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],
  []
]

分析：此题只要将subsets I的方法稍做修改即可。当数组中有duplicates时，在同一层的递归要避免将相同的元素放入到path中。
代码：

class Solution {
public:
    vector<vector<int> > subsetsWithDup(vector<int> &S) {
        sort(S.begin(),S.end());
        vector<vector<int>> res;
        vector<int> path;
        get_subsets(S,res,path,0);
        return res;
    }
    void get_subsets(vector<int> &S, vector<vector<int>> & res, vector<int> & path, int start){
        res.push_back(path);
        for(int i = start; i < S.size(); i++){
            if(i != start && S[i] == S[i-1]) continue;//skip duplicates
            path.push_back(S[i]);
            get_subsets(S,res,path,i+1);
            path.pop_back();
        }
    }
};

 该题的迭代法跟subsets I类似，只是先将结果保存在set中，然后再从set中存入vector<vector<int> >。