Combination Sum

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 · 57 阅读

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原文链接：http://www.cnblogs.com/luckygxf/p/4239854.html

本文介绍了一种基于回溯法的组合求和算法，用于找出候选数集合中所有加和等于目标值的组合。该算法首先对输入数组进行排序，然后通过递归的方式寻找所有可能的组合，并确保组合元素按非递减顺序排列且不包含重复项。

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]

思路

先对数组进行升序排列，然后在用回溯法BS，这里用的递归调用实现的

 1 public class Solution {
 2     public List<List<Integer>> combinationSum(int[] candidates, int target) {
 3         Arrays.sort(candidates);
 4         List<List<Integer>> result = new ArrayList<List<Integer>>();
 5         getResult(result, new ArrayList<Integer>(), candidates, target, 0);
 6         
 7         return result;
 8     }
 9     
10     private void getResult(List<List<Integer>> result, List<Integer> cur, int candidates[], int target, int start){
11         if(target > 0){
12             for(int i = start; i < candidates.length && target >= candidates[i]; i++){
13                 cur.add(candidates[i]);
14                 getResult(result, cur, candidates, target - candidates[i], i);
15                 cur.remove(cur.size() - 1);
16             }//for
17         }//if
18         else if(target == 0 ){
19             result.add(new ArrayList<Integer>(cur));
20         }//else if
21     }
22 }