19.2.3 [LeetCode 39] Combination Sum

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

The same repeated number may be chosen from candidates unlimited number of times.

Note:

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

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

 1 class Solution {
 2 public:
 3     void getsum(vector<vector<int>>&res,vector<int> ans,vector<int>&candidates, int target, int lastidx) {
 4         if (target == 0) {
 5             res.push_back(ans);
 6             return;
 7         }
 8         if (target < candidates[lastidx])return;
 9         for (int i = lastidx; 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>> combinationSum(vector<int>& candidates, int target) {
17         sort(candidates.begin(), candidates.end());
18         vector<vector<int>>res;
19         vector<int>ans;
20         getsum(res, ans, candidates, target, 0);
21         return res;
22     }
23 };

感觉可以用dp，不知道会不会快

原文地址：https://www.cnblogs.com/yalphait/p/10350697.html