LeetCode 239. Sliding Window Maximum

滑动窗口最大值算法

最新推荐文章于 2025-01-17 23:12:23 发布

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本文介绍了一种解决滑动窗口最大值问题的高效算法。通过使用双端队列，可以在线性时间内找到给定数组中大小为k的滑动窗口的最大值。文章详细解释了算法的实现步骤，并提供了两种不同的Python实现方式。

Given an array 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:

Input: nums = [1,3,-1,-3,5,3,6,7], and 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

Note:
You may assume k is always valid, 1 ≤ k ≤ input array's size for non-empty array.

Follow up:
Could you solve it in linear time?

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这题的思路比较容易想到，但是反应比较慢。实际上就三步：1.左边看看是不是要被删了 2. 右边要把比当前元素小等的弹出来，再把最新的放进栈 3. 双端队列最左边是最大的。用Python的list模拟双端口队列。

class Solution:
    def maxSlidingWindow(self, nums, k: int):
        queue,qs,l,res = [],0,len(nums),[]
        for i in range(l):
            if (i >= k and i-k == queue[qs]): #pop
                qs += 1
            ql = len(queue)
            while (ql > qs and nums[queue[-1]] <= nums[i]): #push #bug2: ql >= qs
                queue.pop()
                ql -= 1
            queue.append(i) #bug1: queue.append(nums[i])
            if (i >= k-1):
                res.append(nums[queue[qs]])
        return res

再贴一个来自Stephan的写法：

def maxSlidingWindow(self, nums, k):
    d = collections.deque()
    out = []
    for i, n in enumerate(nums):
        while d and nums[d[-1]] < n:
            d.pop()
        d += i,
        if d[0] == i - k:
            d.popleft()
        if i >= k - 1:
            out += nums[d[0]],
    return out