[leetcode]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],
  []
]

每个元素都有两种选择，加入或不加入集合。所以，深度优先搜索解题。
注意，不能有重合的集合。所以，相同的元素我们只需控制它加入集合的个数即可。

class Solution {
public:
    vector<vector<int> > subsetsWithDup(vector<int> &S) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int>> v;
        vector<int> tmp;
        if(S.size()==0)
            return v;
        sort(S.begin(),S.end());
        DepthFirst(S,v,tmp,0);
        return v;
    }
    void DepthFirst(vector<int> &S , vector<vector<int> > &v ,vector<int> &tmp , int depth)
    {
        if(depth == S.size())
        {
            v.push_back(tmp);
            return;
        }
        int i = depth;
        while(i+1 < S.size() && S[i] == S[i+1])
        {
            i++;
        }
        for(int j = 0;j <= i - depth +1;j++)
        {
            for(int k = 1; k <= j;k++)
                tmp.push_back(S[depth]);
            DepthFirst(S,v,tmp,i+1);
            for(int k = 1; k <= j;k++)
                tmp.pop_back();  
        }
    }
};