# 379.Design-Phone-Directory

## 379. Design Phone Directory

## 题目地址

<https://leetcode.com/problems/design-phone-directory/>

## 题目描述

```
Design a Phone Directory which supports the following operations:

get: Provide a number which is not assigned to anyone.
check: Check if a number is available or not.
release: Recycle or release a number.
Example:

// Init a phone directory containing a total of 3 numbers: 0, 1, and 2.
PhoneDirectory directory = new PhoneDirectory(3);

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

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

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

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

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

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

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

## 代码

### Approach #1

```java
class PhoneDirectory {

    Set<Integer> used = new HashSet<Integer>();
  Queue<Integer> queue = new LinkedList<Integer>();
  int max;
  public PhoneDirectory(int maxNumbers) {
        max = maxNumbers;
    for (int i = 0; i < maxNumbers; i++) {
      queue.offer(i);
    }
  }

  /** Provide a number which is not assigned to anyone.
      @return - Return an available number. Return -1 if none is available. */
  public int get() {
        Integer ret = queue.poll();
    if (ret == null)    return -1;
    used.add(ret);
    return ret;
  }

  /** Check if a number is available or not. */
  public boolean check(int number) {
        if (number >= max || number < 0) {
      return false;
    }
    return !used.contains(number);
  }

  /** Recycle or release a number. */
  public void release(int number) {
        if (used.remove(number)) {
      queue.offer(number);
    }
  }
}

/**
* Your PhoneDirectory object will be instantiated and called as such:
* PhoneDirectory obj = new PhoneDirectory(maxNumbers);
* int param_1 = obj.get();
* boolean param_2 = obj.check(number);
* obj.release(number);
*/
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://wentao-shao.gitbook.io/leetcode/stack/379.design-phone-directory.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.