Combination Sum

最新推荐文章于 2025-05-25 18:32:47 发布

weixin_30662011

最新推荐文章于 2025-05-25 18:32:47 发布

阅读量51

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/Kai-Xing/p/3916538.html

本文针对给定候选数字集合及目标数字，寻找所有可能的数字组合使得这些数字之和等于目标值。通过改进的深度优先搜索算法实现，保证了组合内数字的非递减顺序，并避免重复组合。

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

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

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 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]

分析：这道题类似combinaions和subsets，不同的是，在本题的可行解中可以存在不限数目的相同的数，同样我们用改良的dfs方法求解。因为此题要求每个combination中元素必须是非递减排列的，所以我们先将元素按升序排序。此外由于元素全为正数，当当前path的sum大于target时，即可停止该path的搜素。代码如下：

 1 class Solution {
 2 public:
 3     vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
 4         vector<vector<int>> res;
 5         vector<int> path;
 6         sort(candidates.begin(),candidates.end());
 7         dfs(res,path,candidates,0,0,target);
 8         return res;
 9     }
10     void dfs(vector<vector<int>> &res, vector<int> & path, vector<int> & c, int start, int sum, int target){
11         for(int i = start; i < c.size(); i++){
12             path.push_back(c[i]);
13             if(sum + c[i] == target)res.push_back(path);
14             else if(sum + c[i] < target){
15                 dfs(res,path,c,i,sum+c[i],target);
16             }
17             path.pop_back();
18         }
19     }
20 };