You are given a?0-indexed?integer array?nums. A subarray of?nums?is called?continuous?if:

• Let?i,?i + 1, ...,?j?be the indices in the subarray. Then, for each pair of indices?i <= i1, i2?<= j,?0 <=?|nums[i1] - nums[i2]| <= 2.

Return?the total number of?continuous?subarrays.

A subarray is a contiguous?non-empty?sequence of elements within an array.

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Example 1:

Input: nums = [5,4,2,4]

Output: 8

Explanation:?

Continuous subarray of size 1: [5], [4], [2], [4]. Continuous subarray of size 2: [5,4], [4,2], [2,4]. Continuous subarray of size 3: [4,2,4]. Thereare no subarrys of size 4. Total continuous subarrays = 4 + 3 + 1 = 8. It can be shown that there are no more continuous subarrays.

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Example 2:

Input: nums = [1,2,3]

Output: 6

Explanation: Continuous subarray of size 1: [1], [2], [3]. Continuous subarray of size 2: [1,2], [2,3]. Continuous subarray of size 3: [1,2,3]. Total continuous subarrays = 3 + 2 + 1 = 6.

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Constraints:

• 1 <= nums.length <= 105

• 1 <= nums[i] <= 109

這里利用雙指針以及優(yōu)先隊(duì)列來存儲最大值最小值，只是我考慮的時(shí)候，以為存在相同元素的話，不好實(shí)現(xiàn)，結(jié)果還是能實(shí)現(xiàn)，下面是代碼：（j-i+1是以i為開頭的所有可能的子數(shù)組的數(shù)量）-我有1個(gè)for循環(huán)+while循環(huán)在周賽的時(shí)候也過了呀，過了周賽就TLE了。。。

Runtime:?296 ms, faster than?25.00%?of?Java?online submissions for?Continuous Subarrays.

Memory Usage:?58.4 MB, less than?25.00%?of?Java?online submissions for?Continuous Subarrays.