LeetCode之Number of Recent Calls（Kotlin）

問題： 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.
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方法： 建立一個隊列，每次新來一個ping請求先加入隊列，而後清除全部不在當前ping請求t-3000範圍內的指令，輸出當前隊列長度即爲結果，本題的思路相似請求超時。

具體實現：

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))
}
複製代碼

有問題隨時溝通

具體代碼實現能夠參考Github