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 num…

combination sum I.permutation I.subsets  I 是组合和.全排列.子集的第一种情况,给定数组中没有重复的元素. combination sum II.permutation II.subsets  II 是组合和.全排列.子集的第而种情况,给定数组中有重复元素. combination sum I中元素每个可以被多次使用,所以每次遍历都是从当前元素开始,然后往后面遍历. combination sum II中元素只能被使用一次,所以算下个元素时,只能从当前元…

引言 既上一篇 子集系列(一) 后,这里我们接着讨论带有附加条件的子集求解方法. 这类题目也是求子集,只不过不是返回所有的自己,而往往是要求返回满足一定要求的子集. 解这种类型的题目,其思路可以在上一篇文章的思路略作改进. 例 1,求元素数量为定值的所有子集 Combinations Given two integers n and k, return all possible combinations of k numbers out of 1 ... n. For example,If n …

[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…

[题目] 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 unlimi…

[题目] 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…

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…

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…

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…