packageleetcodeimport"math"funcmaxSubarraySumCircular(nums[]int)int{varmax1,max2,sumint// case: no circulationmax1=int(math.Inf(-1))l:=len(nums)fori:=0;i<l;i++{sum+=nums[i]ifsum>max1{max1=sum}ifsum<1{sum=0}}// case: circlingarr_sum:=0fori:=0;i<l;i++{arr_sum+=nums...
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
total+=num; }returnmax > 0 ? Math.max(total -min, max) : max; } } 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. lee哥真乃人中龙凤 https://leetcode.com/problems/maximum-sum-circular-subarray/discuss/178422/One-Pass...
Input: [-2,-3,-1] Output: -1 Explanation: 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 int maxSubArray(vector<int>& A) { int ans = INT_MIN; int cur = INT_MIN; for (int k ...
package leetcode type MyCircularQueue struct { cap int size int queue []int left int right int } func Constructor(k int) MyCircularQueue { return MyCircularQueue{cap: k, size: 0, left: 0, right: 0, queue: make([]int, k)} } func (this *MyCircularQueue) EnQueue(value int) bool...
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... ...
Explanation: Subarray [-1] has maximum sum -1 Constraints: -30000 <= A[i] <= 30000 <= A.length <= 30000 注意这里的数组是连成环的,因此最大子数组存在两种情况: 在数组内; 在数组的边缘部分,即数组的头尾相连 对于第2种情况,即将数组的和减去最小和的子数组即可。这里会使用到求解最大子数组和...
既然是子数组,则意味着必须是相连的数字,而由于环形数组的存在,说明可以首尾相连,这样的话,最长子数组的范围可以有两种情况,一种是正常的,数组中的某一段子数组,另一种是分为两段的,即首尾相连的,可以参见 [大神 lee215 的帖子](https://leetcode.com/problems/maximum-sum-circular-subarray/discuss/178422/...