[LeetCode] 933. Number of Recent Calls 最近的调用次数

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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:

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

这道题让实现一个 RecentCounter 类，里面有一个 ping 函数，输入给定了一个时间t，让我们求在 [t-3000, t] 时间范围内有多少次 ping。题目中限定了每次的给的时间一定会比上一次的时间大，而且只关心这个大小为 3001 的时间窗口范围内的次数，则利用滑动窗口 Sliding Window 来做就是个很不错的选择。由于数字是不断加入的，可以使用一个 queue，每当要加入一个新的时间点t时，先从队列开头遍历，若前面的时间不在当前的时间窗口内，则移除队列。之后再将当前时间点t加入，并返回队列的长度即可，参见代码如下：

class RecentCounter {
public:
    RecentCounter() {}
    
    int ping(int t) {
        while (!q.empty()) {
            if (q.front() + 3000 >= t) break;
            q.pop();
        }
        q.push(t);
        return q.size();
    }

private:
    queue<int> q;
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/933

参考资料：

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

https://leetcode.com/problems/number-of-recent-calls/discuss/189334/C%2B%2B-Easy-and-Clean-solution-using-queue

https://leetcode.com/problems/number-of-recent-calls/discuss/189239/JavaPython-3-Five-solutions%3A-TreeMap-TreeSet-ArrayList-Queue-Circular-List.

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[LeetCode] 933. Number of Recent Calls 最近的调用次数

标签：使用   int   iss   return   mil   front   ntc   队列   htm   

原文地址：https://www.cnblogs.com/grandyang/p/12359774.html