379. Design Phone Directory

Design a Phone Directory which supports the following operations:

1. get: Provide a number which is not assigned to anyone.
2. check: Check if a number is available or not.
3. 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);

一刷
题解：本来打算只用一个set作为pool, 但是每次get都需要初始化一个iterator, 这样当多次get的时候会出现超时。
采用discuss中的解法，有一个queue中存放数字，一个set中存放已经取出的数字。于是只调用set中add，remove, contains这几个方法。

class PhoneDirectory {

    Set<Integer> used = new HashSet<Integer>();
    Queue<Integer> available = new LinkedList<Integer>();
    int max;
    public PhoneDirectory(int maxNumbers) {
        max = maxNumbers;
        for (int i = 0; i < maxNumbers; i++) {
            available.offer(i);
        }
    }
    
    public int get() {
        Integer ret = available.poll();
        if (ret == null) {
            return -1;
        }
        used.add(ret);
        return ret;
    }
    
    public boolean check(int number) {
        if (number >= max || number < 0) {
            return false;
        }
        return !used.contains(number);
    }
    
    public void release(int number) {
        if (used.remove(number)) {
            available.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);
 */