Combination Sum

最新推荐文章于 2025-05-25 18:32:47 发布

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最新推荐文章于 2025-05-25 18:32:47 发布

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原文链接：http://www.cnblogs.com/FLAGyuri/p/5863499.html

本文介绍了一种解决组合求和问题的算法实现。该算法针对一组候选数和目标数，找出所有可能的组合使得组合内元素之和等于目标数。算法允许候选数可以重复使用，并确保结果中不含重复的组合。

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.
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]
]

 1 public class Solution {
 2     public List<List<Integer>> combinationSum(int[] candidates, int target) {
 3         List<List<Integer>> result = new ArrayList<>();
 4         if (candidates == null || candidates.length == 0) {
 5             return result;
 6         }
 7         List<Integer> list = new ArrayList<>();
 8         Arrays.sort(candidates);
 9         helper(candidates, result, list, 0, target);
10         return result;
11     }
12     
13     private void helper(int[] candidates, List<List<Integer>> result, List<Integer> list, int index, int target) {
14         if (target == 0) {
15             result.add(new ArrayList<Integer>(list));
16         }
17         for (int i = index; i < candidates.length; i++) {
18             if (candidates[i] > target) {
19                 break;
20             }
21             if (i != index && candidates[i] == candidates[i - 1]) {
22                 continue;
23             }
24             list.add(candidates[i]);
25             helper(candidates, result, list, i, target - candidates[i]);
26             list.remove(list.size() - 1);
27         }
28     }
29 }