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 ar... 查看原文 L
package leetcode import "math" func maxSubarraySumCircular(nums []int) int { var max1, max2, sum int // case: no circulation max1 = int(math.Inf(-1)) l := len(nums) for i := 0; i < l; i++ { sum += nums[i] if sum > max1 { max1 = sum } if sum < 1 { sum ...
1classSolution2{3public:4intmaxSumCycle(vector<int>&vec,int&left,int&right)5{6intmaxsum = INT_MIN, curMaxSum =0;7intminsum = INT_MAX, curMinSum =0;8intsum =0;9intbegin_max =0, begin_min =0;10intminLeft, minRight;11for(inti =0; i < vec.size(); i++)12{13sum +=vec...
唯一的 corner case 是如果整个数组都是由负数组成的,那么整个数组的和 sum 跟数组连续的,最小的子数组的和是一样的。最后比较的时候,需要看看到底是 case 1 的结果更大,还是 case 2 里面的 min subarray 更小。 注意代码里全局最大值和全局最小值为什么设置成第一个元素的值而不能设置成 0 是因为如果整个...
Can you solve this real interview question? Maximum Sum Circular Subarray - Given a circular integer array nums of length n, return the maximum possible sum of a non-empty subarray of nums. A circular array means the end of the array connects to the beg
Explanation: Subarray [-1] has maximum sum -1 1. 2. 3. Note: -30000 <= A[i] <= 30000 1 <= A.length <= 30000 分析 题目的意思是:求一个循环数组的子序列和的最大值,代码一是我自己包里破解的,分了三种情况,所有的都为负,所有的都为正,有正也有负。比较死板。
Input:[-2,-3,-1]Output:-1Explanation:Subarray [-1] has maximum sum -1 Note: -30000 <= A[i] <= 30000 1 <= A.length <= 30000 这道题比较好理解,关于最大子数组的解法 可以参考 Loading...leetcode.com/problems/maximum-subarray/ ...
Given acircular integer arraynums of length n, returnthe maximum possible sum of a non-emptysubarrayofnums. Acircular arraymeans the end of the array connects to the beginning of the array. Formally, the next element of nums[i] is nums[(i + 1) % n] and the previous element of nums...
LeetCode 689. Maximum Sum of 3 Non-Overlapping Subarrays 2019-12-09 12:51 −原题链接在这里:https://leetcode.com/problems/maximum-sum-of-3-non-overlapping-subarrays/ 题目: In a given array nums of positive integers, find three... ...
唯一的corner case是如果整个数组都是由负数组成的,那么整个数组的和sum跟数组连续的,最小的子数组的和是一样的。最后比较的时候,需要看看到底是case 1的结果更大,还是case 2里面的min subarray更小。 时间O(n) 空间O(1) Java实现 1classSolution {2publicintmaxSubarraySumCircular(int[] A) {3inttotal =...