933. 最近的请求次数

题目描述

写一个 RecentCounter 类来计算特定时间范围内最近的请求。

请你实现 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 1：

输入：
["RecentCounter", "ping", "ping", "ping", "ping"]
[[], [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

解法

方法一：队列

由题得知，t 是严格递增的，当一个元素不满足 [t - 3000, t] 条件时，在后续的请求当中，它也不可能满足。

对此，需要将其从记录容器中移除，减少无意义的比较。

可以使用队列。每次将 t 进入队尾，同时从队头开始，依次移除小于 t - 3000 的元素。然后返回队列的大小（q.size()）即可。

Python3JavaC++GoTypeScriptRustJavaScriptC#

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 int[] s = new int[10010];
    private int idx;

    public RecentCounter() {
    }

    public int ping(int t) {
        s[idx++] = t;
        return idx - search(t - 3000);
    }

    private int search(int x) {
        int left = 0, right = idx;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (s[mid] >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

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

class RecentCounter {
public:
    queue<int> q;

    RecentCounter() {
    }

    int ping(int t) {
        q.push(t);
        while (q.front() < t - 3000) q.pop();
        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{[]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:]
    }
    return len(this.q)
}

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

class RecentCounter {
    private queue: number[];

    constructor() {
        this.queue = [];
    }

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

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

use std::collections::VecDeque;
struct RecentCounter {
    queue: VecDeque<i32>,
}

/**
 * `&self` means the method takes an immutable reference.
 * If you need a mutable reference, change it to `&mut self` instead.
 */
impl RecentCounter {
    fn new() -> Self {
        Self {
            queue: VecDeque::new(),
        }
    }

    fn ping(&mut self, t: i32) -> i32 {
        self.queue.push_back(t);
        while let Some(&v) = self.queue.front() {
            if v >= t - 3000 {
                break;
            }
            self.queue.pop_front();
        }
        self.queue.len() as i32
    }
}

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)
 */

public class RecentCounter {
    private Queue<int> q = new Queue<int>();

    public RecentCounter() {

    }

    public int Ping(int t) {
        q.Enqueue(t);
        while (q.Peek() < t - 3000)
        {
            q.Dequeue();
        }
        return q.Count;
    }
}

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

方法二：二分查找

t 严格单调递增，非常适合用二分查找来定位 [t-3000, t] 的左右边界。

Python3C++Go

class RecentCounter:
    def __init__(self):
        self.s = []

    def ping(self, t: int) -> int:
        self.s.append(t)
        return len(self.s) - bisect_left(self.s, t - 3000)


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

class RecentCounter {
public:
    vector<int> s;

    RecentCounter() {
    }

    int ping(int t) {
        s.push_back(t);
        return s.size() - (lower_bound(s.begin(), s.end(), t - 3000) - s.begin());
    }
};

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

type RecentCounter struct {
    s []int
}

func Constructor() RecentCounter {
    return RecentCounter{[]int{}}
}

func (this *RecentCounter) Ping(t int) int {
    this.s = append(this.s, t)
    search := func(x int) int {
        left, right := 0, len(this.s)
        for left < right {
            mid := (left + right) >> 1
            if this.s[mid] >= x {
                right = mid
            } else {
                left = mid + 1
            }
        }
        return left
    }
    return len(this.s) - search(t-3000)
}

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

933. 最近的请求次数

题目描述

解法

方法一：队列

方法二：二分查找

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