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 blanck space and then print 1st element then “r”, futher 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 program to print reverse of list #include <stdio.h> #include <stdlib.h> /* Link list node */ struct Node { int data; struct Node* next; }; /* Function to reverse the linked list */ void printReverse( struct Node** head_ref, int n) { int j = 0; struct Node* current = *head_ref; while (current != NULL) { // For each node, print proper number // of spaces before printing it for ( int i = 0; i < 2 * (n - j); i++) printf ( " " ); // use of carriage return to move back // and print. printf ( "%d\r" , current->data); current = current->next; j++; } } /* Function to push a node */ void push( struct Node** head_ref, int new_data) { struct Node* new_node = ( struct Node*) malloc ( sizeof ( struct Node)); new_node->data = new_data; new_node->next = (*head_ref); (*head_ref) = new_node; } /* Function to print linked list and find its length */ int printList( struct Node* head) { // i for finding length of list int i = 0; struct Node* temp = head; while (temp != NULL) { printf ( "%d " , temp->data); temp = temp->next; i++; } return i; } /* Driver program to test above function*/ int main() { /* Start with the empty list */ struct Node* head = NULL; // list nodes are as 6 5 4 3 2 1 push(&head, 1); push(&head, 2); push(&head, 3); push(&head, 4); push(&head, 5); push(&head, 6); printf ( "Given linked list:\n" ); // printlist print the list and // return the size of list int n = printList(head); // print reverse list with help // of carriage return function printf ( "\nReversed Linked list:\n" ); printReverse(&head, n); printf ( "\n" ); return 0; } |
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( "\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 ) { 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( "Reversed 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 |
Output:
Given linked list: 6 5 4 3 2 1 Reversed Linked List: 1 2 3 4 5 6
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 compiler because they do not support anything like carriage return on their console.
Reference :
stackoverflow/Carriage return
This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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