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);                path.pop_back();            }        }    }};