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6.猫狗队列问题

时间:2019-01-26 00:49:31      阅读:195      评论:0      收藏:0      [点我收藏+]

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问题:宠物、狗、猫的类如下

class Pet {
    private String type;

    public Pet(String type) {
        this.type = type;
    }

    public String getType() {
        return this.type;
    }
    public String toString(){
        return this.getType();
    }
}

class Dog extends Pet {
    public Dog(String type) {
        super(type);
    }

    @Override
    public String toString() {
        return super.toString();
    }
}

class Cat extends Pet {
    public Cat(String type) {
        super(type);
    }

    @Override
    public String toString() {
        return super.toString();
    }
}

  实现一个猫狗队列,要求如下:

  1.可以调用add()方法,将cat类和dog类添加到队列中;

  2.可以调用pollAll()方法,将队列中的所有实例按入队列的顺序依次弹出;

  3.可以调用pollCat()方法,将队列中的所有Cat对象实例按入队列的顺序依次弹出;

  4.可以调用pollDog()方法,将队列中的所有Dog对象实例按入队列的顺序依次弹出;

  5.可以调用isEmpty()方法,判断对队列是否为空;

  6.可以调用isCatEmpty()方法,判断队列是否有Cat对象实例;

  7.可以调用isDogEmpty()方法,判断队列是否有Dog对象实例;

 

思路:队列是先进先出,由于要存放两种不同的对象,因此可以用两个队列分别存放Dog对象和Cat对象。由于要维持进入队列的顺序,在不改变原有类的情况下,我们可以构造一个类EnterPetQueue,在类中引入计数变量,来记录对象的入队顺序。

EnterPetQueue构造类

class EnterPetQueue{
    private Pet pet;
    private Integer count;
    public EnterPetQueue(Pet pet,Integer count){
        this.pet=pet;
        this.count=count;
    }
    public Pet getPet(){
        return this.pet;
    }
    public Integer getCount(){
        return this.count;
    }

}

CatAndDogQueue类

public class CatAndDogQueue {
    private LinkedList<EnterPetQueue> catQueue;
    private LinkedList<EnterPetQueue> dogQueue;
    private Integer count=0;
    public CatAndDogQueue() {
        catQueue = new LinkedList<>();
        dogQueue = new LinkedList<>();
    }
    public void add(Pet pet){
        if(pet.getType().contains("cat")){
            catQueue.addLast(new EnterPetQueue(pet,count++));
        }else if(pet.getType().contains("dog")){
            dogQueue.addLast(new EnterPetQueue(pet,count++));
        }else{
            throw new RuntimeException("只能添加Cat和Dog实例,不能添加其他实例");
        }
    }
    public boolean isDogEmpty(){
        return dogQueue.isEmpty();
    }
    public boolean isCatEmpty(){
        return catQueue.isEmpty();
    }
    public boolean isEmpyt(){
        return isCatEmpty()&&isDogEmpty();
    }
    public Pet poolAll(){
        if(isEmpyt()){
            return null;
        }else if(isDogEmpty()){
            return catQueue.pollFirst().getPet();
        }else if(isCatEmpty()){
            return dogQueue.pollFirst().getPet();
        }else{
            if(catQueue.getFirst().getCount()<dogQueue.getFirst().getCount()){
                return catQueue.pollFirst().getPet();
            }else{
                return dogQueue.pollFirst().getPet();
            }
        }

    }
    public Dog pollDog(){
        if(isDogEmpty()){
            return null;
        }else{
            return (Dog)dogQueue.pollFirst().getPet();
        }
    }
    public Cat pollCat(){
        if(isCatEmpty()){
            return null;
        }else{
            return (Cat)catQueue.pollFirst().getPet();
        }
    }
}

测试类Test

public class Test {

    public static void main(String[] args) {
        Cat cat1=new Cat("cat1");
        Cat cat2=new Cat("cat2");
        Cat cat3=new Cat("cat3");

        Dog dog1=new Dog("dog1");
        Dog dog2=new Dog("dog2");
        Dog dog3=new Dog("dog3");

        CatAndDogQueue catAndDogQueue=new CatAndDogQueue();
        catAndDogQueue.add(cat1);
        catAndDogQueue.add(dog3);
        catAndDogQueue.add(cat3);
        catAndDogQueue.add(dog1);
        catAndDogQueue.add(cat2);
        catAndDogQueue.add(dog2);
        while (!catAndDogQueue.isEmpyt()){
            System.out.println(catAndDogQueue.poolAll());
        }
        while(!catAndDogQueue.isCatEmpty()){
            System.out.println(catAndDogQueue.pollCat());
        }
        while(!catAndDogQueue.isDogEmpty()){
            System.out.println(catAndDogQueue.pollDog());
        }
        System.out.println(catAndDogQueue.poolAll());
    }
}

 

6.猫狗队列问题

标签:main   oge   new   etc   contains   思路   构造   ati   xtend   

原文地址:https://www.cnblogs.com/quxiangxiangtiange/p/10322380.html

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