剑指 Offer II 042. 最近请求次数

题目描述

写一个 RecentCounter 类来计算特定时间范围内最近的请求。

请实现 RecentCounter 类：

RecentCounter() 初始化计数器，请求数为 0 。
int ping(int t) 在时间 t 添加一个新请求，其中 t 表示以毫秒为单位的某个时间，并返回过去 3000 毫秒内发生的所有请求数（包括新请求）。确切地说，返回在 [t-3000, t] 内发生的请求数。

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]

解释：
RecentCounter recentCounter = new RecentCounter();
recentCounter.ping(1);     // requests = [1]，范围是 [-2999,1]，返回 1
recentCounter.ping(100);   // requests = [1, 100]，范围是 [-2900,100]，返回 2
recentCounter.ping(3001);  // requests = [1, 100, 3001]，范围是 [1,3001]，返回 3
recentCounter.ping(3002);  // requests = [1, 100, 3001, 3002]，范围是 [2,3002]，返回 3

提示：

1 <= t <= 10⁹
保证每次对 ping 调用所使用的 t 值都 严格递增
至多调用 ping 方法 10⁴ 次

注意：本题与主站 933 题相同： https://leetcode.cn/problems/number-of-recent-calls/

解法

方法一

Python3JavaC++GoTypeScriptJavaScriptSwift

class RecentCounter:
    def __init__(self):
        self.q = deque()

    def ping(self, t: int) -> int:
        self.q.append(t)
        while self.q[0] < t - 3000:
            self.q.popleft()
        return len(self.q)


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

class RecentCounter {
    private Deque<Integer> q;

    public RecentCounter() {
        q = new LinkedList<>();
    }

    public int ping(int t) {
        q.offerLast(t);
        while (q.peekFirst() < t - 3000) {
            q.pollFirst();
        }
        return q.size();
    }
}

/**
 * Your RecentCounter object will be instantiated and called as such:
 * RecentCounter obj = new RecentCounter();
 * int param_1 = obj.ping(t);
 */

class RecentCounter {
public:
    deque<int> q;

    RecentCounter() {
    }

    int ping(int t) {
        q.push_back(t);
        while (q.front() < t - 3000) {
            q.pop_front();
        }
        return q.size();
    }
};

/**
 * Your RecentCounter object will be instantiated and called as such:
 * RecentCounter* obj = new RecentCounter();
 * int param_1 = obj->ping(t);
 */

type RecentCounter struct {
    q []int
}

func Constructor() RecentCounter {
    return RecentCounter{
        q: []int{},
    }
}

func (this *RecentCounter) Ping(t int) int {
    this.q = append(this.q, t)
    for this.q[0] < t-3000 {
        this.q = this.q[1:len(this.q)]
    }
    return len(this.q)
}

/**
 * Your RecentCounter object will be instantiated and called as such:
 * obj := Constructor();
 * param_1 := obj.Ping(t);
 */

class RecentCounter {
    stack: Array<number>;
    cnt: number;
    constructor() {
        this.stack = [];
        this.cnt = 0;
    }

    ping(t: number): number {
        while (this.stack.length && this.stack[0] + 3000 < t) {
            this.cnt--;
            this.stack.shift();
        }
        this.cnt++;
        this.stack.push(t);
        return this.cnt;
    }
}

var RecentCounter = function () {
    this.q = [];
};

/**
 * @param {number} t
 * @return {number}
 */
RecentCounter.prototype.ping = function (t) {
    this.q.push(t);
    while (this.q[0] < t - 3000) {
        this.q.shift();
    }
    return this.q.length;
};

/**
 * Your RecentCounter object will be instantiated and called as such:
 * var obj = new RecentCounter()
 * var param_1 = obj.ping(t)
 */

class RecentCounter {
    private var q: [Int]

    init() {
        q = []
    }

    func ping(_ t: Int) -> Int {
        q.append(t)
        while q.first! < t - 3000 {
            q.removeFirst()
        }
        return q.count
    }
}

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

剑指 Offer II 042. 最近请求次数

题目描述

解法

方法一

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