379. Design Phone Directory
Problem:
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);Analysis:
一开始题意理解错了。get是提供没有用过的。
Use queue and set, set stores the numbers that have been used. The number to be check might less than 0 or greater than max number. So we need to check this as well.
Solution:
class PhoneDirectory { Set<Integer> set; Queue<Integer> queue; int max; /** Initialize your data structure here @param maxNumbers - The maximum numbers that can be stored in the phone directory. */ public PhoneDirectory(int maxNumbers) { queue = new LinkedList<>(); set = new HashSet<>(); max = maxNumbers; for (int i = 0; i < maxNumbers; i++) { queue.offer(i); } } /** Provide a number which is not assigned to anyone. @return - Return an available number. Return -1 if none is available. */ public int get() { if (!queue.isEmpty()) { set.add(queue.peek()); return queue.poll(); } else { return -1; } } /** Check if a number is available or not. */ public boolean check(int number) { if (number > max || number < 0) return false; return !set.contains(number); } /** Recycle or release a number. */ public void release(int number) { if (set.remove(number)) { queue.offer(number); } } }
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