mirror of
https://github.com/chefyuan/algorithm-base.git
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263 lines
9.5 KiB
Java
263 lines
9.5 KiB
Java
> 如果阅读时，发现错误，或者动画不可以显示的问题可以添加我微信好友 **[tan45du_one](https://raw.githubusercontent.com/tan45du/tan45du.github.io/master/个人微信.15egrcgqd94w.jpg)** ，备注 github + 题目 + 问题 向我反馈
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>
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> 感谢支持，该仓库会一直维护，希望对各位有一丢丢帮助。
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>
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> 另外希望手机阅读的同学可以来我的 <u>[**公众号：袁厨的算法小屋**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u> 两个平台同步，想要和题友一起刷题，互相监督的同学，可以在我的小屋点击<u>[**刷题小队**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u>进入。
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#### [1438. 绝对差不超过限制的最长连续子数组](https://leetcode-cn.com/problems/longest-continuous-subarray-with-absolute-diff-less-than-or-equal-to-limit/)
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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 。
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如果不存在满足条件的子数组，则返回 0 。
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示例
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> 输入：nums = [10,1,2,4,7,2], limit = 5
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> 输出：4
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> 解释：满足题意的最长子数组是 [2,4,7,2]，其最大绝对差 |2-7| = 5 <= 5 。
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**提示：**
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- 1 <= nums.length <= 10^5
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- 1 <= nums[i] <= 10^9
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- 0 <= limit <= 10^9
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**题目解析**
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我们结合题目，示例，提示来看，这个题目也可以使用滑动窗口的思想来解决。我们需要判断某个子数组是否满足最大绝对差不超过限制值。
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那么我们应该怎么解决呢？
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我们想一下，窗口内的最大绝对差，如果我们知道窗口的最大值和最小值，最大值减去最小值就能得到最大绝对差。
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所以我们这个问题就变成了获取滑动窗口内的最大值和最小值问题，哦？滑动窗口的最大值，是不是很熟悉，大家可以先看一下[滑动窗口的最大值](https://leetcode-cn.com/problems/hua-dong-chuang-kou-de-zui-da-zhi-lcof/solution/yi-shi-pin-sheng-qian-yan-shuang-duan-du-mbga/)这个题目，那我们完全可以借助刚才题目的思想来解决这个题目。啪的一下我就搞懂了。
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滑动窗口的最大值，我们当时借助了双端队列，来维护一个单调递减的双端队列，进而得到滑动窗口的最大值
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那么我们同样可以借助双端队列，来维护一个单调递增的双端队列，来获取滑动窗口的最小值。既然知道了最大值和最小值，我们就可以判断当前窗口是否符合要求，如果符合要求则扩大窗口，不符合要求则缩小窗口，循环结束返回最大的窗口值即可。
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下面我们来看一下我们的动画模拟，一下就能看懂！
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<img src="https://img-blog.csdnimg.cn/20210320092423565.gif" style="zoom:150%;" />
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其实，我们只要把握两个重点即可，我们的 maxdeque 维护的是一个单调递减的双端队列，头部为当前窗口的最大值， mindeque 维护的是一个单调递增的双端队列，头部为窗口的最小值，即可。好啦我们一起看代码吧。
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Java Code:
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```java
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class Solution {
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public int longestSubarray(int[] nums, int limit) {
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Deque<Integer> maxdeque = new LinkedList<>();
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Deque<Integer> mindeque = new LinkedList<>();
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int len = nums.length;
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int right = 0, left = 0, maxwin = 0;
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while (right < len) {
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while (!maxdeque.isEmpty() && maxdeque.peekLast() < nums[right]) {
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maxdeque.removeLast();
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}
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while (!mindeque.isEmpty() && mindeque.peekLast() > nums[right]) {
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mindeque.removeLast();
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}
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//需要更多视频解算法，可以来我的公众号：袁厨的算法小屋
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maxdeque.addLast(nums[right]);
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mindeque.addLast(nums[right]);
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while (maxdeque.peekFirst() - mindeque.peekFirst() > limit) {
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if (maxdeque.peekFirst() == nums[left]) maxdeque.removeFirst();
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if (mindeque.peekFirst() == nums[left]) mindeque.removeFirst();
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left++;
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}
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//保留最大窗口
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maxwin = Math.max(maxwin,right-left+1);
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right++;
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}
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return maxwin;
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}
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}
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```
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Python Code:
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```python
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from typing import List
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import collections
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class Solution:
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def longestSubarray(self, nums: List[int], limit: int)->int:
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maxdeque = collections.deque()
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mindeque = collections.deque()
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leng = len(nums)
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right = 0
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left = 0
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maxwin = 0
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while right < leng:
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while len(maxdeque) != 0 and maxdeque[-1] < nums[right]:
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maxdeque.pop()
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while len(mindeque) != 0 and mindeque[-1] > nums[right]:
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mindeque.pop()
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maxdeque.append(nums[right])
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mindeque.append(nums[right])
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while (maxdeque[0] - mindeque[0]) > limit:
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if maxdeque[0] == nums[left]:
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maxdeque.popleft()
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if mindeque[0] == nums[left]:
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mindeque.popleft()
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left += 1
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# 保留最大窗口
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maxwin = max(maxwin, right - left + 1)
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right += 1
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return maxwin
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```
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Swift Code
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Swift：数组模拟，超时（58 / 61 个通过测试用例）
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```swift
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class Solution {
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func longestSubarray(_ nums: [Int], _ limit: Int) -> Int {
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var maxQueue:[Int] = []
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var minQueue:[Int] = []
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let len = nums.count
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var right = 0, left = 0, maxWin = 0
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while right < len {
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while !maxQueue.isEmpty && (maxQueue.last! < nums[right]) {
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maxQueue.removeLast()
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}
