[Swift]LeetCode918. 环形子数组的最大和 | Maximum Sum Circular Subarray

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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给定一个由整数数组 A 表示的环形数组 C，求 C 的非空子数组的最大可能和。

在此处，环形数组意味着数组的末端将会与开头相连呈环状。（形式上，当0 <= i < A.length 时 C[i] = A[i]，而当 i >= 0 时 C[i+A.length] = C[i]）

此外，子数组最多只能包含固定缓冲区 A 中的每个元素一次。（形式上，对于子数组 C[i], C[i+1], ..., C[j]，不存在 i <= k1, k2 <= j 其中 k1 % A.length = k2 % A.length）

示例 1：

输入：[1,-2,3,-2]
输出：3
解释：从子数组 [3] 得到最大和 3

示例 2：

输入：[5,-3,5]
输出：10
解释：从子数组 [5,5] 得到最大和 5 + 5 = 10

示例 3：

输入：[3,-1,2,-1]
输出：4
解释：从子数组 [2,-1,3] 得到最大和 2 + (-1) + 3 = 4

示例 4：

输入：[3,-2,2,-3]
输出：3
解释：从子数组 [3] 和 [3,-2,2] 都可以得到最大和 3

示例 5：

输入：[-2,-3,-1]
输出：-1
解释：从子数组 [-1] 得到最大和 -1

提示：

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

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96 ms

 1 class Solution {
 2     func maxSubarraySumCircular(_ A: [Int]) -> Int {
 3         if A == nil || A.count == 0 {return 0}
 4         var preSumMin:Int = 0
 5         var preSumMax:Int = 0
 6         var preSum = 0
 7         var sumMin = Int.max
 8         var sumMax = Int.min
 9         let count = A.count
10         for i in 0..<count
11         {
12             preSum += A[i]
13             sumMax = max(preSum - preSumMin, sumMax)
14             if i != (count - 1)
15             {
16                 sumMin = min(preSum - preSumMax, sumMin)
17             }
18             preSumMin = min(preSumMin, preSum)
19             preSumMax = max(preSumMax, preSum)
20         }
21         return max(sumMax, preSum - sumMin)
22     }
23 }