# -*- 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…

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…