【Java】动物收容所

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有家动物收容所只收容狗与猫，且严格遵守“先进先出”原则。在收养该收容所的动物时，收养人只能收养进入收容所时间最长的动物，或者，挑选猫或狗中收养时间最长的。

请创建适合这个系统的数据结构，实现各种操作方法，比如enqueue，dequeueAny， dequeDog, dequeueCat等，允许使用java内置的LinkedList数据结构

不能使用Queue类，Queue是接口，实例化接口要实例化接口的每个方法

分别实现两个queue存放dog和 cat，并用一个变量记录下每个动物进入队列的时间，需要收养时，则将时间最长的出队

import java.util.LinkedList;public class shourongsuo {public abstract class Animal {//kinda timestamp to mark how long did the animal benn in the zooint order;protected String name;public Animal(String n) {name = n;}public void setOrder(int ord) {order = ord;}public int getOrder() {return order;}public boolean isOlderThan(Animal a) {return this.order < a.getOrder();}}public class Dog extends Animal {public Dog(String n) {super(n);}}public class Cat extends Animal {public Cat(String n) {super(n);}}public class AnimalQueue {LinkedList<Dog> dogs = new LinkedList<Dog> ();LinkedList<Cat> cats = new LinkedList<Cat> ();private int order = 0; //time stamppublic void enqueue(Animal a) {a.setOrder(order);order++;if ( a instanceof Dog) dogs.addLast((Dog)a);else if (a instanceof Cat) cats.addLast((Cat)a);}public Animal dequeueAny() {if (dogs.size() == 0) {return dequeueCats();}else if (cats.size() == 0) {return dequeueDogs();}Dog dog = dogs.peek();Cat cat = cats.peek();if ( dog.isOlderThan(cat) ) {return dequeueDogs();}else {return dequeueCats();}}public Dog dequeueDogs() {return dogs.poll();}public Cat dequeueCats() {return cats.poll();}}}

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