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
     
       recycle, flag;};
     

 

参考资料：