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> result;
public:
void dfs(vector&S, int i, vector 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 > subsetsWithDup(vector &S) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
result.clear();
vector tmp;
dfs(S, 0, tmp);
set>::iterator it;
vector> ret;
for(it = result.begin(); it!=result.end(); it++){
ret.push_back(*it);
}
return ret;
}
};