Given a collection of integers that might contain duplicates, S, return all possible subsets.
Note:
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：
1 class Solution { 2 public: 3 vector<vector<int> > subsetsWithDup(vector<int> &S) { 4 sort(S.begin(),S.end()); 5 vector<vector<int>> res; 6 vector<int> path; 7 get_subsets(S,res,path,0); 8 return res; 9 } 10 void get_subsets(vector<int> &S, vector<vector<int>> & res, vector<int> & path, int start){ 11 res.push_back(path); 12 for(int i = start; i < S.size(); i++){ 13 if(i != start && S[i] == S[i-1]) continue;//skip duplicates 14 path.push_back(S[i]); 15 get_subsets(S,res,path,i+1); 16 path.pop_back(); 17 } 18 } 19 };
原文：http://www.cnblogs.com/Kai-Xing/p/3915650.html