An interesting method to print reverse of a linked list

We are given a linked list, we need to print the linked list in reverse order.
Examples: 

Input : list : 5-> 15-> 20-> 25 
Output : Reversed Linked list : 25-> 20-> 15-> 5

Input : list : 85-> 15-> 4-> 20 
Output : Reversed Linked list : 20-> 4-> 15-> 85

Input : list : 85
Output : Reversed Linked list : 85

For printing a list in reverse order, we have already discussed Iterative and Recursive Methods to Reverse.
In this post, an interesting method is discussed, that doesn’t require recursion and does no modifications to list. The function also visits every node of linked list only once.

Trick : Idea behind printing a list in reverse order without any recursive function or loop is to use Carriage return (“r”). For this, we should have knowledge of length of list. Now, we should print n-1 blank space and then print 1st element then “r”, further again n-2 blank space and 2nd node then “r” and so on.. 
Carriage return (“r”) : It commands a printer (cursor or the display of a system console), to move the position of the cursor to the first position on the same line.

C/C++

C++

#include <iostream>
#include <cstring>

using namespace std;

// Represents node of a linkedlist
class Node {
  public:
      int data;
      Node *next;
      Node(int val)
      {
          data = val;
          next = nullptr;
      }
};

void printReverse(Node *head, int n)
{
        int j = 0;
        Node *current = head;
        while (current != nullptr) {

           // For each node, print proper number
           // of spaces before printing it
           for (int i = 0; i < 2 * (n - j); i++)
                cout << " ";

           // use of carriage return to move back
           // and print.
           cout << "\r" << current->data;

           current = current->next;
           j++;
        }
}

Node *push(Node *head, int data)
{
        Node *new_node = new Node(data);
        new_node->next = head;
        head = new_node;

        return head;
}

int printList(Node *head)
{
        // i for finding length of list
        int i = 0;
        Node *temp = head;
        while (temp != nullptr)
        {
               cout << temp->data << " ";
               temp = temp->next;
               i++;
        }

        return i;
}

// Driver code
int main()
{
      /* Start with the empty list */
      Node *head = nullptr;

      // list nodes are as 6 5 4 3 2 1
      head = push(head, 1);
      head = push(head, 2);
      head = push(head, 3);
      head = push(head, 4);
      head = push(head, 5);
      head = push(head, 6);

      cout << "Given linked list: " << endl;
      // printlist print the list and
      // return the size of list
      int n = printList(head);

      // print reverse list with help
      // of carriage return function
      cout << "Reversed Linked list: " << endl;
      printReverse(head, n);
      cout << endl;

      return 0;
}
//this code is contributed by snehalsalokhe

Java

// Java program to print reverse of list
import java.io.*;
import java.util.*;

// Represents node of a linkedlist
class Node {
    int data;
    Node next;
    Node(int val)
    {
        data = val;
        next = null;
    }
}

public class GFG {

    /* Function to reverse the linked list */
    static void printReverse(Node head, int n)
    {
        int j = 0;
        Node current = head;
        while (current != null) {

            // For each node, print proper number
            // of spaces before printing it
            for (int i = 0; i < 2 * (n - j); i++)
                System.out.print(" ");

            // use of carriage return to move back
            // and print.
            System.out.print(" " + current.data);

            current = current.next;
            j++;
        }
    }

    /* Function to push a node */
    static Node push(Node head, int data)
    {
        Node new_node = new Node(data);
        new_node.next = head;
        head = new_node;

        return head;
    }

    /* Function to print linked list and find its
    length */
    static int printList(Node head)
    {
        // i for finding length of list
        int i = 0;
        Node temp = head;
        while (temp != null) {
            System.out.print(temp.data + " ");
            temp = temp.next;
            i++;
        }

        return i;
    }

    // Driver code
    public static void main(String args[])
    {
        /* Start with the empty list */
        Node head = null;

        // list nodes are as 6 5 4 3 2 1
        head = push(head, 1);
        head = push(head, 2);
        head = push(head, 3);
        head = push(head, 4);
        head = push(head, 5);
        head = push(head, 6);

        System.out.println("Given linked list: ");

        // printlist print the list and
        // return the size of list
        int n = printList(head);

        // print reverse list with help
        // of carriage return function
        System.out.println("\nReversed Linked list: ");
        printReverse(head, n);
        System.out.println();
    }
}

// This code is contributed by rachana soma

C#

// C# program to print reverse of list
using System;

// Represents node of a linkedlist
public class Node {
      public int data;
      public Node next;
      public Node(int val)
      {
          data = val;
          next = null;
      }
}

public class GFG
{

   /* Function to reverse the linked list */
   static void printReverse(Node head, int n)
   {
          int j = 0;
          Node current = head;
          while (current != null) {

             // For each node, print proper number
             // of spaces before printing it
             for (int i = 0; i < 2 * (n - j); i++)
                  Console.Write(" ");

