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Java AbstractSequentialList | ListIterator()

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AbstractSequentialList listIterator(): method in Java is used to get a listIterator over this list. It returns a list iterator over the elements in this list (in proper sequence).

Syntax:

public abstract ListIterator listIterator(int index)

Parameters: This method takes a parameter index which is the index of first element to be returned from the list iterator (by a call to the next method)

Returns: This method returns a list iterator over the elements in this list (in proper sequence).

Exceptions: This method throws IndexOutOfBoundsException, if the index is out of range (index size())

Below is the code to illustrate ListIterator():

Program 1:




// Java program to demonstrate
// add() method
  
import java.util.*;
  
public class GfG {
  
    public static void main(String[] args)
    {
        // Creating an instance of the AbstractSequentialList
        AbstractSequentialList<Integer> absl = new LinkedList<>();
  
        // adding elements to absl
        absl.add(5);
        absl.add(6);
        absl.add(7);
        absl.add(2, 8);
        absl.add(2, 7);
        absl.add(1, 9);
        absl.add(4, 10);
  
        // Getting ListIterator
        ListIterator<Integer> Itr = absl.listIterator(2);
  
        // Traversing elements
        while (Itr.hasNext()) {
            System.out.print(Itr.next() + " ");
        }
    }
}


Output:

6 7 10 8 7

Program 2: To demonstrate IndexOutOfBoundException




// Java code to show IndexOutofBoundException
  
import java.util.*;
  
public class GfG {
  
    public static void main(String[] args)
    {
  
        // Creating an instance of the AbstractSequentialList
        AbstractSequentialList<Integer> absl = new LinkedList<>();
  
        // adding elements to absl
        absl.add(5);
        absl.add(6);
        absl.add(7);
        absl.add(2, 8);
        absl.add(2, 7);
        absl.add(1, 9);
        absl.add(4, 10);
  
        // Printing the list
        System.out.println(absl);
  
        try {
            // showing IndexOutOfBoundsException
            // Getting ListIterator
            ListIterator<Integer> Itr = absl.listIterator(15);
  
            // Traversing elements
            while (Itr.hasNext()) {
                System.out.print(Itr.next() + " ");
            }
        }
        catch (Exception e) {
            System.out.println("Exception: " + e);
        }
    }
}


Output:

[5, 9, 6, 7, 10, 8, 7]
Exception: java.lang.IndexOutOfBoundsException: Index: 15, Size: 7


Last Updated : 26 Nov, 2018
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