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);

又是一道设计题，让我们设计一个电话目录管理系统，可以分配电话号码，查询某一个号码是否已经被使用，释放一个号码，需要注意的是，之前释放的号码下一次应该被优先分配。这题对C++解法的时间要求非常苛刻，尝试了好几种用set，或者stack/queue，或者使用vector的push_back等等，都TLE了，终于找到了一种可以通过OJ的解法。这里用两个一维数组recycle和flag，分别来保存被回收的号码和某个号码的使用状态，还有变量max_num表示最大数字，next表示下一个可以分配的数字，idx表示recycle数组中可以被重新分配的数字的位置，然后在get函数中，没法分配的情况是，当next等于max_num并且index小于等于0，此时返回-1。否则我们先看recycle里有没有数字，有的话先分配recycle里的数字，没有的话再分配next。记得更新相对应的flag中的使用状态，参见代码如下：

class PhoneDirectory {
public:
    /** Initialize your data structure here
        @param maxNumbers - The maximum numbers that can be stored in the phone directory. */
    PhoneDirectory(int maxNumbers) {
        max_num = maxNumbers;
        next = idx = 0;
        recycle.resize(max_num);
        flag.resize(max_num, 1);
    }
    
    /** Provide a number which is not assigned to anyone.
        @return - Return an available number. Return -1 if none is available. */
    int get() {
        if (next == max_num && idx <= 0) return -1;
        if (idx > 0) {
            int t = recycle[--idx];
            flag[t] = 0;
            return t;
        }
        flag[next] = false;
        return next++;
    }
    
    /** Check if a number is available or not. */
    bool check(int number) {
        return number >= 0 && number < max_num && flag[number];
    }
    
    /** Recycle or release a number. */
    void release(int number) {
        if (number >= 0 && number < max_num && !flag[number]) {
            recycle[idx++] = number;
            flag[number] = 1;
        }
    }
private:
    int max_num, next, idx;
    vector<int> recycle, flag;
};

参考资料：

https://discuss.leetcode.com/topic/53136/all-c-solutions-got-lte/2