public Set<Object> findDuplicates(List<Object> list) {
        Set<Object> items = new HashSet<Object>();
        Set<Object> duplicates = new HashSet<Object>();
        for (Object item : list) {
            if (items.contains(item)) {
                duplicates.add(item);
                } else { 
                    items.add(item);
                    } 
            } 
        return duplicates;
        }

        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]

填充数组列表时，为每个元素使用一个条件。例如:

    ArrayList< Integer > al = new ArrayList< Integer >(); 

    // fill 1 
    for ( int i = 0; i <= 5; i++ ) 
        if ( !al.contains( i ) ) 
            al.add( i ); 

    // fill 2 
    for (int i = 0; i <= 10; i++ ) 
        if ( !al.contains( i ) ) 
            al.add( i ); 

    for( Integer i: al )
    {
        System.out.print( i + " ");     
    }

我们将得到一个数组{0,1,2,3,4,5,6,7,8,9,10}

虽然将ArrayList转换为HashSet可以有效地删除重复项，但如果您需要保留插入顺序，我宁愿建议您使用这个变体

// list is some List of Strings
Set<String> s = new LinkedHashSet<>(list);

然后，如果需要返回List引用，可以再次使用转换构造函数。

像这样的东西会更好吗?

public static void removeDuplicates(ArrayList<String> list) {
    Arraylist<Object> ar     = new Arraylist<Object>();
    Arraylist<Object> tempAR = new Arraylist<Object>();
    while (list.size()>0){
        ar.add(list(0));
        list.removeall(Collections.singleton(list(0)));
    }
    list.addAll(ar);
}

这应该保持顺序，并且在运行时也不是二次的。

    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);