Permutations II 去掉重复的全排列

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].

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Backtracking

class Solution {
private:
    vector<vector<int> > ret;
public:
    void perm(vector<int> &num,int i,int len){
        if(i==len){
            ret.push_back(num);
            return;
        }
        for(int j=i;j<len;++j){
            if(!isSwap(num,i,j))
                continue;
            swap(num[i],num[j]);
            perm(num,i+1,len);
            swap(num[j],num[i]);
        }
    }
    bool isSwap(vector<int>& num, int i, int j) {
        while (num[i] != num[j] && i < j) i++;
        if (i == j) return true;
        else return false;
    }
    vector<vector<int> > permuteUnique(vector<int> &num) {
        int len=num.size();
        ret.clear();
        sort(num.begin(), num.end());
        perm(num,0,len);
        return ret;
    }
};

参考http://blog.csdn.net/morewindows/article/details/7370155