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, r] 内每个数字的出现次数 nu...
you want to choose a subarray and optionally delete one element from it so that there is still at least one element left and the sum of the remaining elements is maximum possible.
you want to choose a subarray and optionally delete one element from it so that there is still at least one element left and the sum of the remaining elements is maximum possible.
Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.A subarray is a contiguous part of an array. 英文版地址 leetcode.com/problems/m 中文版描述 给你一个整数数组 nums ,请你找出一个具有最大和的连续子数组(...
如果每次得到新的candidate都和全局的maxSum进行比较,那么必然能找到最大的max sum subarray. 在循环过程中,用maxSum记录历史最大的值。从nums[0]到nums[n-1]一步一步地进行。 思路二: 遍历array,对于每一个数字,我们判断,(之前的sum + 这个数字) 和 (这个数字) 比大小,如果(这个数字)自己就比 (之前的su...
[Leetcode][python]Maximum Subarray/最大子序和 题目大意 由N 个整数元素组成的一维数组 (A[0], A[1],…,A[n-1], A[n]),这个数组有很多连续子数组,那么其中数组之和的最大值是什么呢? 子数组必须是连续的。 不需要返回子数组的具体位置。
class MaximumSubarrayPrefixSum { public int maxSubArray(int[] nums) { int len = nums.length; int maxSum = Integer.MIN_VALUE; int sum = 0; for (int i = 0; i < len; i++) { sum = 0; for (int j = i; j < len; j++) { sum += nums[j]; maxSu...
Explanation: The subarray[1, -1, 5, -2]sums to 3 and is the longest. 1. 1. 1. 首先计算出数组的前缀和,数组变为[1, 0, 5,3,6]。接着遍历这个前缀和的数组,用hashmap记录<prefix sum, i> -每个不同的前缀和和他们出现的位置。如果此时发现nums[i] - k在hashmap中,则res = Math.max(...
publicclassLeetCode_053 {publicstaticintmaxSubArray(int[] nums) {intmax=nums[], sum=nums[];for (inti=1; i<nums.length; i++) {if (sum<=) {sum=nums[i]; } else {sum=sum+nums[i]; }max=Math.max(max, sum); }returnmax; }publicstaticvoidmain(String[] args) {int[...
classSolution:defmaxSubArray(self,nums):sum=0max=nums[0]foriinnums:ifsum+i>0and sum>=max:sum+=i max=sumreturnmax 但是如果示例是和小于0呢? 比如样例是 [-2, -1] 的情况下, 上述的代码就覆盖不了了. 因此还需判断和小于0的情况, 小于0时直接替换, 并于当前最大值比较即可. ...