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], [] ]
思考:直接set去重会不会太暴力。。
class Solution { private: vector<vector<int> > res; vector<int> ans; set<vector<int> > m; public: void DFS(vector<int> S,int n,int start,int dep,int k) { if(dep==k) { set<vector<int> >::iterator iter=m.find(ans); if(iter==m.end()) { res.push_back(ans); m.insert(ans); } return; } for(int i=start;i<n-k+1;i++) { ans.push_back(S[i+dep]); DFS(S,n,i,dep+1,k); ans.pop_back(); } } vector<vector<int> > subsetsWithDup(vector<int> &S) { res.clear(); ans.clear(); int n=S.size(); sort(S.begin(),S.end()); for(int k=0;k<=n;k++) { DFS(S,n,0,0,k); } return res; } };