> For the complete documentation index, see [llms.txt](https://imhuay.gitbook.io/studies/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://imhuay.gitbook.io/studies/algorithms/problems/2022/01/jian-zhi-offer5901-kun-nan-hua-dong-chuang-kou-de-zui-da-zhi.md).
# 滑动窗口的最大值
 [](https://imhuay.gitbook.io/studies/algorithms/problems/2022/01/pages/R5NyOzkn3qAZy7wCx1pS#困难) [](https://imhuay.gitbook.io/studies/algorithms/problems/2022/01/pages/R5NyOzkn3qAZy7wCx1pS#剑指offer) [](https://imhuay.gitbook.io/studies/algorithms/problems/2022/01/pages/R5NyOzkn3qAZy7wCx1pS#滑动窗口) [](https://imhuay.gitbook.io/studies/algorithms/problems/2022/01/pages/R5NyOzkn3qAZy7wCx1pS#单调栈单调队列)
**问题简述**
```
给定一个数组 nums 和滑动窗口的大小 k，请找出所有滑动窗口里的最大值。
```
详细描述
```
给定一个数组 nums 和滑动窗口的大小 k，请找出所有滑动窗口里的最大值。
示例:
输入: nums = [1,3,-1,-3,5,3,6,7], 和 k = 3
输出: [3,3,5,5,6,7]
解释:
滑动窗口的位置 最大值
--------------- -----
[1 3 -1] -3 5 3 6 7 3
1 [3 -1 -3] 5 3 6 7 3
1 3 [-1 -3 5] 3 6 7 5
1 3 -1 [-3 5 3] 6 7 5
1 3 -1 -3 [5 3 6] 7 6
1 3 -1 -3 5 [3 6 7] 7
提示：
你可以假设 k 总是有效的，在输入数组不为空的情况下，1 ≤ k ≤ 输入数组的大小。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/hua-dong-chuang-kou-de-zui-da-zhi-lcof
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
```
**思路**
* 使用单调队列维护一个最大值序列，每次滑动窗口前，更新单调队列，使队首元素为下一个窗口中的最大值，详见参考链接或具体代码；
> [滑动窗口的最大值（单调队列，清晰图解）](https://leetcode-cn.com/problems/hua-dong-chuang-kou-de-zui-da-zhi-lcof/solution/mian-shi-ti-59-i-hua-dong-chuang-kou-de-zui-da-1-6/)
Python
* 跟[官方写法](https://leetcode-cn.com/problems/hua-dong-chuang-kou-de-zui-da-zhi-lcof/solution/hua-dong-chuang-kou-de-zui-da-zhi-by-lee-ymyo/)的区别：
* 官方的单调队列维护的是数组下标，通过判断下标位置来确定是否移除队首元素；因此可以使用**严格单调队列**；而下面的写法中使用值来判断是否移除队首，因此使用的是非严格单调队列（相关代码段：`if q[0] == nums[i - k]: q.popleft()`）
```python
class Solution:
def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
from collections import deque
if not nums: return []
# 初始化单调队列，对任意 i > j，有 q[i] >= q[j]
q = deque()
for x in nums[:k]:
while q and q[-1] < x: # 注意这里是非严格单调的
q.pop()
q.append(x)
# print(q)
ret = [q[0]] #
for i in range(k, len(nums)):
if q[0] == nums[i - k]: # 因为是通过值判断，所以需要保留所有相同的最大值，所以队列是非严格单调的
q.popleft()
while q and q[-1] < nums[i]:
q.pop()
q.append(nums[i])
ret.append(q[0])
# print(q)
return ret
```
---
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