如果sum比k小,那么就需要扩大窗口,right右移(注意这里可能会到达最右端的判断);如果sum==k,满足条件,窗口缩小,left右移;如果sum>k,窗口缩小,left右移。 整理代码如下: 1publicintsubarraySum(int[] nums,intk) {2intleft = 0;3intright = 0;4intsum = nums[0];5intcount = 0;6while( right < nu...
Example 1: Input:nums = [1,1,1], k =2Output:2 解题思路 一、思路一 暴力解决 publicintsubarraySum(int[] nums,intk){intres=0;intlen=nums.length;for(inti=0;i<len;i++){intsum=nums[i];if(sum == k) res++;for(intj=i+1;j<len;j++){ sum += nums[j];if(sum == k) res +...
Given an array of integersnumsand an integerk, returnthe total number of subarrays whose sum equals tok. A subarray is a contiguousnon-emptysequence of elements within an array. Example 1: Input: nums = [1,1,1], k = 2 Output: 2 Example 2: Input: nums = [1,2,3], k = 3 Outp...
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/* * @lc app=leetcode id=560 lang=javascript * * [560] Subarray Sum Equals K *//** * @param {number[]} nums * @param {number} k * @return {number} */var subarraySum = function (nums, k) { const hashmap = {}; let acc = 0; let count = 0; for (let i = 0;...
Given an array nums and a target value k, find the maximum length of a subarray that sums to k. If there isn’t one, return 0 instead. Hashmap is also used in this problem. Iterate the array and calculate the sum, if map[sum-k] in the hashmap, update the maxlength=max(maxlength...
Leetcode 560. Subarray Sum Equals K Subarray Sum Equals K 题目大意:给定一个数组,要求找到不同的连续子序列,他们的和为k,问这样的连续子序列有多少个 题目思路:首先根据原始的想法,我们可以想到连续子序列的问题可以通过前缀和的方式进行解决,可以通过n^2的问题解决,但是这个做法并不太优,所以我们...
LeetCode Top 100 Liked Questions 560. Subarray Sum Equals K (Java版; Medium) 题目描述 AI检测代码解析 Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.
Path Sum III *//** * Definition for a binary tree node. * function TreeNode(val) { * this.val = val; * this.left = this.right = null; * } */function helper(root, acc, target, hashmap) { // see also : https://leetcode.com/problems/subarray-sum-equals-k/ if ...
classSolution {public:intmaxSubArrayLen(vector<int>& nums,intk) {if(nums.empty())return0;intres =0; unordered_map<int, vector<int>>m; m[nums[0]].push_back(0); vector<int> sum =nums;for(inti =1; i < nums.size(); ++i) { ...