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], [] ]
这个曾经让我很捉鸡的题目。。。为什么这次用了个很烂的dfs,一次性通过,而且是80ms过大集合。。。。太讽刺了。。。
class Solution { set<vector<int>> result; public: void dfs(vector<int>&S, int i, vector<int> tmp){ if(i == S.size()){ sort(tmp.begin(), tmp.end()); result.insert(tmp); return; } dfs(S, i+1, tmp); tmp.push_back(S[i]); dfs(S, i+1, tmp); } vector<vector<int> > subsetsWithDup(vector<int> &S) { // Start typing your C/C++ solution below // DO NOT write int main() function result.clear(); vector<int> tmp; dfs(S, 0, tmp); set<vector<int>>::iterator it; vector<vector<int>> ret; for(it = result.begin(); it!=result.end(); it++){ ret.push_back(*it); } return ret; } };