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一、问题描述
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]
二、问题分析
1、采用递归算法；
2、详见代码注释。
三、算法代码
public class Solution {public List<List<Integer>> combinationSum(int[] candidates, int target) {Arrays.sort(candidates);ArrayList<Integer> cur = new ArrayList<Integer>();List<List<Integer>> result = new ArrayList<List<Integer>>();dfs(candidates, target, 0, cur, result);return result;}public void dfs(int[] candidates, int gap, int start, ArrayList<Integer> cur, List<List<Integer>> result){if(gap == 0){result.add(cur);return;}for(int i = start; i < candidates.length; i++){if(gap < candidates[i]){return;}cur.add(candidates[i]);//做两点说明//1、因为元素可以重复，所以第三个参数是i而不是i + 1//2、第四个参数是new ArrayList(cur)，而不是cur，需要每次递归都创建一个保存中间结果的容器dfs(candidates, gap - candidates[i], i, new ArrayList(cur), result);cur.remove(cur.size() - 1);}}
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