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] = C[i] when i >= 0.)
Also, a subarray may only include each element of the fixed buffer A at most once. (Formally, for a subarray C[i], C[i+1], ..., C[j], there does not exist i <= k1, k2 <= j with k1 % A.length = k2 % A.length.)
Example 1:
Input: [1,-2,3,-2]
Output: 3
Explanation: Subarray [3] has maximum sum 3
Example 2:
Input: [5,-3,5]
Output: 10
Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10
Example 3:
Input: [3,-1,2,-1]
Output: 4
Explanation: Subarray [2,-1,3] has maximum sum 2 + (-1) + 3 = 4
Example 4:
Input: [3,-2,2,-3]
Output: 3
Explanation: Subarray [3] and [3,-2,2] both have maximum sum 3
Example 5:
Input: [-2,-3,-1]
Output: -1
Explanation: Subarray [-1] has maximum sum -1
Solution
如果最大子数组出现在数组的中间部分，那么我们用当前最大和全局最大的方式得到。
如果最大子数据是在数组的两头，我们的思路是用数组的和减去全局最小。
还有一种情况，数组全是负数，我么取全局最大。
class Solution:
def maxSubarraySumCircular(self, A: List[int]) -> int:
cur_max = glo_max = cur_min = glo_min = s =A[0]
for i in range(1,len(A)):
cur_max = max(A[i], A[i] + cur_max)
glo_max = max(glo_max, cur_max)
cur_min = min(A[i], A[i] + cur_min)
glo_min = min(glo_min, cur_min)
s +=A[i]
if s==glo_min:
return glo_max
return max(glo_max, s-glo_min)
