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:
-30000 <= A[i] <= 300001 <= A.length <= 30000
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Answer
有两种情况。
第一，子数组只有中间部分,我们知道如何找到最大子数组求和。
第二，是子数组头阵的一部分,尾巴数组的一部分。
最大的结果等于总和减去最小值子数组只子数组求和。
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;
}
原文地址：https://www.cnblogs.com/thougr/p/10206912.html