题目链接: Maximum Sum of Distinct Subarrays With Length K: https://leetcode.com/problems/maximum-sum-of-distinct-subarrays-with-length-k/ 长度为 K 子数组中的最大和: https://leetcode.cn/problems/maximum-sum-of-distinct-subarrays-with-length-k/ LeetCode 日更第318天,感谢阅读至此的你 欢迎...
[LeetCode] 1695. Maximum Erasure Value You are given an array of positive integersnumsand want to erase a subarray containing unique elements. The score you get by erasing the subarray is equal to the sum of its elements. Returnthe maximum score you can get by erasing exactly one subarray....
Given therootof a binary tree, the level of its root is1, the level of its children is2, and so on. Return the smallest levelXsuch that the sum of all the values of nodes at levelXis maximal. Example 1: Input:[1,7,0,7,-8,null,null] Output:2 Explanation: Level 1 sum = 1. ...
题目不算太难的。 classSolution{publiclongmaxMatrixSum(int[][]matrix){long sum=0;int counter=0;int min=Integer.MAX_VALUE;int n=matrix.length;for(int i=0;i<n;i++){for(int j=0;j<matrix[0].length;j++){if(matrix[i][j]<0)counter++;matrix[i][j]=Math.abs(matrix[i][j]);if(...
1191 K-Concatenation Maximum Sum K 次串联后最大子数组之和 Description: Given an integer array arr and an integer k, modify the array by repeating it k times. For example, if arr = [1, 2] and k = 3 then the modified array will be [1, 2, 1, 2, 1, 2]. ...
题目leetcode 918 Maximum Sum Circular Subarray 给一个环形数组,也就是数组的尾部可以和头部相连,然后求子数组的最大和 比如 Input: [5,-3,5] Output: 10 Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10 分析和解法 做这个题目首先我们需要熟悉普通数组的最大子数组和的解法,思路是动态规划,...
1和two sum类似,建立一个hashtable,hashtable中累积和是key,index是value 2 如果当前的累积和acc减k存在在hashtable中,则求得长度,由于最后是求最长的subarray,所以其中会有个比较的过程 3 这题和560. Subarray Sum Equals K的区别在于,这题要求最大size,所以hashmap中存的是累积和和index了,560存的是累积和...
Return the maximum number of non-empty non-overlapping subarrays such that the sum of values in each subarray is equal to target. Example 1: Input: nums = [1,1,1,1,1], target = 2 Output: 2 Explanation: There are 2 non-overlapping subarrays [1,1,1,1,1] with sum equals to targ...
假设我们可以让n台电脑同时运行x分钟,那么对于电量大于x的电池,其只能被使用x分钟。因此每个电池的使用时间至多为min(batteries[i],x),我们将其累加起来,记作sum。那么要让n台电脑同时运行x分钟,必要条件是n⋅x≤sum。 下面证明该条件是充分的,即当n⋅x≤sum成立时,必然可以让n台电脑同时运行x...
(nums, middle+1, high); // 计算中间的情况 // 从middle开始,向两边扩散: 先找贴着middle左边的最大值,再找贴着middle右边的最大值 // left // int sum = nums[middle]; int sum = 0; int middle_max_l = INT_MIN; int middle_max_r = INT_MIN; // for(int i = middle-1; ...