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 su
Maximum Sum Subarray of Size K (easy) 问题描述 解决方法 代码 课后回顾 sliding window pattern 滑窗应用场景 滑动窗口模式(sliding window pattern)是用于在给定数组或链表的特定窗口大小上执行所需的操作,比如寻找包含所有 1 的最长子数组。从第一个元素开始滑动窗口并逐个元素地向右滑,并根据你所求解的问题调整...
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Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and returnits sum. Example 1: Input: nums=[-2,1,-3,4,-1,2,1,-5,4] Output:6 Explanation: [4,-1,2,1] has the largest sum=6. 1. 2. 3. Exampl...
* @param k: a target value * @return: the maximum length of a subarray that sums to k*/intmaxSubArrayLen(vector<int> &nums,intk) {//Write your code hereunordered_map<int,int>m; m[0] = -1;intres =0;intsum =0;for(inti =0;i < nums.size();i++){ ...
Given an arraynumsand a target valuek, find the maximum length of a subarray that sums tok. If there isn't one, return 0 instead. Note: The sum of the entirenumsarray is guaranteed to fit within the 32-bit signed integer range. ...
LeetCode53 Maximum sum of subarray classic dp: dp[i] represents the maximum sum of subarray which ends in nums[i], and dp[i] = Math.max(nums[i], dp[i - 1] + nums[i]). and since we have to include nums[i] due to it’s on the defination of dp[i], and when dp[i-1]<...
Given an integer array nums and two integers firstLen and secondLen, return the maximum sum of elements in two non-overlapping subarrays with lengths firstLen and secondLen. The array with length firstLen could occur before or after the array with length secondLen, but they have to be non...
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in prefixSum array, find the maxisum size subarray sum equals k// sum of subarray (i, j) = prefixSum [j] - prefixSum[i]Map<Integer,Integer>prefixSumVsIndex=newHashMap<>();prefixSumVsIndex.put(0,-1);intmaxSize=0;intprefixSum=0;for(intj=0;j<nums.length;j++){prefixSum+=nums[...