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

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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

---

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

---

80 ms

class Solution {
    func maxSubarraySumCircular(_ A: [Int]) -> Int {
        if A == nil || A.count == 0 {return 0}
        var preSumMin:Int = 0
        var preSumMax:Int = 0
        var preSum = 0
        var sumMin = Int.max
        var sumMax = Int.min
        let count = A.count
        for i in 0..<count
        {
            preSum += A[i]
            sumMax = max(preSum - preSumMin, sumMax)
            if i != (count - 1)
            {
                sumMin = min(preSum - preSumMax, sumMin)
            }
            preSumMin = min(preSumMin, preSum)
            preSumMax = max(preSumMax, preSum)
        }
        return max(sumMax, preSum - sumMin)
    }
}

---

300ms

class Solution {
    func maxSubarraySumCircular(_ A: [Int]) -> Int {
        
        var maxSum = A.max()!
        
        var simpleA: [Int] = []
        simpleA.reserveCapacity(A.count)
        var isPos = A.first! > 0
        var sum = 0
        for i in 0..<A.count {
            if A[i] > 0 || A[i] < 0{
                if isPos == (A[i] > 0) {
                    sum += A[i]
                } else {
                    simpleA.append(sum)
                    sum = A[i]
                    isPos = A[i] > 0
                }
            }
        }
        simpleA.append(sum)
        
        let AA = simpleA + simpleA
        
        iCycle: for i in 0..<simpleA.count {
            if simpleA[i] < 0 {
                continue iCycle
            }
            var sum = simpleA[i]
            maxSum = max(sum, maxSum)
            jCycle: for j in (i+1)..<(i+simpleA.count) {
                sum += AA[j]
                if sum < 0 {
                    continue iCycle
                }
                maxSum = max(sum, maxSum)
            }
        }
        return maxSum
    }
}

---

704ms

class Solution {
    func maxSubarraySumCircular(_ A: [Int]) -> Int {
        if A == nil || A.count == 0 {return 0}
        var preSumMin:Int = 0
        var preSumMax:Int = 0
        var preSum = 0
        var sumMin = Int.max
        var sumMax = Int.min
        let len = A.count
        for i in 0..<len
        {
            preSum += A[i]
            sumMax = max(preSum - preSumMin, sumMax)
            if i != (len - 1)
            {
                sumMin = min(preSum - preSumMax, sumMin)
            }
            preSumMin = min(preSumMin, preSum)
            preSumMax = max(preSumMax, preSum)
        }
        return max(sumMax, preSum - sumMin)
    }
}

---

836ms

class Solution {
    func maxSubarraySumCircular(_ A: [Int]) -> Int {
        guard A.count > 0 else {
            return 0
        }
        let s = A.reduce(0, +)
        var M = A[0]
        var m = A[0]
        var lastM = A[0]
        var lastm = A[0]
        for i in 1 ..< A.count {
            let a = A[i]
            lastM = max(lastM+a, a)
            lastm = min(lastm+a, a)
            M = max(M, lastM)
            m = min(m, lastm)
        }
        return M > 0 ? max(M, s-m) : M
    }
}

---

948ms

class Solution {
    func maxSubarraySumCircular(_ A: [Int]) -> Int {
        var sum = 0
        var curMax = 0
        var curMin = 0
        var maxSum = -30000
        var minSum = 30000
        for i in 0..<A.count {
            sum += A[i]
            curMax = max(curMax + A[i], A[i])
            maxSum = max(maxSum, curMax)
            curMin = min(curMin + A[i], A[i])
            minSum = min(minSum, curMin)
        }
        return maxSum > 0 ? max(sum - minSum, maxSum) : maxSum
    }
}