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LeetCode: 933. 最近的请求次数

1、题目描述

写一个 RecentCounter 类来计算最近的请求。

它只有一个方法:ping(int t) ,其中 t 代表以毫秒为单位的某个时间。

返回从 3000 毫秒前到现在的 ping 数。

任何处于 [t - 3000, t] 时间范围之内的 ping 都将会被计算在内,包括当前(指 t 时刻)的 ping

保证每次对 ping 的调用都使用比之前更大的 t 值。

示例:

输入:inputs = ["RecentCounter","ping","ping","ping","ping"], 
     inputs = [[],[1],[100],[3001],[3002]]
输出:[null,1,2,3,3]

提示:

  • 每个测试用例最多调用 10000ping

  • 每个测试用例会使用严格递增的 t 值来调用 ping

  • 每次调用 ping 都有 $ 1 <= t <= 10^9$ 。

2、解题思路

class RecentCounter:
    def __init__(self):
        self.buff = []

    def ping(self, t: int) -> int:
        self.buff.append(t)
        start_time = t - 3000 if t >= 3000 else 0
        pos = 0
        for index, i in enumerate(self.buff):
            if i < start_time:
                continue
            else:
                pos = index
                break
        self.buff = self.buff[pos:]
        return len(self.buff)

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


class RecentCounter:

    def __init__(self):
        self.buff = deque(maxlen=10000)

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

        self.buff.append(t)
        start_time = t - 3000 if t >= 3000 else 0
        while self.buff[0] < start_time:
            self.buff.popleft()
        return len(self.buff)

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