List、Set、Map的foreach与迭代

List

有序且不唯一

public static void main(String[] args) {
        List<Person> list=new ArrayList<Person>();
        Person p1=new Person("LL", 12, "Beijing");
        Person p2=new Person("KK", 22, "Shanghai");
        Person p3=new Person("JJ", 32, "Wuhan");
        list.add(p1);
        list.add(p2);
        list.add(p3);

        for (Person p : list) {
            System.out.println(p.toString());
        }

        Iterator<Person> ir=list.iterator();
        while(ir.hasNext()){
            Person p=ir.next();
            System.out.println(p.toString());
        }

    }

运行结果
Person [Name=LL, Age=12, Address=Beijing]
Person [Name=KK, Age=22, Address=Shanghai]
Person [Name=JJ, Age=32, Address=Wuhan]
Person [Name=LL, Age=12, Address=Beijing]
Person [Name=KK, Age=22, Address=Shanghai]
Person [Name=JJ, Age=32, Address=Wuhan]

Set

无序且唯一

public static void main(String[] args) {
        Set<Person> set=new HashSet<Person>();
        Person p1=new Person("LL", 12, "Beijing");
        Person p2=new Person("KK", 22, "Shanghai");
        Person p3=new Person("JJ", 32, "Wuhan");
        set.add(p1);
        set.add(p2);
        set.add(p3);

        for (Person p : set) {
            System.out.println(p.toString());
        }

        Iterator<Person> ir=set.iterator();
        while(ir.hasNext()){
            Person p=ir.next();
            System.out.println(p.toString());
        }

    }


Person [Name=JJ, Age=32, Address=Wuhan]
Person [Name=LL, Age=12, Address=Beijing]
Person [Name=KK, Age=22, Address=Shanghai]
Person [Name=JJ, Age=32, Address=Wuhan]
Person [Name=LL, Age=12, Address=Beijing]
Person [Name=KK, Age=22, Address=Shanghai]