Removing Duplicate Values from ArrayList

You can create a LinkedHashSet from the list. The LinkedHashSet will contain each element only once, and in the same order as the List. Then create a new List from this LinkedHashSet. So effectively, it's a one-liner:

list = new ArrayList<String>(new LinkedHashSet<String>(list))

Any approach that involves List#contains or List#remove will probably decrease the asymptotic running time from O(n) (as in the above example) to O(n^2).

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EDIT For the requirement mentioned in the comment: If you want to remove duplicate elements, but consider the Strings as equal ignoring the case, then you could do something like this:

Set<String> toRetain = new TreeSet<String>(String.CASE_INSENSITIVE_ORDER);
toRetain.addAll(list);
Set<String> set = new LinkedHashSet<String>(list);
set.retainAll(new LinkedHashSet<String>(toRetain));
list = new ArrayList<String>(set);

It will have a running time of O(n*logn), which is still better than many other options. Note that this looks a little bit more complicated than it might have to be: I assumed that the order of the elements in the list may not be changed. If the order of the elements in the list does not matter, you can simply do

Set<String> set = new TreeSet<String>(String.CASE_INSENSITIVE_ORDER);
set.addAll(list);
list = new ArrayList<String>(set);

Java 8 stream function

You could use the distinct function like above to get the distinct elements of the list,

stringList.stream().distinct();

From the documentation,

Returns a stream consisting of the distinct elements (according to Object.equals(Object)) of this stream.

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Another way, if you do not wish to use the equals method is by using the collect function like this,

stringList.stream()  
    .collect(Collectors.toCollection(() -> 
        new TreeSet<String>((p1, p2) -> p1.compareTo(p2)) 
));  

From the documentation,

Performs a mutable reduction operation on the elements of this stream using a Collector.

Hope that helps.

You can make use of Google Guava utilities, as shown below

 list = ImmutableSet.copyOf(list).asList(); 

This is probably the most efficient way of eliminating the duplicates from the list and interestingly, it preserves the iteration order as well.

UPDATE

But, in case, you don't want to involve Guava then duplicates can be removed as shown below.

ArrayList<String> list = new ArrayList<String>();
    list.add("Krishna");
    list.add("Krishna");
    list.add("Kishan");
    list.add("Krishn");
    list.add("Aryan");
    list.add("Harm");

System.out.println("List"+list);
HashSet hs = new HashSet();
hs.addAll(list);
list.clear();
list.addAll(hs);

But, of course, this will destroys the iteration order of the elements in the ArrayList.

Shishir

if you want to use only arraylist then I am worried there is no better way which will create a huge performance benefit. But by only using arraylist i would check before adding into the list like following

void addToList(String s){
  if(!yourList.contains(s))
       yourList.add(s);
}

In this cases using a Set is suitable.