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

Let i, i + 1, ..., jbe 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.

(资料图片仅供参考)

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

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 continuous subarrays = 4 + 3 + 1 = can be shown that there are no more continuous subarrays.

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.

Constraints:

1 <= <= 105

1 <= nums[i] <= 109

这里利用双指针以及优先队列来存储最大值最小值，只是我考虑的时候，以为存在相同元素的话，不好实现，结果还是能实现，下面是代码：（j-i+1是以i为开头的所有可能的子数组的数量）-我有1个for循环+while循环在周赛的时候也过了呀，过了周赛就TLE了。。。

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

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