379.Design-Phone-Directory
379. Design Phone Directory
题目地址
https://leetcode.com/problems/design-phone-directory/
题目描述
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);
代码
Approach #1
class PhoneDirectory {
Set<Integer> used = new HashSet<Integer>();
Queue<Integer> queue = new LinkedList<Integer>();
int max;
public PhoneDirectory(int maxNumbers) {
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() {
Integer ret = queue.poll();
if (ret == null) return -1;
used.add(ret);
return ret;
}
/** Check if a number is available or not. */
public boolean check(int number) {
if (number >= max || number < 0) {
return false;
}
return !used.contains(number);
}
/** Recycle or release a number. */
public void release(int number) {
if (used.remove(number)) {
queue.offer(number);
}
}
}
/**
* 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);
*/
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