Five Methods to Remove Duplicates from a Java ArrayList

The article demonstrates five different ways to delete duplicate entries from a Java ArrayList, providing both explanation and runnable code for each method.

1. Using LinkedHashSet – A LinkedHashSet removes duplicates while preserving insertion order. The example creates a list of integers, converts it to a LinkedHashSet, then back to a list.

import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedHashSet;

public class ArrayListExample {
    public static void main(String[] args) {
        ArrayList<Integer> numbersList = new ArrayList<>(Arrays.asList(1, 1, 2, 3, 3, 3, 4, 5, 6, 6, 6, 7, 8));
        System.out.println(numbersList);
        LinkedHashSet<Integer> hashSet = new LinkedHashSet<>(numbersList);
        ArrayList<Integer> listWithoutDuplicates = new ArrayList<>(hashSet);
        System.out.println(listWithoutDuplicates);
    }
}

Output:

[1, 1, 2, 3, 3, 3, 4, 5, 6, 6, 6, 7, 8]
[1, 2, 3, 4, 5, 6, 7, 8]

2. Using Java 8 Stream distinct() – The Stream API’s distinct() method returns a stream of unique elements, which can be collected back into a list.

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.stream.Collectors;

public class ArrayListExample {
    public static void main(String[] args) {
        ArrayList<Integer> numbersList = new ArrayList<>(Arrays.asList(1, 1, 2, 3, 3, 3, 4, 5, 6, 6, 6, 7, 8));
        System.out.println(numbersList);
        List<Integer> listWithoutDuplicates = numbersList.stream().distinct().collect(Collectors.toList());
        System.out.println(listWithoutDuplicates);
    }
}

Output:

[1, 1, 2, 3, 3, 3, 4, 5, 6, 6, 6, 7, 8]
[1, 2, 3, 4, 5, 6, 7, 8]

3. Using HashSet with order check – Because a plain HashSet does not guarantee order, this method adds elements to a HashSet only if they are not already present, building a new list that keeps the original order.

private static void removeDuplicate(List<String> list) {
    HashSet<String> set = new HashSet<String>(list.size());
    List<String> result = new ArrayList<String>(list.size());
    for (String str : list) {
        if (set.add(str)) {
            result.add(str);
        }
    }
    list.clear();
    list.addAll(result);
}

4. Using List.contains() – Iterate through the original list and add each element to a new list only if the new list does not already contain it.

private static void removeDuplicate(List<String> list) {
    List<String> result = new ArrayList<String>(list.size());
    for (String str : list) {
        if (!result.contains(str)) {
            result.add(str);
        }
    }
    list.clear();
    list.addAll(result);
}

5. Double for‑loop removal – A nested loop compares each pair of elements and removes the duplicate when found.

for (int i = 0; i < list.size(); i++) {
    for (int j = 0; j < list.size(); j++) {
        if (i != j && list.get(i) == list.get(j)) {
            list.remove(list.get(j));
        }
    }
}

Each technique offers a trade‑off between simplicity, performance, and order preservation, allowing developers to choose the most suitable approach for their specific use case.