滑动窗口最大值leetcode & nowcoder（python）

方法一：暴力   时间O(nk)

class Solution:
    def maxInWindows(self, num, size):
        # write code here
        if len(num)*size == 0:
            return []
        res=[]
        for i in range(len(num)-size+1):
            res.append(max(num[i:i+size]))
        return res

方法二：双端队列

双端队列保存当前窗口的最大值

from collections import deque
class Solution:
    def maxSlidingWindow(self, nums: 'List[int]', k: 'int') -> 'List[int]':
        # base cases
        n = len(nums)
        if n * k == 0:
            return []
        if k == 1:
            return nums
        
        def clean_deque(i):
            # remove indexes of elements not from sliding window
            if deq and deq[0] == i - k:
                deq.popleft()
                
            # remove from deq indexes of all elements 
            # which are smaller than current element nums[i]
            while deq and nums[i] > nums[deq[-1]]:
                deq.pop()
        
        # init deque and output
        deq = deque()
        max_idx = 0
        for i in range(k):
            clean_deque(i)
            deq.append(i)
            # compute max in nums[:k]
            if nums[i] > nums[max_idx]:
                max_idx = i
        output = [nums[max_idx]]
        
        # build output
        for i in range(k, n):
            clean_deque(i)          
            deq.append(i)
            output.append(nums[deq[0]])
        return output

复杂度分析

时间复杂度：O(N)，每个元素被处理两次- 其索引被添加到双向队列中和被双向队列删除。

空间复杂度：O(N)，输出数组使用了O(N−k+1) 空间，双向队列使用了O(k)。

2020-01-08 20:53:02