40. 组合总和 II

 1 class Solution 
 2 {
 3     vector<vector<int>> res;
 4     int sum = 0;
 5 public:
 6     void helper(vector<int>& nums, int start,int target, vector<int>& out)
 7     {
 8         if(sum == target)
 9         {
10             res.push_back(out);
11         }
12         
13         if(sum > target) return;
14 
15         for (int i = start; i < nums.size(); ++i)
16         {
17             //去重
18             if(i != start && nums[i] == nums[i - 1]) continue;
19 
20             out.push_back(nums[i]);
21             sum += nums[i];
22             helper(nums, i + 1,target,out);//每个数字在每个组合中只能使用一次，所以是 i+1 
23 
24             out.pop_back();
25             sum -= nums[i];
26         }
27     }
28 
29     vector<vector<int>> combinationSum2(vector<int>& nums, int target) 
30     {
31         vector<int> out;
32         sort(nums.begin(),nums.end());
33         helper(nums,0,target,out);
34         return res;
35     }
36 };