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);

分析

1 一个set放released, 一个int记录当前已达数字（++--）

2 一个set装满所有available，加加减减.cons 初始要O(N)

3 java BitSet 和 index。 index就是当前数，release 就clear and set index. get 就 return index and set index

4 segment tree

Design Phone Directory - LeetCodeLeetCode

Design Phone Directory - LeetCodeLeetCode

方法1:

release要有判断，max不可达

class PhoneDirectory:

    def __init__(self, maxNumbers: int):
        """
        Initialize your data structure here
        @param maxNumbers - The maximum numbers that can be stored in the phone directory.
        """
        self.next = 0
        self.released = set()
        self.max = maxNumbers
        

    def get(self) -> int:
        """
        Provide a number which is not assigned to anyone.
        @return - Return an available number. Return -1 if none is available.
        """
        res = 0
        if self.next < self.max:
            res = self.next
            self.next += 1
            return res
        if self.released:
            return self.released.pop()
        
        return -1
        
            
        

    def check(self, number: int) -> bool:
        """
        Check if a number is available or not.
        """
        if number in self.released or self.next <= number < self.max :
            return True
        return False
        

    def release(self, number: int) -> None:
        """
        Recycle or release a number.
        """
        if number < self.next: #记得要判断这里，而且不是<max
            self.released.add(number)
        


# Your PhoneDirectory object will be instantiated and called as such:
# obj = PhoneDirectory(maxNumbers)
# param_1 = obj.get()
# param_2 = obj.check(number)
# obj.release(number)

方法2：

class PhoneDirectory:

    def __init__(self, maxNumbers: int):
        """
        Initialize your data structure here
        @param maxNumbers - The maximum numbers that can be stored in the phone directory.
        """
        self.avail = set(range(maxNumbers))
        

    def get(self) -> int:
        """
        Provide a number which is not assigned to anyone.
        @return - Return an available number. Return -1 if none is available.
        """
        if self.avail:
            return self.avail.pop()
        return -1
        

    def check(self, number: int) -> bool:
        """
        Check if a number is available or not.
        """
        return number in self.avail
        

    def release(self, number: int) -> None:
        """
        Recycle or release a number.
        """
        self.avail.add(number)
            
        


# Your PhoneDirectory object will be instantiated and called as such:
# obj = PhoneDirectory(maxNumbers)
# param_1 = obj.get()
# param_2 = obj.check(number)
# obj.release(number)

3 bitset

Space is much efficient: O(c), wouldn’t say it is O(1) because we still need max_size number of bits

这里bit的长度 = max number,某个数字就是设置这个位上的bit

nextClearBit(num)注意此处需要参数

class PhoneDirectory {

    /** Initialize your data structure here
        @param maxNumbers - The maximum numbers that can be stored in the phone directory. */
    BitSet bs;
    final int max;
    int nextAvail;
    public PhoneDirectory(int maxNumbers) {
        bs = new BitSet(maxNumbers);
        max = maxNumbers;
        nextAvail = 0;      
    }
    
    /** Provide a number which is not assigned to anyone.
        @return - Return an available number. Return -1 if none is available. */
    public int get() {
        if (nextAvail >= max) {
            return -1;
        }
        int res = nextAvail;
        bs.set(res); 
        nextAvail = bs.nextClearBit(res);//注意nextClearBit用法，需要传入num
        return res;
        
        
    }
    
    /** Check if a number is available or not. */
    public boolean check(int number) {
        return number < max &&  !bs.get(number);
        
    }
    
    /** Recycle or release a number. */
    public void release(int number) {
        if(number >= 0 && number < max){
            bs.clear(number);
        }
        nextAvail = Math.min(number,nextAvail);
        
    }
}

/**
 * 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);
 */

4 segment Tree

O(log n) time in allocate and release

总共2*max个节点，max个叶节点》max，max个内节点 《max。线段树是Bool arr

get取第一个超过max的未赋值，也就是叶节点。 每次使用叶节点或者释放，都要更新内节点。pushup

class PhoneDirectory:

    def __init__(self, maxNumbers: int):
        """
        Initialize your data structure here
        @param maxNumbers - The maximum numbers that can be stored in the phone directory.
        """
        self.max = maxNumbers
        self.segTree = [True]*(self.max << 1)
    
    def pushup(self, i) -> None:
        i >>=1
        while i > 0:
            self.segTree[i] =  self.segTree[i<<1] or self.segTree[i<<1|1]
            i>>=1

    def get(self) -> int:
        """
        Provide a number which is not assigned to anyone.
        @return - Return an available number. Return -1 if none is available.
        """
        if not self.segTree[1]:
            return -1
        i = 1
        while i < self.max:
            if (i << 1) < (self.max << 1) and  self.segTree[i<<1]:
                i = i<<1
            if i<< 1|1 < self.max << 1 and  self.segTree[i<<1|1]:
                i = i << 1|1
        self.segTree[i] = False   
        res = i - self.max
        self.pushup(i)
        return res
            
        
        
        

    def check(self, number: int) -> bool:
        """
        Check if a number is available or not.
        """
        if 0<=number < self.max:
            return self.segTree[number + self.max]
        return False
        

    def release(self, number: int) -> None:
        """
        
        Recycle or release a number.
        """
        i = number+self.max
        self.segTree[i] = True
        self.pushup(i)
        
        


# Your PhoneDirectory object will be instantiated and called as such:
# obj = PhoneDirectory(maxNumbers)
# param_1 = obj.get()
# param_2 = obj.check(number)
# obj.release(number)