Question

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:

1. -30000 <= A[i] <= 30000
2. 1 <= A.length <= 30000

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Answer

　　有兩種情況。
　　
　　第一，子陣列只有中間部分,我們知道如何找到最大子陣列求和。
　　第二，是子陣列頭陣的一部分,尾巴陣列的一部分。
　　最大的結果等於總和減去最小值子陣列只子陣列求和。

    int maxSubarraySumCircular(vector<int>& A) {
        int total = 0, maxSum = -30000, curMax = 0, minSum = 30000, curMin = 0;
        for (int a : A) {
            curMax = max(curMax + a, a);
            maxSum = max(maxSum, curMax);
            curMin = min(curMin + a, a);
            minSum = min(minSum, curMin);
            total += a;
        }
        return maxSum > 0 ? max(maxSum, total - minSum) : maxSum;
    }