LinkedHashSet可以做到这一点。

String[] arr2 = {"5","1","2","3","3","4","1","2"};
Set<String> set = new LinkedHashSet<String>(Arrays.asList(arr2));
for(String s1 : set)
    System.out.println(s1);

System.out.println( "------------------------" );
String[] arr3 = set.toArray(new String[0]);
for(int i = 0; i < arr3.length; i++)
     System.out.println(arr3[i].toString());

/ /输出:5、1、2、3、4

你可以在下面使用嵌套循环:

ArrayList<Class1> l1 = new ArrayList<Class1>();
ArrayList<Class1> l2 = new ArrayList<Class1>();

        Iterator iterator1 = l1.iterator();
        boolean repeated = false;

        while (iterator1.hasNext())
        {
            Class1 c1 = (Class1) iterator1.next();
            for (Class1 _c: l2) {
                if(_c.getId() == c1.getId())
                    repeated = true;
            }
            if(!repeated)
                l2.add(c1);
        }

Set<String> strSet = strList.stream().collect(Collectors.toSet());

是删除副本的最简单方法。

如前所述，应该使用实现Set接口的类而不是List来确保元素的唯一性。如果必须保持元素的顺序，则可以使用SortedSet接口;TreeSet类实现了该接口。

    ArrayList<String> list = new ArrayList<String>();
    HashSet<String> unique = new LinkedHashSet<String>();
    HashSet<String> dup = new LinkedHashSet<String>();
    boolean b = false;
    list.add("Hello");
    list.add("Hello");
    list.add("how");
    list.add("are");
    list.add("u");
    list.add("u");

    for(Iterator iterator= list.iterator();iterator.hasNext();)
    {
        String value = (String)iterator.next();
        System.out.println(value);

        if(b==unique.add(value))
            dup.add(value);
        else
            unique.add(value);


    }
    System.out.println(unique);
    System.out.println(dup);

        List<String> result = new ArrayList<String>();
        Set<String> set = new LinkedHashSet<String>();
        String s = "ravi is a good!boy. But ravi is very nasty fellow.";
        StringTokenizer st = new StringTokenizer(s, " ,. ,!");
        while (st.hasMoreTokens()) {
            result.add(st.nextToken());
        }
         System.out.println(result);
         set.addAll(result);
        result.clear();
        result.addAll(set);
        System.out.println(result);

output:
[ravi, is, a, good, boy, But, ravi, is, very, nasty, fellow]
[ravi, is, a, good, boy, But, very, nasty, fellow]