集合框架（二)熟悉Collection接口的通用方法

1、AbstractCollection类是提供Collection接口部分实现的便利类，除了size方法和iterator方法之外，它实现了Collection接口中的所有方法。
2、所有这些方法都返回boolean值，如果执行方法会改变这个集合，则返回true。
3、Iterator接口提供了对不同集合中的元素进行遍历的统一方法。

设计注意：Collection接口中的有些方法是不能在具体子类中实现的。在这种情况下，这些方法会抛出异常java.lang.UnsupportedOperationException，它是RuntimeException异常类的一个子类。如果一个方法在子类中没有意义，可以按如下方式实现它：
public void someMethod(){
throw new UnsupportedOperationException
(“Method not supported”);
}

探究Collection接口中的方法的例子
TestMethodInCollection.java

public class TestMethodsInCollection{
public static void main(String[] args){
//Create set1
java.util.Set set1 = new java.util.HashSet();

    //Add strings to set1
    set1.add("London");
    set1.add("Paris");
    set1.add("New York");
    set1.add("San Francisco");
    set1.add("Beijing");

    System.out.println("set1 is " + set1);
    System.out.println(set1.size() + " elements in set1");

    //Delete a string from set1
    set1.remove("London");
    System.out.println("\nset1 is " + set1);
    System.out.println(set1.size() + " elements in set1");

    //Create set2
    java.util.Set set2 = new java.util.HashSet();

    //Add strings to set2
    set2.add("London");
    set2.add("Shanghai");
    set2.add("Paris");
    System.out.println("\nset2 is " + set2);
    System.out.println("set2.size() + " elements in set2");

    System.out.println("\nIs Taipei in set2? " + set2.contains("Taipei"));

    set1.addAll(set2);
    System.out.println("\nAfter adding set2 to set1, set1 is " + set1);

    set1.removeAll(set2);
    System.out.println("After removing set2 from set1, set1 is " + set1);

    set1.retainAll(set2);
    System.out.println("After removing common elements in set2  from set1, set1 is " + set1);
    //
}

}

结果：
set1 is [San Francisco, New York, Paris, Beijing, London]
5 elements in set1
set1 is [San Francisco, New York, Paris, Beijing]
4 elements in set1
set2 is [Shanghai, Paris, London]
3 elements in set2
Is Taipei in set2? false
After adding set2 to set1, set1 is
[San Francisco, New York, Shanghai, Paris, Beijing, London]
After removing set2 from set1, set1 is
[San Francisco, New York, Beijing]
After removing common elements in set2 from set1, set1 is []