Leetcode_subsets-ii

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

 Given a collection of integers that might contain duplicates, S, return all possible subsets.

Note:

• Elements in a subset must be in non-descending order.
• The solution set must not contain duplicate subsets.

For example,
If S = [1,2,2], a solution is:

[  [2],  [1],  [1,2,2],  [2,2],  [1,2],  []]

思路：dfs+回溯。关于如何去重，有几种方式，笔者这种是比较耗费时间的，可以用条件判断是否继续dfs来去重。这个还要研究一下，因为更节约时间。AC耗时136ms

参考代码：

class Solution {public:    void dfs(vector<vector<int>>&ans, vector<int>&S, vector<bool>&visited, vector<int>vec, int st)    {        if(st == S.size())        {            if(find(ans.begin(), ans.end(), vec)==ans.end())                ans.push_back(vec);            return;        }        dfs(ans, S, visited, vec, st+1);        if(visited[st])            return;        vec.push_back(S[st]);        visited[st] = true;        dfs(ans, S, visited, vec, st+1);        visited[st] = false;        vec.pop_back();    }    vector<vector<int> > subsetsWithDup(vector<int> &S) {        vector<vector<int>>ans;        if(S.empty())            return ans;        sort(S.begin(), S.end());        vector<int>vec;        vector<bool>visited(S.size(), false);        dfs(ans, S, visited, vec, 0);        return ans;    }};