1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit

题目

Given an array of integers nums and an integer limit, return the size of the longest non-empty subarray such that the absolute difference between any two elements of this subarray is less than or equal to limit.

Example 1:

Input: nums = [8,2,4,7], limit = 4
Output: 2 
Explanation: All subarrays are: 
[8] with maximum absolute diff |8-8| = 0 <= 4.
[8,2] with maximum absolute diff |8-2| = 6 > 4. 
[8,2,4] with maximum absolute diff |8-2| = 6 > 4.
[8,2,4,7] with maximum absolute diff |8-2| = 6 > 4.
[2] with maximum absolute diff |2-2| = 0 <= 4.
[2,4] with maximum absolute diff |2-4| = 2 <= 4.
[2,4,7] with maximum absolute diff |2-7| = 5 > 4.
[4] with maximum absolute diff |4-4| = 0 <= 4.
[4,7] with maximum absolute diff |4-7| = 3 <= 4.
[7] with maximum absolute diff |7-7| = 0 <= 4. 
Therefore, the size of the longest subarray is 2.

Example 2:

Input: nums = [10,1,2,4,7,2], limit = 5
Output: 4 
Explanation: The subarray [2,4,7,2] is the longest since the maximum absolute diff is |2-7| = 5 <= 5.

Example 3:

Input: nums = [4,2,2,2,4,4,2,2], limit = 0
Output: 3

Constraints:

1 <= nums.length <= 10^5
1 <= nums[i] <= 10^9
0 <= limit <= 10^9

题目大意

给你一个整数数组 nums ，和一个表示限制的整数 limit，请你返回最长连续子数组的长度，该子数组中的任意两个元素之间的绝对差必须小于或者等于 limit 。如果不存在满足条件的子数组，则返回 0 。

解题思路

最开始想到的思路是利用滑动窗口遍历一遍数组，每个窗口内排序，取出最大最小值。滑动窗口遍历一次的时间复杂度是 O(n)，所以此题时间复杂度是否高效落在了排序算法上了。由于前后 2 个窗口数据是有关联的，仅仅只变动了 2 个数据（左窗口移出的数据和右窗口移进的数据），所以排序没有必要每次都重新排序。这里利用二叉排序树来排序，添加和删除元素时间复杂度是 O(log n)，这种方法总的时间复杂度是 O(n log n)。空间复杂度 O(n)。
二叉排序树的思路是否还有再优化的空间？答案是有。二叉排序树内维护了所有结点的有序关系，但是这个关系是多余的。此题只需要找到最大值和最小值，并不需要除此以外节点的有序信息。所以用二叉排序树是大材小用了。可以换成 2 个单调队列，一个维护窗口内的最大值，另一个维护窗口内的最小值。这样优化以后，时间复杂度降低到 O(n)，空间复杂度 O(n)。具体实现见代码。
单调栈的题还有第 42 题，第 84 题，第 496 题，第 503 题，第 739 题，第 856 题，第 901 题，第 907 题，第 1130 题，第 1425 题，第 1673 题。

代码

package leetcode

func longestSubarray(nums []int, limit int) int {
	minStack, maxStack, left, res := []int{}, []int{}, 0, 0
	for right, num := range nums {
		for len(minStack) > 0 && nums[minStack[len(minStack)-1]] > num {
			minStack = minStack[:len(minStack)-1]
		}
		minStack = append(minStack, right)
		for len(maxStack) > 0 && nums[maxStack[len(maxStack)-1]] < num {
			maxStack = maxStack[:len(maxStack)-1]
		}
		maxStack = append(maxStack, right)
		if len(minStack) > 0 && len(maxStack) > 0 && nums[maxStack[0]]-nums[minStack[0]] > limit {
			if left == minStack[0] {
				minStack = minStack[1:]
			}
			if left == maxStack[0] {
				maxStack = maxStack[1:]
			}
			left++
		}
		if right-left+1 > res {
			res = right - left + 1
		}
	}
	return res
}