Leetcode solution (0933): number of recent requests (Python)

subject ：​ ​ Original link ​​（ Simple ）

solution

Time complexity

Spatial complexity

Execution time

Ans 1 (Python)

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344ms (74.43%)

Ans 2 (Python)

–

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376ms (42.68%)

Ans 3 (Python)

LeetCode Of Python Execution time , As long as there is no significant difference in time complexity , The execution time is generally of the same order of magnitude , For reference only .

Solution 1 （ Use arrays to implement ）：

class RecentCounter:



 def __init__(self):

 self.queue = []



 def ping(self, t: int) -> int:

 self.queue.append(t)

 while self.queue[0] < t - 3000:

 self.queue.pop(0)

 return len(self.queue)


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Solution 2 （ Use collections.deque Realization ）

class RecentCounter:



 def __init__(self):

 self.queue = collections.deque()



 def ping(self, t: int) -> int:

 self.queue.append(t)

 while self.queue[0] < t - 3000:

 self.queue.popleft()

 return len(self.queue)


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