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 [2,3] has maximum sum 2 + (-1) + 3 = 4
Example 4:
Input: [3,-3]
Output: 3 Explanation: Subarray [3] and [3,2] both have maximum sum 3
Example 5:
Input: [-2,-1]
Output: -1 Explanation: Subarray [-1] has maximum sum -1
Note:
-30000 <= A[i] <= 30000
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[j]
,不存在 i <= k1,k2 <= j
其中 k1 % A.length = k2 % A.length
)
示例 1:
输入:[1,-2] 输出:3 解释:从子数组 [3] 得到最大和 3
示例 2:
输入:[5,5] 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] 输出:4 解释:从子数组 [2,3] 得到最大和 2 + (-1) + 3 = 4
示例 4:
输入:[3,-3] 输出:3 解释:从子数组 [3] 和 [3,2] 都可以得到最大和 3
示例 5:
输入:[-2,-1] 输出:-1 解释:从子数组 [-1] 得到最大和 -1
提示:
-30000 <= A[i] <= 30000
1 <= A.length <= 30000
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 }