LeetCode 6911. Continuous Subarrays

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.

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.

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 <= nums.length <= 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 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.