379. Design Phone Directory

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题目

Design a phone directory that initially has maxNumbers empty slots that can store numbers. The directory should store numbers, check if a certain slot is empty or not, and empty a given slot.

Implement the PhoneDirectory class:

• PhoneDirectory(int maxNumbers) Initializes the phone directory with the number of available slots maxNumbers.
• int get() Provides a number that is not assigned to anyone. Returns -1 if no number is available.
• bool check(int number) Returns true if the slot number is available and false otherwise.
• void release(int number) Recycles or releases the slot number.

Example 1:

Input

["PhoneDirectory", "get", "get", "check", "get", "check", "release", "check"]

[[3], [], [], [2], [], [2], [2], [2]]

Output

[null, 0, 1, true, 2, false, null, true]

Explanation

PhoneDirectory phoneDirectory = new PhoneDirectory(3);

phoneDirectory.get(); // It can return any available phone number. Here we assume it returns 0.

phoneDirectory.get(); // Assume it returns 1.

phoneDirectory.check(2); // The number 2 is available, so return true.

phoneDirectory.get(); // It returns 2, the only number that is left.

phoneDirectory.check(2); // The number 2 is no longer available, so return false.

phoneDirectory.release(2); // Release number 2 back to the pool.

phoneDirectory.check(2); // Number 2 is available again, return true.

Constraints:

• 1 <= maxNumbers <= 10^4
• 0 <= number < maxNumbers
• At most 2 * 10^4 calls will be made to get, check, and release.

题目大意

要求设计一个电话目录管理系统，包含以下两个操作：

1. get()：分配一个未被使用的电话号码，返回其编号，如果没有未被使用的号码，则返回 -1。
2. check(number)：检查指定的电话号码是否被使用。
3. release(number)：释放一个已经被使用的电话号码。

解题思路

这个问题可以通过维护一个可用号码集合和已使用号码集合来实现，可以使用两个数据结构：

1. 一个 Set 存储可用的号码，初始时包含所有可能的号码。
2. 一个 Set 存储已经被使用的号码，初始时为空。

通过两个 Set 来分别管理可用和已使用的号码，实现了分配、检查和释放号码的功能。

代码

class PhoneDirectory {
  constructor(maxNumbers) {
    this.available = new Set();
    this.used = new Set();
    for (let i = 0; i < maxNumbers; i++) {
      this.available.add(i);
    }
  }
  get() {
    if (this.available.size == 0) return -1;
    const num = this.available.values().next().value;
    this.available.delete(num);
    this.used.add(num);
    return num;
  }
  check(number) {
    return !this.used.has(number);
  }
  release(number) {
    if (this.used.has(number)) {
      this.used.delete(number);
      this.available.add(number);
    }
  }
}

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