[leetcode] 239. Sliding Window Maximum
作者：互联网
Description
You are given an array of integers nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position.
Return the max sliding window.
Example 1:
Input: nums = [1,3,-1,-3,5,3,6,7], k = 3
Output: [3,3,5,5,6,7]
Explanation:
Window position Max
--------------- -----
[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
Example 2:
Input: nums = [1], k = 1
Output: [1]
Example 3:
Input: nums = [1,-1], k = 1
Output: [1,-1]
Example 4:
Input: nums = [9,11], k = 2
Output: [11]
Example 5:
Input: nums = [4,-2], k = 2
Output: [4]
Constraints:
- 1 <= nums.length <= 105
- -104 <= nums[i] <= 104
- 1 <= k <= nums.length
分析
题目的意思是：求出滑动窗口内的最大值。这道题用队列存储遍历的索引，我只能写出简单的暴力破解版本，发现AC不了。学习一下别人的思路：
- 如果双端队列非空，存储的值超过了windows大小，则把双端队列队头的值移除；+ 如果当前的队列非空，移除小于当前加入元素值的索引；
- 如果当前的索引大于等于k-1，则更新最大值
代码
class Solution:
def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
n=len(nums)
q=deque()
res=[]
for i in range(n):
if(q and q[0]<=i-k): # 移除队头
q.popleft()
while(q and nums[i]>nums[q[-1]]): # 移除小于当前要加入元素的值的索引
q.pop()
q.append(i)
if(i>=k-1): # 更新最大值
res.append(nums[q[0]])
return res
参考文献
标签：nums,最大值,leetcode,window,Example,Window,Output,Input,Sliding 来源： https://blog.csdn.net/w5688414/article/details/111188291