Remove repeated elements from ArrayList in Java

Prerequisite: ArrayList in Java

Given an ArrayList, the task is to remove repeated elements of the ArrayList in Java.

Examples:



Input: ArrayList = [1, 2, 2, 3, 4, 4, 4] 
Output: [1, 2, 3, 4] 

Input: ArrayList = [12, 23, 23, 34, 45, 45, 45, 45, 57, 67, 89] 
Output: [12, 23, 34, 45, 57, 67, 89]

Below are the various methods to remove repeated elements an ArrayList in Java:

  • Using a Set: Since Set is a collection which do not includes any duplicate elements. Hence the solution can be achieved with the help of a Set.

    Approach:

    1. Get the ArrayList with repeated elements.
    2. Convert the ArrayList to Set.
    3. Now convert the Set back to ArrayList. This will remove all the repeated elements.

    Below is the implementation of the above approach:

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    // Java code to illustrate remove duolicate
    // of ArrayList using hashSet<> method
      
    import java.util.*;
      
    public class GFG {
        public static void main(String args[])
        {
      
            // create a ArrayList String type
            ArrayList<String>
                gfg = new ArrayList<String>();
      
            // Initialize an ArrayList
            gfg.add("Geeks");
            gfg.add("for");
            gfg.add("Geeks");
      
            // print ArrayList
            System.out.println("Original ArrayList : "
                               + gfg);
      
            // -----Using LinkedHashSet-----
            System.out.println("\nUsing LinkedHashSet:\n");
      
            // create a set and copy all value of list
            Set<String> set = new LinkedHashSet<>(gfg);
      
            // create a list and copy all value of set
            List<String> gfg1 = new ArrayList<>(set);
      
            // print ArrayList
            System.out.println("Modified ArrayList : "
                               + gfg1);
      
            // -----Using HashSet-----
            System.out.println("\nUsing HashSet:\n");
      
            // create a set and copy all value of list
            Set<String> set1 = new HashSet<>(gfg);
      
            // create a list and copy all value of set
            List<String> gfg2 = new ArrayList<>(set);
      
            // print ArrayList
            System.out.println("Modified ArrayList : "
                               + gfg2);
        }
    }

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    Output:

    Original ArrayList : [Geeks, for, Geeks]
    
    Using LinkedHashSet:
    
    Modified ArrayList : [Geeks, for]
    
    Using HashSet:
    
    Modified ArrayList : [Geeks, for]
    
  • Using Java 8 Lambdas:

    Approach:

    1. Get the ArrayList with repeated elements.
    2. Convert the ArrayList to Stream using stream() method.
    3. Set the filter condition to be distinct using distinct() method.
    4. Collect the filtered values as List using collect() method. This list will be with repeated elements removed

    Below is the implementation of the above approach:

    Example:

    filter_none

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    // Java code to illustrate remove duolicate
    // of ArrayList using hashSet<> method
      
    import java.util.*;
    import java.util.stream.Collectors;
      
    public class GFG {
        public static void main(String args[])
        {
            // create a ArrayList String type
            ArrayList<String>
                gfg = new ArrayList<String>();
      
            // Initialize an ArrayList
            gfg.add("Geeks");
            gfg.add("for");
            gfg.add("Geeks");
      
            // print ArrayList
            System.out.println("Original ArrayList : "
                               + gfg);
      
            // create a list and copy all distinct value of list
            List<String> gfg1 = gfg.stream()
                                    .distinct()
                                    .collect(Collectors.toList());
      
            // print modified list
            System.out.println("Modified List : " + gfg1);
        }
    }

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    Output:

    Original ArrayList : [Geeks, for, Geeks]
    Modified List : [Geeks, for]
    


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