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Java Program For Moving Last Element To Front Of A Given Linked List

  • Last Updated : 14 Dec, 2021

Write a function that moves the last element to the front in a given Singly Linked List. For example, if the given Linked List is 1->2->3->4->5, then the function should change the list to 5->1->2->3->4.
Algorithm:
Traverse the list till the last node. Use two pointers: one to store the address of the last node and the other for the address of the second last node. After the end of the loop do the following operations.

  1. Make second last as last (secLast->next = NULL).
  2. Set next of last as head (last->next = *head_ref).
  3. Make last as head ( *head_ref = last).

Java




/* Java Program to move last element to 
   front in a given linked list */
class LinkedList
{
    // Head of list
    Node head;  
   
    // Linked list Node
    class Node
    {
        int data;
        Node next;
        Node(int d) 
        {
            data = d; 
            next = null
        }
    }
  
    void moveToFront()
    {
        /* If linked list is empty or 
           it contains only one node 
           then simply return. */
           if(head == null || 
              head.next == null
              return;
  
        /* Initialize second last and 
           last pointers */
        Node secLast = null;
        Node last = head;
  
        /* After this loop secLast contains 
           address of second last  node and 
           last contains address of last node 
           in Linked List */
        while (last.next != null)  
        {
           secLast = last;
           last = last.next; 
        }
  
        // Set the next of second last as null 
        secLast.next = null;
  
        // Set the next of last as head 
        last.next = head;
  
        // Change head to point to 
        // last node. 
        head = last;
    }              
  
     // Utility functions 
    /* Inserts a new Node at front 
       of the list. */
    public void push(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                  Put in the data*/
        Node new_node = new Node(new_data);
   
        // 3. Make next of new Node as head 
        new_node.next = head;
   
        // 4. Move the head to point to 
        // new Node 
        head = new_node;
    }
  
    // Function to print linked list 
    void printList()
    {
        Node temp = head;
        while(temp != null)
        {
           System.out.print(temp.data + " ");
           temp = temp.next;
        }  
        System.out.println();
    }
  
    // Driver code
    public static void main(String args[])
    {
        LinkedList llist = new LinkedList();
  
        /* Constructed Linked List is 
           1->2->3->4->5->null */
        llist.push(5);
        llist.push(4);
        llist.push(3);
        llist.push(2);
        llist.push(1);
          
        System.out.println(
               "Linked List before moving last to front ");
        llist.printList();
          
        llist.moveToFront();
          
        System.out.println(
               "Linked List after moving last to front ");
        llist.printList();
    }
// This code is contributed by Rajat Mishra 


Output:

Linked list before moving last to front 
1 2 3 4 5 
Linked list after removing last to front 
5 1 2 3 4

Time Complexity: O(n) where n is the number of nodes in the given Linked List.

Please refer complete article on Move last element to front of a given Linked List for more details!


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