LeetCode 47.Permutations II

题目：

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].

分析与解答：

这个题比上个题目更难了，问题的关键在于如何判断重复。我自己的方法是先排序，然后遇到重复进栈的数就跳过。

class Solution {
public:
    vector<vector<int> > result;
    vector<vector<int> > permuteUnique(vector<int> &num) {
        sort(num.begin(),num.end());
        vector<int> tempresult;
        vector<int> path;//存放选过数的下标
        dfs(num,tempresult,path);
        return result;
    }
    void dfs(vector<int> &num,vector<int> &tempresult,vector<int> &path){
        if(tempresult.size() == num.size()){
            result.push_back(tempresult);
            return;
        }
        int last = INT_MAX;
        for(int i = 0;i < num.size();++i){
            auto index = find(path.begin(),path.end(),i);
            if(index == path.end() && num[i] != last){//这个数没被选过
                tempresult.push_back(num[i]);
                path.push_back(i);
                dfs(num,tempresult,path);
                tempresult.pop_back();
                path.pop_back();
                last = num[i];
            }
            
        }
    }
};

网上有一个更简洁的版本：

class Solution {
public:
    void recursion(vector<int> num, int i, int j, vector<vector<int> > &res) {
        if (i == j-1) {
            res.push_back(num);
            return;
        }
        for (int k = i; k < j; k++) {
            if (i != k && num[i] == num[k]) continue;
            swap(num[i], num[k]);
            recursion(num, i+1, j, res);
        }
    }
    vector<vector<int> > permuteUnique(vector<int> &num) {
        sort(num.begin(), num.end());
        vector<vector<int> >res;
        recursion(num, 0, num.size(), res);
        return res;
    }
};