# 918.maximum sum circular subarray

## Description

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
``````

### Note:

• -30000 <= A[i] <= 30000
• 1 <= A.length <= 30000

1. updating

## 参考代码

``````class Solution(object):
def maxSubarraySumCircular(self, A):
N = len(A)

ans = cur = None
for x in A:
cur = x + max(cur, 0)
ans = max(ans, cur)

# ans is the answer for 1-interval subarrays.
# Now, let's consider all 2-interval subarrays.
# For each i, we want to know
# the maximum of sum(A[j:]) with j >= i+2

# rightsums[i] = sum(A[i:])
rightsums = [None] * N
rightsums[-1] = A[-1]
for i in xrange(N-2, -1, -1):
rightsums[i] = rightsums[i+1] + A[i]

# maxright[i] = max_{j >= i} rightsums[j]
maxright = [None] * N
maxright[-1] = rightsums[-1]
for i in xrange(N-2, -1, -1):
maxright[i] = max(maxright[i+1], rightsums[i])

leftsum = 0
for i in xrange(N-2):
leftsum += A[i]
ans = max(ans, leftsum + maxright[i+2])
return ans
``````