代码如下: classSolution(object):defcheckSubarraySum(self, nums, k):""":type nums: List[int] :type k: int :rtype: bool"""continuousZero=Falseforiinxrange(len(nums) - 1):ifnums[i] + nums[i + 1] ==0: continuousZero=T
bool checkSubarraySum(vector<int>& nums, int k) { for (int i = 0; i < nums.size(); ++i) { int sum = nums[i]; for (int j = i + 1; j < nums.size(); ++j) { sum += nums[j]; if (sum == k) return true; if (k != 0 && sum % k == 0) return true; } }...
力扣leetcode-cn.com/problems/continuous-subarray-sum/ 题目描述 给你一个整数数组 nums 和一个整数 k ,编写一个函数来判断该数组是否含有同时满足下述条件的连续子数组: 子数组大小 至少为 2 ,且 子数组元素总和为 k 的倍数。 如果存在,返回 true ;否则,返回 false 。 如果存在一个整数 n ,令整数 x...
力扣leetcode-cn.com/problems/continuous-subarray-sum/ 题目描述 给定一个包含 非负数 的数组和一个目标 整数 k,编写一个函数来判断该数组是否含有连续的子数组,其大小至少为 2,总和为 k 的倍数,即总和为 n*k,其中 n 也是一个 整数。 示例1: 输入: [23,2,4,6,7], k = 6 输出: True 解释:...
classSolution {public:boolcheckSubarraySum(vector<int>& nums,intk) {intn = nums.size(), sum =0; unordered_map<int,int> m{{0,-1}};for(inti =0; i < n; ++i) { sum+=nums[i];intt = (k ==0) ? sum : (sum %k);if(m.count(t)) {if(i - m[t] >1)returntrue; ...
Given an integer array, you need to find one continuous subarray that if you only sort this subarray in ascending order, then the whole array will be sorted in ascending order, too. You need to find the shortest such subarray and output its length. Example 1: Input: [2, 6, 4, 8, ...
Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k. Example 1: Input:nums = [1,1,1], k = 2 Output: 2 1. 2. Constraints: The length of the array is in range [1, 20,000]. ...
Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k. Example 1: Input:nums = [1,1,1], k = 2 Output: 2 1. 2. Note: The length of the array is in range [1, 20,000]. ...
523Continuous Subarray SumPythonO(n) solution using dict() 538Convert BST to Greater TreePythonJavaRight first DFS with a variable recording sum of node.val and right.val. 1. Recursive. 2. Stack 3. Reverse Morris In-order Traversal
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