You are given a?0-indexed?integer array?nums?and a positive integer?x.

You are?initially?at position?0?in the array and you can visit other positions according to the following rules:

• If you are currently in position?i, then you can move to?any?position?j?such that?i < j.

• For each position?i?that you visit, you get a score of?nums[i].

• If you move from a position?i?to a position?j?and the?parities?of?nums[i]?and?nums[j]?differ, then you lose a score of?x.

Return?the?maximum?total score you can get.

Note?that initially you have?nums[0]?points.

?

Example 1:

Input: nums = [2,3,6,1,9,2], x = 5Output: 13Explanation: We can visit the following positions in the array: 0 -> 2 -> 3 -> 4. The corresponding values are 2, 6, 1 and 9. Since the integers 6 and 1 have different parities, the move 2 -> 3 will make you lose a score of x = 5. The total score will be: 2 + 6 + 1 + 9 - 5 = 13.

Example 2:

Input: nums = [2,4,6,8], x = 3Output: 20Explanation: All the integers in the array have the same parities, so we can visit all of them without losing any score. The total score is: 2 + 4 + 6 + 8 = 20.

?

Constraints:

• 2 <= nums.length <= 105

• 1 <= nums[i], x <= 106

--------------------------------------------------

給定一個(gè) 0 索引的整數(shù)數(shù)組 nums 和一個(gè)正整數(shù) x。

您最初位于數(shù)組中的位置 0，您可以根據(jù)以下規(guī)則訪問(wèn)其他位置：

如果您當(dāng)前處于位置 i，那么您可以移動(dòng)到任意位置 j，使得 i < j。

對(duì)于您訪問(wèn)的每個(gè)位置 i，您將獲得 nums[i] 分?jǐn)?shù)。

如果你從位置 i 移動(dòng)到位置 j 并且 nums[i] 和 nums[j] 的奇偶性不同，那么你會(huì)失去 x 的分?jǐn)?shù)。

返回您可以獲得的最高總分。

請(qǐng)注意，最初您有 nums[0] 點(diǎn)。

------------------------------------------

用2個(gè)變量依次去存儲(chǔ)每次到該位置時(shí)候的odd，even的值，最后max返回；

Runtime:?11 ms, faster than?100.00%?of?Java?online submissions for?Visit Array Positions to Maximize Score.

Memory Usage:?55.3 MB, less than?100.00%?of?Java?online submissions for?Visit Array Positions to Maximize Score.