             // use of carriage return to move back
             // and print.
             Console.Write("\r" + current.data);

             current = current.next;
             j++;
          }
   }

   /* Function to push a node */
   static Node push(Node head, int data)
   {
          Node new_node = new Node(data);
          new_node.next = head;
          head = new_node;

          return head;
   }

   /* Function to print linked list and find its
   length */
   static int printList(Node head)
   {
          // i for finding length of list
          int i = 0;
          Node temp = head;
          while (temp != null)
          {
                 Console.Write(temp.data + " ");
                 temp = temp.next;
                 i++;
          }

          return i;
   }

   // Driver code
   public static void Main(String []args)
   {
      /* Start with the empty list */
      Node head = null;

      // list nodes are as 6 5 4 3 2 1
      head = push(head, 1);
      head = push(head, 2);
      head = push(head, 3);
      head = push(head, 4);
      head = push(head, 5);
      head = push(head, 6);

      Console.WriteLine("Given linked list: ");
      // printlist print the list and
      // return the size of list
      int n = printList(head);

      // print reverse list with help
      // of carriage return function
      Console.WriteLine("Reversed Linked list: ");
      printReverse(head, n);
      Console.WriteLine();
   }
}

// This code is contributed by Arnab Kundu

Python3

# Python3 program to print reverse of list

# Link list node
class Node:
    def __init__(self):
        self.data=  0
        self.next=None

# Function to reverse the linked list
def printReverse( head_ref, n):

    j = 0
    current = head_ref
    while (current != None):
        i = 0

        # For each node, print proper number
        # of spaces before printing it
        while ( i < 2 * (n - j) ):
            print(end=" ")
            i = i + 1

        # use of carriage return to move back
        # and print.
        print( current.data, end = "\r")

        current = current.next
        j = j + 1

 # Function to push a node
def push( head_ref, new_data):

    new_node = Node()

    new_node.data = new_data
    new_node.next = (head_ref)
    (head_ref) = new_node
    return head_ref;

# Function to print linked list and find its
#  length
def printList( head):

    # i for finding length of list
    i = 0
    temp = head
    while (temp != None):
        print( temp.data,end = " ")
        temp = temp.next
        i = i + 1

    return i

# Driver program to test above function

# Start with the empty list
head = None

# list nodes are as 6 5 4 3 2 1
head = push(head, 1)
head = push(head, 2)
head = push(head, 3)
head = push(head, 4)
head = push(head, 5)
head = push(head, 6)

print("Given linked list:")

# printlist print the list and
# return the size of list
n = printList(head)

# print reverse list with help
# of carriage return function
print("\nReversed Linked list:")
printReverse(head, n)
print()

# This code is contributed by Arnab Kundu

Javascript

<script>

// Represents node of a linkedlist
class Node {
  constructor(val) {
    this.data = val;
    this.next = null;
  }
}

function printReverse(head, n) {
  let j = 0;
  let current = head;
  while (current !== null) {
    // For each node, print proper number
    // of spaces before printing it
    for (let i = 0; i < 2 * (n - j); i++) {
      process.stdout.write(" ");
    }
    // use of carriage return to move back
    // and print.
    process.stdout.write("\r" + current.data);
    current = current.next;
    j++;
  }
}

function push(head, data) {
  const new_node = new Node(data);
  new_node.next = head;
  head = new_node;
  return head;
}

function printList(head) {
  // i for finding length of list
  let i = 0;
  let temp = head;
  while (temp !== null) {
    process.stdout.write(temp.data + " ");
    temp = temp.next;
    i++;
  }
  return i;
}

// Driver code
(function main() {
  /* Start with the empty list */
  let head = null;

  // list nodes are as 6 5 4 3 2 1
  head = push(head, 1);
  head = push(head, 2);
  head = push(head, 3);
  head = push(head, 4);
  head = push(head, 5);
  head = push(head, 6);

  console.log("Given linked list: ");

  // printlist print the list and
  // return the size of list
  let n = printList(head);

  // print reverse list with help
  // of carriage return function
  console.log("\nReversed Linked list: ");
  printReverse(head, n);
  console.log();
})();


// this code is contributed by bhardwajji

</script>

Output:

Given linked list:
6 5 4 3 2 1
Reversed Linked List:
1 2 3 4 5 6

Time Complexity: O(N).
Auxiliary Space: O(1),

Input and Output Illustration :
Input: 6 5 4 3 2 1
1st Iteration _ _ _ _ _ 6
2nd Iteration _ _ _ _ 5 6
3rd Iteration _ _ _ 4 5 6
4th Iteration _ _ 3 4 5 6
5th Iteration _ 2 3 4 5 6
Final Output 1 2 3 4 5 6
NOTE: Above program may not work on online compilers because they do not support anything like carriage return on their console.
Reference : 
StackOverflow/Carriage return
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