Input: candidates = [10,1,2,7,6,1,5], target = 8, A solution set is: [ [1, 7], [1, 2, 5], [2, 6], [1, 1, 6] ] Example 2: Input: candidates = [2,5,2,1,2], target = 5, A solution set is: [ [1,2,2], [5] ] LeetCode 39. Combination Sum —— 组合总...
Example 2: Input: candidates = [2,5,2,1,2], target = 5, A solution set is: [ [1,2,2], [5] ] 与Combination Sum的区别在于,本题每次递归需要考虑重复元素,用while循环递归重复元素出现的次数;而Combination Sum每次递归只需要考虑两种情况,即放入该元素,或不放入该元素。 classSolution {publicLis...
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:
要求C中的每一个元素在一个组合中只能被使用一次。 For example, 输入候选数字集 [10, 1, 2, 7, 6, 1, 5] 和目标数字 8, 结果集应当是 [ [1, 7], [1, 2, 5], [2, 6], [1, 1, 6] ] 想法 首先这道题和39题CombinationSum非常的相像。唯一的差别就在于这道题要求,每一个元素只能被使...
translated from Java solution no need to use another var to keep track of used value class Solution: def combinationSum(self, candidates, target): self.nums = sorted(candidates) self.results = [] self.dfs(0, target, []) return self.results def dfs(self, start, target, result): if tar...
Combination Sum II 题目解析 给一组数和一个目标值,求和为目标值的组合。数组中的每个数最多只能取一次。 解题思路 本题与上一题LeetCode 39. Combination Sum十分相似,改变的条件是:数组中每个元素最多只能取一次。 在上一题中,我们是通过可以同一个起点(start)反复取同一个数得到结果,本题中为了避免重复,...
题目链接:https://leetcode.com/problems/combination-sum-ii/?tab=Description 给定数组,数组中的元素均为正数,target也是正数。(数组中的元素可能有重复) 求出所有的满足求和等于terget的组合。 数组中的元素只能使用一次。(数组中重复的元素可以最多使用重复次数) ...
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 target) will be positive integers. ...
[leetcode] Combination Sum and Combination SumII httpsjava网络安全 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. 全栈程序员站长 2022/07/06
(int i=start;i<candidates.length;i++){if(candidates[i]<target){for(List<Integer>ar:combinationSum(candidates,target-candidates[i],i)){ar.add(0,candidates[i]);res.add(ar);}}elseif(candidates[i]==target){List<Integer>lst=newArrayList<>();lst.add(candidates[i]);res.add(lst);}else...