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[学习笔记] Leetcode 239. Sliding Window Maximum

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发表于 2020-9-23 03:43:03 | 显示全部楼层 |阅读模式

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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
from collections import deque
class Solution:
    def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
        queue = deque()
        res = []
        if len(nums) == 0:
            return res
        for i in range(k - 1):
            self.indeque(queue, nums[i])
        for i in range(k - 1, len(nums)):
            self.indeque(queue, nums[i])
            res.append(queue[0])
            self.outdeque(queue, nums[i - k + 1])
        return res
    
    def indeque(self, queue: List[int], num: int) -> None:
        while len(queue) != 0 and queue[-1] < num:
            queue.pop()
        queue.append(num)
    
    def outdeque(self, queue: List[int], num: int) -> None:
        if queue[0] == num:
            queue.popleft()

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