Combination Sum II

最新推荐文章于 2025-05-25 18:32:47 发布

转载最新推荐文章于 2025-05-25 18:32:47 发布 · 51 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/amazingzoe/p/4850635.html

本文介绍了一种使用递归方法解决组合总和问题的算法，通过C++代码实现了给定候选数集合和目标数求所有可能组合的解决方案。文章详细解释了算法逻辑、代码实现及示例应用。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

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

Note:

All numbers (including target) will be positive integers.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
The solution set must not contain duplicate combinations.

For example, given candidate set 10,1,2,7,6,1,5 and target 8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]

 1 class Solution {
 2 public:
 3     vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
 4         sort(candidates.begin(), candidates.end());
 5         vector<vector<int> > result;
 6         vector<int> temp;
 7         
 8         dfs(candidates, target, result, temp, 0);
 9         return result;
10     }
11     
12     void dfs(vector<int>& candidates, int diff, vector<vector<int> >& result, vector<int>& temp, int start){
13         if(diff == 0){
14             result.push_back(temp);
15             return;
16         }
17         
18         for(int i = start; i < candidates.size(); i++){
19             if(diff < candidates[i]) return;
20             
21             //only add the first number into the vector if there are many
22             if(i > start && candidates[i] == candidates[i - 1]) continue;
23             
24             temp.push_back(candidates[i]);
25             dfs(candidates, diff - candidates[i], result, temp, i + 1);
26             temp.pop_back();
27         }
28         
29     }
30 };

转载于:https://www.cnblogs.com/amazingzoe/p/4850635.html