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…

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…

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…

# -*- 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…

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…

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…

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…

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…