摘要：題目鏈接直接用一個，結果了看了加了個，不過感覺沒什么必要加，反正保存的都一樣，只是的時間大于，用可以保證。看了題目條件是可以隨便返回一個值，但是不讓這么做。很無語啊如果這道題要求要求的是的，那就和一樣了。

題目鏈接：https://leetcode.com/problems...

直接用一個set，結果tle了= =

public class PhoneDirectory {

    /** Initialize your data structure here
        @param maxNumbers - The maximum numbers that can be stored in the phone directory. */
    Set set = new HashSet();
    int maxNum;
    public PhoneDirectory(int maxNumbers) {
        for(int i = 0; i < maxNumbers; i++) set.add(i);
        maxNum = maxNumbers;
    }
    
    /** Provide a number which is not assigned to anyone.
        @return - Return an available number. Return -1 if none is available. */
    public int get() {
        // no available numbers left
        if(set.size() == 0) return -1;
        // return 1 number
        else {
            int number = set.iterator().next();
            set.remove(number);
            return number;
        }
    }
    
    /** Check if a number is available or not. */
    public boolean check(int number) {
        return set.contains(number);
    }
    
    /** Recycle or release a number. */
    public void release(int number) {
        if(number >= 0 && number < maxNum) set.add(number);
    }
}

看了discussion加了個queue，不過感覺沒什么必要加q，反正保存的都一樣，只是set.iterator().next()的時間大于O(1)，用q可以保證O(1)。看了題目條件是get可以隨便返回一個值，但是test case不讓這么做。很無語啊

public class PhoneDirectory {

    /** Initialize your data structure here
        @param maxNumbers - The maximum numbers that can be stored in the phone directory. */
    Set set = new HashSet();
    int maxNum;
    Queue available = new LinkedList();
    public PhoneDirectory(int maxNumbers) {
        for(int i = 0; i < maxNumbers; i++) available.add(i);
        maxNum = maxNumbers;
    }
    
    /** Provide a number which is not assigned to anyone.
        @return - Return an available number. Return -1 if none is available. */
    public int get() {
        // no available numbers left
        if(available.size() == 0) return -1;
        // return 1 number
        else {
            int number = available.poll();
            set.add(number);
            return number;
        }
    }
    
    /** Check if a number is available or not. */
    public boolean check(int number) {
        return !set.contains(number);
    }
    
    /** Recycle or release a number. */
    public void release(int number) {
        if(number >= 0 && number < maxNum) {
            if(set.remove(number)) available.add(number);
        }
    }
}

如果這道題要求get要求的是random的，那就和380. Insert Delete GetRandom O(1)一樣了。
https://segmentfault.com/a/11...