[leetcode] 933. Number of Recent Calls @ python

原题

Write a class RecentCounter to count recent requests.

It has only one method: ping(int t), where t represents some time in milliseconds.

Return the number of pings that have been made from 3000 milliseconds ago until now.

Any ping with time in [t - 3000, t] will count, including the current ping.

It is guaranteed that every call to ping uses a strictly larger value of t than before.

Example 1:

Input: inputs = [“RecentCounter”,“ping”,“ping”,“ping”,“ping”], inputs = [[],[1],[100],[3001],[3002]]
Output: [null,1,2,3,3]

Note:

Each test case will have at most 10000 calls to ping.
Each test case will call ping with strictly increasing values of t.
Each call to ping will have 1 <= t <= 10^9.

解法

构造列表, 当列表不为空且t与列表的第一个时间相差大于3000时, 一直删除第一个元素. 返回列表的长度即可.

代码

class RecentCounter(object):

    def __init__(self):
        self.data = []
        

    def ping(self, t):
        """
        :type t: int
        :rtype: int
        """ 
        while self.data and t - self.data[0] > 3000:
            self.data.pop(0)
        self.data.append(t)
        return len(self.data)


# Your RecentCounter object will be instantiated and called as such:
# obj = RecentCounter()
# param_1 = obj.ping(t)