class Solution: def maximumSubarraySum(self, nums: List[int], k: int) -> int: # ans 维护所有长度为 k 且数字各不相同的子数组中,子数组和的最大值 ans: int = 0 # sum 维护当前滑动窗口 [l, r] 内的数字和 sum: int = 0 # num_to_cnt 表示滑动窗口 [l,
Given a circular array C of integers represented by A, find the maximum possible sum of a non-empty subarray of C. Here, a circular array means the end of the array connects to the beginning of the array. (Formally, C[i] = A[i] when 0 <= i < A.length, and C[i+A.length] ...
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(...
1)和maxSum( 2)两种情况; 当sum > 0 时, 有 maxSum(2) + (k-2)*sum. Java代码classSolution{publicintkConcatenationMaxSum(int[] arr,intk){if( arr.length ==0|| arr ==null)return0;if(k <3)return(int) (maxSum(arr, k)%(1e9+7));longsum =0;for(intnum:arr) sum += num;lon...
then get an empty subarray to make the sum equals to 0. Constraints: 1 <= arr.length <= 10^5 -10^4 <= arr[i] <= 10^4 解题思路:假设删除arr[i]后可以获得有最大和的子数组,那么这个子数组的的和就相当于被arr[i]分成了两部分,只要找出这样的 x: i-1 >x>=0,y: i+1 <y < len...
Can you solve this real interview question? Maximum Matrix Sum - You are given an n x n integer matrix. You can do the following operation any number of times: * Choose any two adjacent elements of matrix and multiply each of them by -1. Two elements
[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....
LeetCode 325. Maximum Size Subarray Sum Equals k 若谷 追求卓越,成功就会在不经意间追上你。 来自专栏 · 今日事 题目: 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. Note:The sum of the entire ...
lSum); int rSum = max(r.rSum, r.iSum + l.rSum); int mSum = max(max(l.mSum, r.mSum), l.rSum + r.lSum); return (Status) {lSum, rSum, mSum, iSum}; }; Status get(vector<int> &a, int l, int r) { if (l == r) { return (Status) {a[l], a[l], a[l], a[...
private int maxValue = Integer.MIN_VALUE; public int maxPathSum(TreeNode root) { maxPathSumHelper(root); return maxValue; } public int maxPathSumHelper(TreeNode root) { if (root == null) return 0; //左子节点的值 int left = maxPathSumHelper(root.left); //右子节点的值 int right...