Linked List is a linear data structure, in which the elements are not stored at the contiguous memory locations. Here, the task is to get the elements of a LinkedList.
1. We can use get(int variable) method to access an element from a specific index of LinkedList:
In the given example, we have used the get(i) method. Here, the method returns the element which is at the i th index.
Syntax:
LinkedList.get(int index)
Parameters: The parameter index is of integer data type that specifies the position or index of the element to be fetched from the LinkedList.
Return Value: The method returns the element present at the position specified by the parameter index.
Java
// Java program to get the elements of Linkedlist import java.io.*; import java.util.LinkedList; class GFG { public static void main(String[] args) { // Creating LinkedList LinkedList<String> gfg = new LinkedList<String>(); // Adding values gfg.add( "GEEKS" ); gfg.add( "FOR" ); gfg.add( "GEEKS" ); System.out.println( "LinkedList Elements : " ); for ( int i = 0 ; i < gfg.size(); i++) { // get(i) returns element present at index i System.out.println( "Element at index " + i + " is: " + gfg.get(i)); } } } |
LinkedList Elements : Element at index 0 is: GEEKS Element at index 1 is: FOR Element at index 2 is: GEEKS
2. We can use the iterator() method
- To use this method we have to import java.util.Iterator package.
- In this method, we can iterate over the LinkedList and then extract the element at the given index accordingly.
Java
// Java program to iterate over linkedlist // to extract elements of linkedlist import java.io.*; import java.util.LinkedList; import java.util.Iterator; class GFG { public static void main(String[] args) { LinkedList<String> gfg = new LinkedList<String>(); // Adding elements gfg.add( "GEEKS" ); gfg.add( "FOR" ); gfg.add( "GEEKS" ); // Create an object of Iterator Iterator<String> i = gfg.iterator(); System.out.print( "The elements of the input LinkedList: \n" ); int j = 0 ; // has.next() returns true if there is a next // element while (i.hasNext()) { System.out.print( "The element at the index " + j + " " ); // next() returns the next element String str = i.next(); System.out.print(str); System.out.print( " \n" ); ++j; } } } |
The elements of the input LinkedList: The element at the index 0 GEEKS The element at the index 1 FOR The element at the index 2 GEEKS
3. We can use ListIterator() method.
- ListIterator() is a subinterface of Iterator() method.
- It provides us with the function to access the elements of a list.
- It is bidirectional that means it allows us to iterate elements of a list in the both the direction.
- To use this method we have to import java.util.ListIterator.
Java
// Java program to iterate over the // linkedlist using listIterator() import java.io.*; import java.util.LinkedList; import java.util.ListIterator; class GFG { public static void main(String[] args) { LinkedList<String> gfg = new LinkedList<String>(); // Adding elements gfg.add( "GEEKS" ); gfg.add( "FOR" ); gfg.add( "GEEKS" ); // Create an object of ListIterator ListIterator<String> li = gfg.listIterator(); System.out.print( "The elements of the LinkedList: \n" ); // hasNext() returns true if there is next element int j = 0 ; while (li.hasNext()) { // next() returns the next element System.out.print( "The element at the index " + j + " " ); System.out.print(li.next()); System.out.print( "\n" ); ++j; } --j; // Now to show that ListIterator() can traverse in // both the direction System.out.print( "\nThe elements of the LinkedList in Reverse order: \n" ); // hasprevious() checks if there is a previous // element while (li.hasPrevious()) { System.out.print( "The element at the index " + j + " " ); // previous() returns the previous element System.out.print(li.previous()); System.out.print( "\n" ); --j; } } } |
The elements of the LinkedList: The element at the index 0 GEEKS The element at the index 1 FOR The element at the index 2 GEEKS The elements of the LinkedList in Reverse order: The element at the index 2 GEEKS The element at the index 1 FOR The element at the index 0 GEEKS