.size();
         solvePermute(num,0,ret);
         return ret;
    }
};