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提示

给定一个长度为 n环形整数数组 nums ,返回 nums 的非空 子数组 的最大可能和 

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, nums[i] 的下一个元素是 nums[(i + 1) % n]nums[i] 的前一个元素是 nums[(i - 1 + n) % n]

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

 

示例 1:

输入:nums = [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:nums = [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:nums = [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

 

提示:

  • n == nums.length
  • 1 <= n <= 3 * 104
  • -3 * 104 <= nums[i] <= 3 * 104​​​​​​​
通过次数
117,211/261.3K
通过率
44.9%


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提示 1
For those of you who are familiar with the Kadane's algorithm, think in terms of that. For the newbies, Kadane's algorithm is used to finding the maximum sum subarray from a given array. This problem is a twist on that idea and it is advisable to read up on that algorithm first before starting this problem. Unless you already have a great algorithm brewing up in your mind in which case, go right ahead!

提示 2
What is an alternate way of representing a circular array so that it appears to be a straight array? Essentially, there are two cases of this problem that we need to take care of. Let's look at the figure below to understand those two cases:

提示 3
The first case can be handled by the good old Kadane's algorithm. However, is there a smarter way of going about handling the second case as well?

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nums =
[1,-2,3,-2]
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