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90. Mos.

90. Mos.

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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90. Mos. 

 

37 sloc) 1.05 KB 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. For example, Given nums = [1,3,-1,-3,5,3,6,7], and k = 3. Window position [1 3 -1] -3 5 367 3 1 [3 -1 -3] 5 367 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 [367] 7 Therefore, return the max sliding window as [3,3,5,5,6,7]. import collections def max_sliding_window(nums, k): :type nums: List[int] :type k: int :rtype: List[int] if not nums: Max return nums queue = collections.deque() res = [] for num in nums: if len(queue) <k: queue.append(num) else: res.append(max(queue)) queue.popleft() queue.append(num) res.append(max(queue)) return res
Transcribed Image Text:37 sloc) 1.05 KB 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. For example, Given nums = [1,3,-1,-3,5,3,6,7], and k = 3. Window position [1 3 -1] -3 5 367 3 1 [3 -1 -3] 5 367 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 [367] 7 Therefore, return the max sliding window as [3,3,5,5,6,7]. import collections def max_sliding_window(nums, k): :type nums: List[int] :type k: int :rtype: List[int] if not nums: Max return nums queue = collections.deque() res = [] for num in nums: if len(queue) <k: queue.append(num) else: res.append(max(queue)) queue.popleft() queue.append(num) res.append(max(queue)) return res
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