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.
#include <iostream> #include <cstring> using namespace std;
// Represents node of a linked list class Node {
public :
int data;
Node* next;
Node( int val)
{
data = val;
next = nullptr;
}
}; void printReverse(Node* head)
{ if (!head)
return ;
printReverse(head->next);
cout << head->data << " " ;
} 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 the length of the 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 the list
int n = printList(head);
// print reverse list
cout << "\nReversed Linked list: " << endl;
printReverse(head);
cout << endl;
return 0;
} |
#include <stdio.h> #include <stdlib.h> // Represents node of a linked list struct Node {
int data;
struct Node* next;
}; void printReverse( struct Node* head) {
if (!head)
return ;
printReverse(head->next);
printf ( "%d " , head->data);
} struct Node* push( struct Node* head, int data) {
struct Node* new_node = ( struct Node*) malloc ( sizeof ( struct Node));
new_node->data = data;
new_node->next = head;
head = new_node;
return head;
} int printList( struct Node* head) {
// i for finding the length of the list
int i = 0;
struct Node* temp = head;
while (temp != NULL) {
printf ( "%d " , temp->data);
temp = temp->next;
i++;
}
return i;
} // Driver code int main() {
/* Start with the empty list */
struct 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);
printf ( "Given linked list:\n" );
// printlist print the list and
// return the size of the list
int n = printList(head);
// print reverse list
printf ( "\nReversed Linked list:\n" );
printReverse(head);
printf ( "\n" );
return 0;
} |
class Node {
int data;
Node next;
Node( int val) {
data = val;
next = null ;
}
} public class ReversePrint {
static void printReverse(Node head) {
if (head == null )
return ;
printReverse(head.next);
System.out.print(head.data + " " );
}
static Node push(Node head, int data) {
Node new_node = new Node(data);
new_node.next = head;
head = new_node;
return head;
}
static int printList(Node head) {
int i = 0 ;
Node temp = head;
while (temp != null ) {
System.out.print(temp.data + " " );
temp = temp.next;
i++;
}
return i;
}
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 the list
int n = printList(head);
// print reverse list
System.out.println( "\nReversed Linked list:" );
printReverse(head);
System.out.println();
}
} |
class Node:
def __init__( self , val):
self .data = val
self . next = None
def print_reverse(head):
if not head:
return
print_reverse(head. next )
print (head.data, end = " " )
def push(head, data):
new_node = Node(data)
new_node. next = head
head = new_node
return head
def print_list(head):
i = 0
temp = head
while temp is not None :
print (temp.data, end = " " )
temp = temp. next
i + = 1
return i
# Driver code if __name__ = = "__main__" :
# 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 the list
n = print_list(head)
# print reverse list
print ( "\nReversed Linked list:" )
print_reverse(head)
print ()
|
using System;
// Represents node of a linked list public class Node {
public int data;
public Node next;
public Node( int val) {
data = val;
next = null ;
}
} public class ReversePrint {
static void PrintReverse(Node head) {
if (head == null )
return ;
PrintReverse(head.next);
Console.Write(head.data + " " );
}
static Node Push(Node head, int data) {
Node new_node = new Node(data);
new_node.next = head;
head = new_node;
return head;
}
static void PrintList(Node head) {
Node temp = head;
while (temp != null ) {
Console.Write(temp.data + " " );
temp = temp.next;
}
}
public static void Main() {
/* 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 prints the list
PrintList(head);
// print reverse list
Console.WriteLine( "\nReversed Linked list:" );
PrintReverse(head);
Console.WriteLine();
}
} |
class Node { constructor(val) {
this .data = val;
this .next = null ;
}
} function printReverse(head) {
if (head === null )
return ;
printReverse(head.next);
console.log(head.data + " " );
} function push(head, data) {
let new_node = new Node(data);
new_node.next = head;
head = new_node;
return head;
} function printList(head) {
let temp = head;
while (temp !== null ) {
console.log(temp.data + " " );
temp = temp.next;
}
} // Driver code 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 prints the list printList(head); // print reverse list console.log( "\nReversed Linked list:" );
printReverse(head); console.log(); |
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