Given an integer array nums that may contain duplicates, return all possible subsets (the power set).
The solution set must not contain duplicate subsets. Return the solution in any order.
Solution
如何处理相同的子集：先把 /(vector/) /(sort/) 一下，然后在 /(ans/) 里面 /(find/)， 依旧记得擦除标记
点击查看代码
class Solution {
private:
vector<vector<int>> ans;
vector<int> res;
void dfs(vector<int> nums, vector<int>&res, int pos){
for(int i=pos; i<nums.size();i++){
res.push_back(nums[i]);
vector<int> tmp = res;
sort(res.begin(), res.end());
if(find(ans.begin(), ans.end(), res) == ans.end()){
// not repeated
ans.push_back(res);
}
dfs(nums, res, i+1);
res = tmp;
res.pop_back();
}
}
public:
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
ans.push_back(res);
dfs(nums, res, 0);
return ans;
}
};
原创文章，作者：ItWorker，如若转载，请注明出处：https://blog.ytso.com/279305.html