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);
Python Solution:
class PhoneDirectory(object):
def __init__(self, maxNumbers):
"""
Initialize your data structure here
@param maxNumbers - The maximum numbers that can be stored in the phone directory.
:type maxNumbers: int
"""
self.valid = set()
self.begin = 0
self.end = maxNumbers
def get(self):
"""
Provide a number which is not assigned to anyone.
@return - Return an available number. Return -1 if none is available.
:rtype: int
"""
if self.begin < self.end:
self.begin += 1
return self.begin - 1
if not self.valid:
return -1
val = self.valid.pop()
return val
def check(self, number):
"""
Check if a number is available or not.
:type number: int
:rtype: bool
"""
return self.begin <= number < self.end or number in self.valid
def release(self, number):
"""
Recycle or release a number.
:type number: int
:rtype: void
"""
if number < self.begin:
self.valid.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)
Summary:
- It's much easier to use python.
- Is this related to linked list?
- 125ms, 80.65%
LeetCode: 379. Design Phone Directory





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