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74 lines
2.1 KiB
Markdown
74 lines
2.1 KiB
Markdown
# [933. Number of Recent Calls](https://leetcode.com/problems/number-of-recent-calls/)
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## 题目
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Write a class `RecentCounter` to count recent requests.
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It has only one method: `ping(int t)`, where t represents some time in milliseconds.
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Return the number of `ping`s that have been made from 3000 milliseconds ago until now.
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Any ping with time in `[t - 3000, t]` will count, including the current ping.
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It is guaranteed that every call to `ping` uses a strictly larger value of `t` than before.
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**Example 1**:
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```
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Input: inputs = ["RecentCounter","ping","ping","ping","ping"], inputs = [[],[1],[100],[3001],[3002]]
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Output: [null,1,2,3,3]
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```
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**Note**:
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1. Each test case will have at most `10000` calls to `ping`.
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2. Each test case will call `ping` with strictly increasing values of `t`.
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3. Each call to ping will have `1 <= t <= 10^9`.
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## 题目大意
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写一个 RecentCounter 类来计算最近的请求。它只有一个方法：ping(int t)，其中 t 代表以毫秒为单位的某个时间。返回从 3000 毫秒前到现在的 ping 数。任何处于 [t - 3000, t] 时间范围之内的 ping 都将会被计算在内，包括当前（指 t 时刻）的 ping。保证每次对 ping 的调用都使用比之前更大的 t 值。
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提示：
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- 每个测试用例最多调用 10000 次 ping。
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- 每个测试用例会使用严格递增的 t 值来调用 ping。
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- 每次调用 ping 都有 1 <= t <= 10^9。
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## 解题思路
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- 要求设计一个类，可以用 `ping(t)` 的方法，计算 [t-3000, t] 区间内的 ping 数。t 是毫秒。
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- 这一题比较简单，`ping()` 方法用二分搜索即可。
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## 代码
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```go
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type RecentCounter struct {
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list []int
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}
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func Constructor933() RecentCounter {
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return RecentCounter{
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list: []int{},
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}
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}
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func (this *RecentCounter) Ping(t int) int {
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this.list = append(this.list, t)
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index := sort.Search(len(this.list), func(i int) bool { return this.list[i] >= t-3000 })
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if index < 0 {
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index = -index - 1
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}
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return len(this.list) - index
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}
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/**
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* Your RecentCounter object will be instantiated and called as such:
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* obj := Constructor();
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* param_1 := obj.Ping(t);
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*/
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``` |