代码 1 class Solution { 2 public: 3 vector<vector<int>> combinationSum3(int k, int n) { 4 vector<vector<int>> res; 5 if(k <= 0 || n <= 0) return res; 6 vector<int> nums, temp; 7 for(int i = 0; i < 9; i++) 8 nums.push_back(i + 1); 9 combine(nums, 0, 0...
Find all possible combinations ofknumbers that add up to a numbern, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers. Example 1: Input:k= 3,n= 7 Output: [[1,2,4]] Example 2: Input:k= 3,n= 9 Output: [[1,2,6], [1...
虽然这个解法在leetcode上是超时的,但是我们把这道题放在这里,并且用backtracking的方法进行解答,主要的目的是介绍在不同条件下,我们是如何应对的combination sum这一系列题目的。 classSolution:defcombinationSum4(self,nums:List[int],target:int)->int:res=[]nums.sort()self.dfs(nums,target,[],res)returnlen...
def combination_sum3(k, n) if (k > 9) or (k < 1) or (n > 9) or (n < 1) return $res end f = $f + 1 if (n < (2 * f + k - 1 ) * k / 2) && (k != 1 ) return $res end if k != 1 for i in f .. n/k do $f = i $tmp << i combination_sum3(...
[LeetCode] 216. Combination Sum III Find all possible combinations ofknumbers that add up to a numbern, 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....
LeetCode: 216. Combination Sum III 题目描述 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 numbe...
示例3: 输入: candidates = [2], target = 1 输出: [] 提示: 1 <= candidates.length <= 30 2 <= candidates[i] <= 40 candidates 的所有元素 互不相同 1 <= target <= 40 题目难度:中等 通过次数:1.2M 提交次数:1.6M 贡献者:LeetCode 相关标签 相似题目 C++...
你这题保熟吗 Leetcode 1074. Number of Submatrices That Sum to Target 39 -- 12:02 App 你这题保熟吗 Leetcode 1473. Paint House III 34 -- 2:39 App 你这题保熟吗 Leetcode 80. Remove Duplicates from Sorted Array II 31 -- 7:14 App 你这题保熟吗 Leetcode 258. Add Digits 69 -...
leetcode39黑板上排列组合你舍得解开吗[Combination Sum][Python], 视频播放量 187、弹幕量 0、点赞数 9、投硬币枚数 5、收藏人数 2、转发人数 0, 视频作者 咖啡猫啡咖, 作者简介 ,相关视频:leetcode2小学加法练习题[Add Two Numbers],leetcode18得寸进尺的四数之和[4Sum
递进关系就是,sum 代表要求的和,如果想求 opt [ sum ] ,就遍历给定的数组 nums,然后分两种情况。 如果sum 刚好等于 nums [ i ],那么就直接把 nums [ i ] 加到 list 里,算作一种情况。 例如nums = [ 2, 3, 6, 7 ] , target = 7。 当求sum = 3 的时候,也就是求 opt [ 3 ] 的时候,此...