So I change the format of the sub problem into something like:maxSubArray(int A[], int i), which means the maxSubArray for A[0:i ] which must has A[i] as the end element. Note that now the sub problem's format is less flexible and less powerful than the previous one because the...
问题简介 本文将介绍计算机算法中的经典问题——最大子数组问题(maximum subarray problem)。所谓的最大子数组问题,指的是:给定一个数组A,寻找A的和最大的非空连续子数组。比如,数组 A = [-2, -3, 4, -1, -2…
LeetCode - The World's Leading Online Programming Learning Platformleetcode.com/problems/maximum-gcd-sum-of-a-subarray/description/ 简单介绍一下题目大意,更详细的描述可以见链接中的原题描述及样例数据: 给定一个长度为n的正整数数组nums。对于nums的任意一个子数组(连续、非空的一段),定义s为子数组...
唯一的边界条件是如果全部为负数,那么minsubarray = sum,这个时候直接返回max即可。 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 public int maxSubarraySumCircular(int[] A) { int n = A.length; int sum = 0; for (int num : A) sum += num; int max = A[0]; int[]...
本文将介绍计算机算法中的经典问题——最大子数组问题(maximum subarray problem)。所谓的最大子数组问题,指的是:给定一个数组A,寻找A的和最大的非空连续...
Given a circular array C of integers represented byA, find the maximum possible sum of a non-empty subarray of C. Here, acircular arraymeans the end of the array connects to the beginning of the array. (Formally,C[i] = A[i]when0 <= i < A.length, andC[i+A.length] = C[i]wh...
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]<...
When an overlapping property needs to be observed, previous algorithms for the maximum sum are not directly applicable. We designed an O(K * n) algorithm for the K maximum subsequences problem. This was then modified to solve the K maximum subarrays problem in O(K * n3) time. Finally, ...
I am getting WA on test 28 ofthisproblem. Would really appreciate all the help. My Approach Basically, I am partitioning each segment[l…r][l…r]into three partitions,p1,p2,p3p1,p2,p3where p2p2= maximum subarray sum in[l…r][l…r] ...
(5kyu) The maximum sum subarray problem consists in finding the maximum sum of a contiguous subsequence in an array or list of integers: maxSequence([-2 , 1, -3, 4, -1 , 2, 1, -5, 4]) # should be 6: [4, -1, 2, 1] ...