[Leetcode] Combination Sum 组合数之和 Combination Sum I 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 (i...
Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations incandidateswhere the candidate numbers sums totarget. The same repeated number may be chosen fromcandidatesunlimited number of times. Note: All numbers (includingtarget) wi...
class Solution { public: vector<vector<int>> combinationSum(vector<int>& candidates, int target) { vector<vector<int>> res; vector<int> curr; sort(candidates.begin(), candidates.end()); dfs(res, curr, target, 0, candidates); return res; } void dfs(vector<vector<int>>& res, vector...
import java.util.List; import java.util.ArrayList; import java.util.HashSet; class Solution { public List<List<Integer>> combinationSum(int[] candidates, int target) { List<List<Integer>> res = new ArrayList<>(); if(candidates==null || candidates.length==0) return res; // Core(candida...
链接:https://leetcode-cn.com/problems/combination-sum-iii 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 2,解题思路 采用dfs递归+剪枝的方法,借助于堆栈来实现; sum记录当前的累加和,depth记录当前数组中元素数目,tem记录当前存储的元素(dfs的路径),ans最终答案; ...
请一键三连, 非常感谢LeetCode 力扣题解377. 组合总和 Ⅳ377. Combination Sum IV帮你深度理解 记忆化搜索算法 深度优先搜索 dfs bfs 广度优先搜索 回溯 暴力枚举, 视频播放量 284、弹幕量 0、点赞数 3、投硬币枚数 2、收藏人数 1、转发人数 0, 视频作者 程序员写代码, 作者
如何针对这四组变量进行解题,我从leetcode中找了combination I, II, III, IV四题,非常有代表性。 Combination Sum[1] Given asetof candidate numbers (candidates)(without duplicates)and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target. ...
publicclassSolution{publicList<List<Integer>>combinationSum(int[]candidates,int target){Arrays.sort(candidates);solve(candidates,0,target,newLinkedList<Integer>());returnans;}privateList<List<Integer>>ans=newLinkedList<>();privatevoidsolve(int[]candidates,int start,int target,LinkedList<Integer>current...
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给定一个无重复元素的数组 candidates 和一个目标数 target ,找出 candidates 中所有可以使数字和为 target 的组合。 candidates 中的数字可以无限制重复被选取。 说明: 所有数字(包括 target)都是正整数。 解集不能包含重复的组合。 示例: 输入: candidates = [2,3,6,7], target = 7, ...