代码运行次数:0 publicbooleancheckSubarraySum(int[]nums,int k){Map<Integer,Integer>map=newHashMap<Integer,Integer>();map.put(0,-1);//为了处理nums=[0,0] k=-1这样的情况int sum=0;for(int i=0;i<nums.length;i++){sum+=nums[i];if(k!=0)sum%=k;Integer prev=map.get(sum);if(prev!
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; } }...
Can you solve this real interview question? Continuous Subarray Sum - Given an integer array nums and an integer k, return true if nums has a good subarray or false otherwise. A good subarray is a subarray where: * its length is at least two, and * t
[leetcode]523. Continuous Subarray Sum 从数组中找到子串的和是给定值得倍数 哈希表法的精髓就是,到ab两个位置的和对target取余结果一样的话,ab之间的和肯定是target的整数倍 publicbooleancheckSubarraySum(int[] nums,intk) {intl =nums.length;/*两种方法,动态规划和哈希表 动态规划就是把长度为2-数组size...
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:...
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: AI检测代码解析 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
523 Continuous Subarray Sum 21.3 Medium 522 Longest Uncommon Subsequence II 28.10% Medium 521 Longest Uncommon Subsequence I 50.70% Easy 520 Detect Capital 54.20% Easy 519 Random Flip Matrix 32.20% Medium 518 Coin Change 2 33.20% Medium 517 Super Washing Machines 34.60% Hard 516 Longest Palindromic...