You are given an integer array nums with length n.
The cost of a subarray nums[l..r], where 0 <= l <= r < n, is defined as:
cost(l, r) = nums[l] - nums[l + 1] + ... + nums[r] * (−1)r − l
Your task is to split nums into subarrays such that the total cost of the subarrays is maximized, ensuring each element belongs to exactly one subarray.
Formally, if nums is split into k subarrays, where k > 1, at indices i1, i2, ..., ik − 1, where 0 <= i1 < i2 < ... < ik - 1 < n - 1, then the total cost will be:
cost(0, i1) + cost(i1 + 1, i2) + ... + cost(ik − 1 + 1, n − 1)
Return an integer denoting the maximum total cost of the subarrays after splitting the array optimally.
\n\nNote: If nums is not split into subarrays, i.e. k = 1, the total cost is simply cost(0, n - 1).
\n
Example 1:
\n\nInput: nums = [1,-2,3,4]
\n\nOutput: 10
\n\nExplanation:
\n\nOne way to maximize the total cost is by splitting [1, -2, 3, 4] into subarrays [1, -2, 3] and [4]. The total cost will be (1 + 2 + 3) + 4 = 10.
Example 2:
\n\nInput: nums = [1,-1,1,-1]
\n\nOutput: 4
\n\nExplanation:
\n\nOne way to maximize the total cost is by splitting [1, -1, 1, -1] into subarrays [1, -1] and [1, -1]. The total cost will be (1 + 1) + (1 + 1) = 4.
Example 3:
\n\nInput: nums = [0]
\n\nOutput: 0
\n\nExplanation:
\n\nWe cannot split the array further, so the answer is 0.
\nExample 4:
\n\nInput: nums = [1,-1]
\n\nOutput: 2
\n\nExplanation:
\n\nSelecting the whole array gives a total cost of 1 + 1 = 2, which is the maximum.
\n
Constraints:
\n\n1 <= nums.length <= 105-109 <= nums[i] <= 109给你一个长度为 n 的整数数组 nums。
子数组 nums[l..r]（其中 0 <= l <= r < n）的 成本 定义为：
cost(l, r) = nums[l] - nums[l + 1] + ... + nums[r] * (−1)r − l
你的任务是将 nums 分割成若干子数组，使得所有子数组的成本之和 最大化，并确保每个元素 正好 属于一个子数组。
具体来说，如果 nums 被分割成 k 个子数组，且分割点为索引 i1, i2, ..., ik − 1（其中 0 <= i1 < i2 < ... < ik - 1 < n - 1），则总成本为：
cost(0, i1) + cost(i1 + 1, i2) + ... + cost(ik − 1 + 1, n − 1)
返回在最优分割方式下的子数组成本之和的最大值。
\n\n注意：如果 nums 没有被分割，即 k = 1，则总成本即为 cost(0, n - 1)。
\n\n
示例 1：
\n\n输入： nums = [1,-2,3,4]
\n\n输出： 10
\n\n解释：
\n\n一种总成本最大化的方法是将 [1, -2, 3, 4] 分割成子数组 [1, -2, 3] 和 [4]。总成本为 (1 + 2 + 3) + 4 = 10。
示例 2：
\n\n输入： nums = [1,-1,1,-1]
\n\n输出： 4
\n\n解释：
\n\n一种总成本最大化的方法是将 [1, -1, 1, -1] 分割成子数组 [1, -1] 和 [1, -1]。总成本为 (1 + 1) + (1 + 1) = 4。
示例 3：
\n\n输入： nums = [0]
\n\n输出： 0
\n\n解释：
\n\n无法进一步分割数组，因此答案为 0。
\n示例 4：
\n\n输入： nums = [1,-1]
\n\n输出： 2
\n\n解释：
\n\n选择整个数组，总成本为 1 + 1 = 2，这是可能的最大成本。
\n\n
提示：
\n\n1 <= nums.length <= 105-109 <= nums[i] <= 109dp[i][0/1] be the largest sum we can get for prefix nums[0..i], where dp[i][0] is the maximum if the ith element wasn't flipped, and dp[i][1] is the maximum if the ith element was flipped.",
"Based on the restriction:dp[i][0] = min(dp[i - 1][0], dp[i - 1][1]) + nums[i]dp[i][1] = dp[i - 1][0] - nums[i]",
"The initial state is:dp[1][0] = nums[0] + nums[1]dp[1][1] = nums[0] - nums[1]max(dp[n - 1][0], dp[n - 1][1]).",
"Can you optimize the space complexity?"
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