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_379.java
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_379.java
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/**
* LeetCode 379 - Design Phone Directory
*
* This problem is a bit weird. I don't see why the function check() is particularly useful.
* Ideally, one would like to do the following two steps in a row.
*
* if (check(number)) get(number);
*
* But, the get() function in this problem is not restricted and thus can return any available numbers.
*
* The problem will become more interesting if it asks for implementing one more method:
*
* bool checkAndGet(int number)
* @return - Return true if number is available and reserve it on the fly.
* - Return false if number is already requested, indicating the failure of this attempt.
*
*/
public class _379 {
int[] heap;
boolean[] visited;
int heapPtr;
/**
* Initialize your data structure here
*
* @param maxNumbers - The maximum numbers that can be stored in the phone directory.
*/
public _379(int maxNumbers) {
heap = new int[maxNumbers];
visited = new boolean[maxNumbers];
for (int i = 0; i < maxNumbers; i++) heap[i] = i;
heapPtr = maxNumbers;
}
/**
* Provide a number which is not assigned to anyone.
*
* @return - Return an available number. Return -1 if none is available.
*/
public int get() {
if (heapPtr <= 0) return -1;
int num = heap[--heapPtr];
visited[num] = true;
return num;
}
/**
* Check if a number is available or not.
*/
public boolean check(int number) {
return !visited[number];
}
/**
* Recycle or release a number.
*/
public void release(int number) {
if (visited[number]) {
visited[number] = false;
heap[heapPtr++] = 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);
*/