首页 > 代码库 > Leetcode: Combination Sum

Leetcode: Combination Sum

2024-07-08 18:51:49   232人阅读

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]

典型的recursion方法，找符合要求的path，存入result的ArrayList中。所以方法还是建立一个ArrayList<ArrayList<Integer>> result, 建立一个ArrayList<Integer> path，用recursion当找到符合条件的path时，存入result中。

我在做这道题时遇到了一个问题：添加 path 进入 result 中时，需要这样res.add(new ArrayList<Integer>(path)); 如果直接res.add(path); 会出错

比如我遇到的错误是：Input:[1], 1    Output:[[]]     Expected:[[1]]，没能够把path: [1] 添加到res里面去，没有成功。（具体我现在也不知道为什么）

 1 public class Solution { 2     public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) { 3         ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>(); 4         ArrayList<Integer> path = new ArrayList<Integer>(); 5         java.util.Arrays.sort(candidates); 6         findpaths(res, path, candidates, target); 7         return res; 8     } 9     10     public void findpaths(ArrayList<ArrayList<Integer>> res, ArrayList<Integer> path, int[] candidates, int remain) {11         if (remain == 0) {12             res.add(new ArrayList<Integer>(path));13             return;14         }15         if (remain < 0) {16             return;17         }18         for (int c : candidates) {19             if (path.size() != 0 && c < path.get(path.size() - 1)) continue;20             path.add(c);21             remain = remain - c;22             findpaths(res, path, candidates, remain);23             remain = remain + c;24             path.remove(path.size() - 1);25         }26     }27 }

 


联系
我们