Question

JAVA 习题, 高手进

Q6: Complete the output when the ex06() method is executed:

private void ex06() {
ArrayList myList = createArrayList();
System.out.println( "1. LIST IS: " + myList );
Object item1 = myList.get(0);
System.out.println("2. Element at position 2 " + myList.get(2));
int pos2 = myList.indexOf( new Point(3,3) );
if( pos2 > -1 ) {
myList.remove(pos2);
}
Point item3 = new Point(3, 3);
myList.add(1, item3);
myList.remove(1);

System.out.print("3. ");
for( int i=0; i<myList.size(); i++ ) {
String item = (myList.get(i)).toString();
if (item.length() == 3 ) {
System.out.print(item.toString() + " ");
}
}
System.out.println();
System.out.println("4. LIST IS: " + myList );
}
private ArrayList createArrayList() {
ArrayList myList = new ArrayList();
myList.add( new String("Uno") );
myList.add( new String("Due") );
myList.add( new Point(3, 3) );
myList.add( new Integer(456) );
myList.add( new Character('c') );
myList.add( new String("Tre") );
myList.add( new Point(3, 3) );
return myList;
}

1. LIST IS: [Uno, Due, java.awt.Point[x=3,y=3], 456, c, Tre, java.awt.Point[x=3,y=3]]
2.
3.
4. LIST IS: [ ]

这是道填空题:
帮我解释下, 谢谢

Answer 1

　　运行结果如下：
　　1. LIST IS: [Uno, Due, java.awt.Point[x=3,y=3], 456, c, Tre, java.awt.Point[x=3,y=3]]
　　2. Element at position 2 java.awt.Point[x=3,y=3]
　　3. Uno Due 456 Tre
　　4. LIST IS: [Uno, Due, 456, c, Tre, java.awt.Point[x=3,y=3]]

　　解释：
　　1.定义了方法createArrayList() ：该方法中创建了一个ArrayList实例，对其复制（依次地添加元素），并返回该实例；
　　2.System.out.println( "1. LIST IS: " + myList )，打印“1.LIST：”以及ArrayList类型的myList实例；
　　3.System.out.println("2. Element at position 2 " + myList.get(2));
　　打印myList中下标为2的元素；
　　4.int pos2 = myList.indexOf( new Point(3,3) );
　　if( pos2 > -1 ) {
　　myList.remove(pos2);
　　}
　　Point item3 = new Point(3, 3);
　　myList.add(1, item3);
　　myList.remove(1);
　　获取元素new Point(3,3)的下标，若存在，则将其删除，并在myList下标为1的位置中插入元素item3；在删除下标为1的元素。
　　5.for( int i=0; i<myList.size(); i++ ) {
　　String item = (myList.get(i)).toString();
　　if (item.length() == 3 ) {
　　System.out.print(item.toString() + " ");
　　}
　　}
　　遍历myList，并打印出其长度为3的元素；
　　6.System.out.println("4. LIST IS: " + myList );
　　打印整个myList

Answer 2

1. LIST IS: [Uno, Due, java.awt.Point[x=3,y=3], 456, c, Tre, java.awt.Point[x=3,y=3]]

2. Element at position 2 java.awt.Point[x=3,y=3]

3. Uno Due 456 Tre

4. LIST IS: [Uno, Due, 456, c, Tre, java.awt.Point[x=3,y=3]]

Answer 3

1. LIST IS: [Uno, Due, java.awt.Point[x=3,y=3], 456, c, Tre, java.awt.Point[x=3,y=3]]
2. Element at position 2 java.awt.Point[x=3,y=3]
3. Uno Due 456 Tre
4. LIST IS: [Uno, Due, 456, c, Tre, java.awt.Point[x=3,y=3]]

Answer 4

1. LIST IS: [Uno, Due, java.awt.Point[x=3,y=3], 456, c, Tre, java.awt.Point[x=3,y=3]]
2. Element at position 2 java.awt.Point[x=3,y=3]
3. Uno Due 456 Tre
4. LIST IS: [Uno, Due, 456, c, Tre, java.awt.Point[x=3,y=3]]

Answer 5

2. Element at position 2 java.awt.Point[x=3,y=3]

3. Uno Due 456 Tre

Answer 6

1. LIST IS: [Uno, Due, java.awt.Point[x=3,y=3], 456, c, Tre, java.awt.Point[x=3,y=3]]
输出整个列表。
2. Element at position 2 java.awt.Point[x=3,y=3]
返回此列表中索引为2的元素.也就是列表中的第3个元素。而第三个元素就是
java.awt.Point[x=3,y=3]
3. Uno Due 456 Tre
输出列表中字符串长度为3的字符串
4. LIST IS: [Uno, Due, 456, c, Tre, java.awt.Point[x=3,y=3]]
因为 int pos2 = myList.indexOf( new Point(3,3) );
if( pos2 > -1 ) {
myList.remove(pos2);
}
把第三个元素给removed掉了。所以结果就这样了。