LeetCode之Number of Recent Calls（

问题：
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.

方法：
创建一个队列，每次新来一个ping请求先加入队列，然后清除所有不在当前ping请求t-3000范围内的指令，输出当前队列长度即为结果，本题的思路类似请求超时。

具体实现：

class NumberOfRecentCalls {
    private val mutableList = mutableListOf<Int>()

    fun ping(t: Int): Int {
        mutableList.add(t)
        while (mutableList[0] < t - 3000) {
            mutableList.removeAt(0)
        }
        return mutableList.size
    }
}

fun main(args: Array<String>) {
    val numberOfRecentCalls = NumberOfRecentCalls()
    print(numberOfRecentCalls.ping(1))
    print(numberOfRecentCalls.ping(100))
    print(numberOfRecentCalls.ping(3001))
    print(numberOfRecentCalls.ping(3002))
}

有问题随时沟通

具体代码实现可以参考Github