Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target. The same repeated number may be chosen from candidates unlimited n…

# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 39: Combination Sumhttps://oj.leetcode.com/problems/combination-sum/ Given a set of candidate numbers (C) and a target number (T),find all unique combinations in C where the candidate numbe…

leetcode - 39. Combination Sum - Medium descrition 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…

39. 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. N…

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…

Question 39. Combination Sum Solution 分析:以candidates = [2,3,5], target=8来分析这个问题的实现,反向思考,用target 8减2,3,5这三个数,等到target为0的时候,所减过的数就是我们要求的一个结果. Java实现: public List<List<Integer>> combinationSum(int[] candidates, int target) { List<List<Integ…

▶ 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 set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target. The same repeated number may be chosen from candidates unlimited n…

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 (includi…

Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target. Example: nums = [1, 2, 3] target = 4 The possible combination ways are: (1, 1, 1, 1) (1, 1, 2) (1,…

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers. Note: All numbers will be positive integers. The solution set must not cont…

题目描述 给定一个无重复的正整数数组 candidates 和一个正整数 target, 求所有和为 target 的 candidates 中数的组合中.其中相同数的不同顺序组合算做同一种组合,candidates 中的数可以重复使用. 算法一 首先想到的方法就是枚举所有的组合可能性,判断其和是否为target.枚举的方法可以使用递归,对candidates中每一个数,有“加入组合”和“不加入组合”两种选择,每一种选择又可以向后面元素的不同选择递归,直到candidate中最后一个元素.可以用…

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)…

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…

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…

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. Note: All numbers (in…

一天一道LeetCode系列 (一)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 (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 (including target) will…

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target. The same repeated number may be chosen from candidates unlimited n…

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 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. Note: All numbers (in…

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…

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. Note: All numbers (in…

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target. The same repeated number may be chosen from candidates unlimited n…

题目链接: https://leetcode.com/problems/combination-sum/?tab=Description   Problem: 给定数组并且给定一个target,求出所有满足求和等于target的数字组合   遍历所有的数组中元素,然后对target进行更新,将该元素添加到tempList中,直到remain等于0时达到条件,可以将该tempList添加到list中   注意:每个元素可以使用多次,因此每次的遍历都要从上次的那个下标开始.   当target更新到…

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target. The same repeated number may be chosen from candidates unlimited n…

思路:先对数据进行排序(看评论给的测试数据好像都是有序数组了,但题目里没有给出这个条件),然后回溯加剪枝即可. class Solution { public: ; vector<vector<int>> ans; void search(int pos, vector<int>& candidates, int target, vector<int>& res, int sums){ if(sums == target){ ans.push…

1. 原题链接 https://leetcode.com/problems/combination-sum/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)数组中的每一个元素可以重复用:(2)数组中不存在重复元素:(3)数组中都是正整数 3. 解题思路 采用迭代的方法检验所有元素组合 4. 代码实现 import java.util.ArrayList; import java…

https://leetcode.wang/leetCode-39-Combination-Sum.html 描述 Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target. The same…