代码还是这一块代码,但是还是写的很慢.. 其中用到了Java对 List的排序.查了很久,发现使用 Collections.sort 很方便. 另外对结果的去重,使用了 Java的HashSet. https://leetcode.com/problems/combination-sum-ii/ package com.company; import java.util.*; class Solution { Map<String, Set<List<Integer>>>…

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…

Combination Sum II 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 t…

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…

# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 40: Combination Sum IIhttps://oj.leetcode.com/problems/combination-sum-ii/ Given a collection of candidate numbers (C) and a target number (T),find all unique combinations in C where the ca…

Combination Sum II 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 ta…

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…

Combination Sum II 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 t…

39. Combination Sum 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 (inc…

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…

Combination sum: 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 (includ…

题目要求:Combination Sum II 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 (includ…

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…

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

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 posi…

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…

这两道题的基本思路和combination那一题是一致的,也是分治的方法. 其中combination Sum复杂一点,因为每个数可能用多次.仔细分析下,本质上也是一样的.原来是每个数仅两种可能.现在每个数有k +1中可能,k = target / i. 所以就是把简单的if else 分支变成for循环: vector<vector<int> > combHelper(vector<int>::iterator first, vector<int>::it…

[题目] Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target. Each number in candidates may only be used once in the combination. Note: A…

▶ 给定一个数组 和一个目标值.从该数组中选出若干项(项数不定),使他们的和等于目标值. ▶ 36. 数组元素无重复 ● 代码,初版,19 ms .从底向上的动态规划,但是转移方程比较智障(将待求数分解为左右两个半段,分别找解,拼在一起,再在接缝上检查是否是重复解). class Solution { public: vector<vector<int>> combinationSum(vector<int>& candidates, int target) {…

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 posi…

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…

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

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 posi…

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. (Medium) Note: All numbers (including target)…

Question 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) wil…

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 posi…

一天一道LeetCode系列 (一)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…

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

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

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…