LeetCode--Subsets II

Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

For example,

If nums = [1,2,2], a solution is:

[

[2],

[1],

[1,2,2],

[2,2],

[1,2],

[]

]

思路：深度优先搜索。

这一题是之前的Subsets的扩展，给的数组数字有重复，那么为了得到的子集不会重复，就有去重的条件判断if(nums[i]!=nums[i-1]||i==start)，而且注意要先排序，其余类似上一题。

class Solution {
public:
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
sort(nums.begin(),nums.end());
vector<vector<int>>result;
vector<int>path;
result.push_back(path);
for(int i=1;i<=nums.size();i++){
dfs(result,path,nums,i,0);
}
return result;
}
void dfs(vector<vector<int>>&result,vector<int>&path,vector<int>&nums,int k,int start){
if(path.size()==k){
result.push_back(path);
return;
}
for(int i=start;i<nums.size();i++){
if(nums[i]!=nums[i-1]||i==start){
path.push_back(nums[i]);
dfs(result,path,nums,k,i+1
4000
);
path.pop_back();
}
}
}
};