933.Number of Recent Calls（最近的请求次数）

Description

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.

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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。

题目链接：https://leetcode.com/problems/number-of-recent-calls/

Difficulty: easy

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.

分析

1. 用list存储所有的t；
2. ping是，删除list中所有小于t-3000的值；
3. 返回list的长度。

参考代码

class RecentCounter:

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


def ping(self, t):
    self.li.append(t)
    index=0
    while(True):
        if(self.li[index]<t-3000):
            index+=1
        else:
            self.li=self.li[index:]
            break
    return len(self.li)