题目: Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. Each number in C may only be used once in the combination. Note: All numbers (including target) will be…

题目 Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. Each number in C may only be used once in the combination. Note: All numbers (including target) will be p…

题目: 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) w…

题目 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) wi…

1. 原题链接 https://leetcode.com/problems/combination-sum-ii/description/ 2. 题目要求 给定一个整型数组candidates[ ]和目标值target,找出数组中累加之后等于target的所有元素组合 注意:(1)每个可能的答案中,数组中的每一个元素只能使用一次:(2)数组存在重复元素:(3)数组中都是正整数:(4)不能存在重复解 3. 解题思路 这与第39题 Combination Sum 看起来很是类似,但一些细节要求完全不…

39. Combination Sum 依旧与subsets问题相似,每次选择这个数是否参加到求和中 因为是可以重复的,所以每次递归还是在i上,如果不能重复,就可以变成i+1 class Solution { public: vector<vector<int>> combinationSum(vector<int>& candidates, int target) { vector<vector<int>> result; vector…

Leetcode之回溯法专题-40. 组合总和 II(Combination Sum II) 给定一个数组 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 中的每个数字在每个组合中只能使用一次. 说明: 所有数字(包括目标数)都是正整数. 解集不能包含重复的组合. 示例 1: 输入: candidates = [10,1,2,7,6,1,5], target = 8, 所求解集为: [ [1, 7…

leetcode - 40. Combination Sum II - Medium descrition Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. Each number in C may only be used once in the combinat…

乘风破浪:LeetCode真题_040_Combination Sum II 一.前言 这次和上次的区别是元素不能重复使用了,这也简单,每一次去掉使用过的元素即可. 二.Combination Sum II 2.1 问题 2.2 分析与解决 通过分析我们可以知道使用递归就可以解决问题,并且这次我们从头遍历一次就不会出现多次使用某一个元素了. class Solution { List<List<Integer>> ans; public List<List<Intege…

1. Combination Sum Given a set of candidate numbers (C) (without duplicates) 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. No…