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while !minQueue.isEmpty && (minQueue.last! > nums[right]) {
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minQueue.removeLast()
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}
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maxQueue.append(nums[right])
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minQueue.append(nums[right])
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while (maxQueue.first! - minQueue.first!) > limit {
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if maxQueue.first! == nums[left] {
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maxQueue.removeFirst()
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}
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if minQueue.first! == nums[left] {
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minQueue.removeFirst()
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}
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left += 1
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}
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maxWin = max(maxWin, right - left + 1)
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right += 1
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}
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return maxWin
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}
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}
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```
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Swift：使用双端队列（击败了 100.00%）
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```swift
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class Solution {
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func longestSubarray(_ nums: [Int], _ limit: Int) -> Int {
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var maxQueue = Deque<Int>.init()
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var minQueue = Deque<Int>.init()
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let len = nums.count
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var right = 0, left = 0, maxWin = 0
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while right < len {
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while !maxQueue.isEmpty && (maxQueue.peekBack()! < nums[right]) {
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maxQueue.dequeueBack()
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}
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while !minQueue.isEmpty && (minQueue.peekBack()! > nums[right]) {
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minQueue.dequeueBack()
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}
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maxQueue.enqueue(nums[right])
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minQueue.enqueue(nums[right])
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while (maxQueue.peekFront()! - minQueue.peekFront()!) > limit {
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if maxQueue.peekFront()! == nums[left] {
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maxQueue.dequeue()
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}
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if minQueue.peekFront()! == nums[left] {
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minQueue.dequeue()
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}
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left += 1
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}
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maxWin = max(maxWin, right - left + 1)
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right += 1
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}
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return maxWin
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}
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// 双端队列数据结构
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public struct Deque<T> {
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private var array: [T?]
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private var head: Int
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private var capacity: Int
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private let originalCapacity: Int
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public init(_ capacity: Int = 10) {
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self.capacity = max(capacity, 1)
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originalCapacity = self.capacity
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array = [T?](repeating: nil, count: capacity)
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head = capacity
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}
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public var isEmpty: Bool {
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return count == 0
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}
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public var count: Int {
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return array.count - head
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}
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public mutating func enqueue(_ element: T) {
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array.append(element)
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}
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public mutating func enqueueFront(_ element: T) {
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if head == 0 {
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capacity *= 2
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let emptySpace = [T?](repeating: nil, count: capacity)
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array.insert(contentsOf: emptySpace, at: 0)
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head = capacity
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}
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head -= 1
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array[head] = element
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}
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public mutating func dequeue() -> T? {
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guard head < array.count, let element = array[head] else { return nil }
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array[head] = nil
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head += 1
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if capacity >= originalCapacity && head >= capacity*2 {
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let amountToRemove = capacity + capacity/2
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array.removeFirst(amountToRemove)
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head -= amountToRemove
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capacity /= 2
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}
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return element
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}
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public mutating func dequeueBack() -> T? {
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if isEmpty {
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return nil
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} else {
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return array.removeLast()
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}
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}
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public func peekFront() -> T? {
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if isEmpty {
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return nil
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} else {
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return array[head]
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}
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}
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public func peekBack() -> T? {
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if isEmpty {
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return nil
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} else {
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return array.last!
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}
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}
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}
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}
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```
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