JAVA基础面试题-3(答案版)

JAVA语言基础笔试题-3

Question 1

Given:

1. public class Team extends java.util.LinkedList {

2. public void addPlayer(Player p) {

3. add(p);

4. }

5. public void compete(Team opponent) { /* more code here */ }

6. }

7. class Player { /* more code here */ }

Which two are true? (Choose two.)

A. This code will compile.

B. This code demonstrates proper design of an is-a relationship.

C. This code demonstrates proper design of a has-a relationship.

D. A Java programmer using the Team class could remove Player

objects from a Team object.

答案：AD

考点：List接口实现，对象间泛化和关联关系

说明：

is a 关系一般用继承来表示，但此题这个继承从逻辑上不是太恰当.

Has a 关系一般来说表示某对象体内有其它对象的存在，该对象体现为属性形态，在

UML中叫做关联关系。本题Team中虽然可以保存Player，但并非显式以属性形式存在。

由于LinkedList自带remove方法，可以通过Team.remove(p) 来删除player对象。

Question 2

Which four are true? (Choose four.)

A. Has-a relationships should never be encapsulated.

B. Has-a relationships should be implemented using inheritance.

C. Has-a relationships can be implemented using instance variables.

D. Is-a relationships can be implemented using the extends keyword.

E. Is-a relationships can be implemented using the implements

keyword.

F. The relationship between Movie and Actress is an example of an is-a relationship.

G. An array or a collection can be used to implement a one-to-many

has-a relationship.

答案：CDEG

考点：集合类型，对象间泛化和关联关系的理解

Has a 关系一般表示为一个类拥有一个属性，属性被封装是常见的事情。

Is a 关系一般用继承来表示，子类体内拥有父类部分。

接口的实现，也适用于is a 关系来理解，因为接口从本质来说，也属于类的形态。

Question 3

Which two are true about has-a and is-a relationships? (Choose two.)

A. Inheritance(继承) represents an is-a relationship.

B. Inheritance represents a has-a relationship.

C. Interfaces must be used when creating a has-a relationship.

D. Instance variables can be used when creating a has-a relationship. 答案：AD

考点：对象间泛化和关联关系的理解

Question 4

Given:

10. interface Jumper { public void jump(); }

......

20. class Animal {}

......

30. class Dog extends Animal {

31. Tail tail;

32. }

......

40. class Beagle extends Dog implements Jumper {

41. public void jump() { }

42. }

.......

50. class Cat implements Jumper {

51. public void jump() { }

52. }

Which three are true? (Choose three.)

A. Cat is-a Animal

B. Cat is-a Jumper

C. Dog is-a Animal

D. Dog is-a Jumper

E. Cat has-a Animal

F. Beagle has-a Tail

G. Beagle has-a Jumper

答案：BCF

考点：对象间泛化和关联关系的理解

Question 5

Given:

1. import java.util.*;

2. public class Example {

3. public static void main(String[] args) {

4. // insert code here

5. set.add(new Integer(2));

6. set.add(new Integer(l));

7. System.out.println(set);

8. }

9. }

Which code, inserted at line 4, guarantees that this program will

output [1, 2]?

A. Set set = new TreeSet();

B. Set set = new HashSet();

C. Set set = new SortedSet();

D. List set = new SortedList();

E. Set set = new LinkedHashSet();

答案：A

考点：集合类型的概念，TreeSet的特性

说明：

HashSet:底层数据结构是hash表，先判断hash值，然后比较值

TreeSet:数据结构二叉排序树，可以对set集合中的元素进行排序，自定义类需要实现comparable接口（排序时，当主要条件相同时，一定要判断次要条件）

此题从输出角度来看，属于天然顺序输出，目前只有TreeSet能达到该效果，同时Integer 类也实现了Comparable接口，所以可以安全的在TreeSet中使用。

Question 6

Given:

1. import java.util.*;

2. public class PQ {

3. public static void main(String[] args) {

4. PriorityQueue pq = new PriorityQueue();

5. pq.add(”carrot”);

6. pq.add(”apple”);

7. pq.add(”banana”);

8. System.out.printl n(pq.poll() +”:” + pq.peek());

9. }

10. }

What is the result?

A. apple:apple

B. carrot:apple

C. apple:banana

D. banana:apple

E. carrot:carrot

F. carrot:banana

答案：C

考点：集合类型，了解Queue接口的使用

说明：

优先级队列，如果不提供Comparable的话，优先级队列中的元素默认按自然顺序排列，也就是数字默认是小的在队列头，字符串则按字典序排列

Poll()从队列头部取值，该值从队列中删除; peek()并未实际取值，而仅是取得队列头部的元素。

Question 7

Given:

1. import java.util.*;

2. public class WrappedString {

3. private String s;

4. public WrappedString(String s) { this.s = s; }

5. public static void main(String[] args) {

6. HashSet