19.2.3 [LeetCode 40] Combination Sum II

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

Each number in candidates may only be used once in the combination.

Note:

• All numbers (including target) will be positive integers.
• The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

Example 2:

Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
  [1,2,2],
  [5]
]

 1 class Solution {
 2 public:
 3     void getsum(set<vector<int>>&res,vector<int> ans,vector<int>&candidates, int target, int lastidx) {
 4         if (target == 0) {
 5             res.insert(ans);
 6             return;
 7         }
 8         if (target < candidates[lastidx])return;
 9         for (int i = lastidx+1; i < candidates.size(); i++) {
10             if (target < candidates[i])break;
11             ans.push_back(candidates[i]);
12             getsum(res, ans, candidates, target - candidates[i], i);
13             ans.pop_back();
14         }
15     }
16     vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
17         sort(candidates.begin(), candidates.end());
18         set<vector<int>>res;
19         vector<int>ans;
20         getsum(res, ans, candidates, target, -1);
21         return vector<vector<int>>(res.begin(),res.end());
22     }
23 };

上一题稍微改了一下

原文地址：https://www.cnblogs.com/yalphait/p/10351182.html