### Introduction

The linked list is one of the most important concepts and data structures to learn while preparing for interviews. Having a good grasp of Linked Lists can be a huge plus point in a coding interview.

### Problem Statement

In this problem, we will be given a linked list and are required to print the reverse of the linked list without using conventional methods.

### Problem Statement Understanding

This is an interesting problem. According to the problem statement, we will have to print the reverse of the given linked list without using any conventional method for reversing the linked list.

If the given linked list: 23 → 12 → 2 → 10

- Then the reverse of the linked list is: 10 → 2 → 12 → 23.
- Our output will be the print of the reverse of the linked list, which will be 10, 2, 12, 23.

If the given linked list is: 1 → 3 → 5 → 7 → 9 → 11 → 13 → 15

- In this case, our output will be 15, 13, 11, 9, 7, 5, 3, 1.

##### Some more examples

Sample Input 1: 1 → 2 → 3 → 4

Sample Output 1: 4, 3, 2, 1

Sample Input 2: 2 → 4 → 6 → 8 → 10

Sample Output 2: 10, 8, 6, 4, 2

**Explanation:** We are printing the reverse of the given linked list.

Well, reversing a linked list is pretty easy, right? But here, we can’t reverse the linked list. So, how should we approach the problem?

- The answer is
**Carriage return**("r"). Whenever we use "r", it commands the printer to move the position of the cursor to the first position on the same line.

Let us have a glance at the approach.

### Approach

The approach is going to be pretty straightforward.

- First, we will find out the length of the linked list.
- Then, we will print
**(n-1)**blank spaces and then print the 1^{st}element then the carriage return "r", further again we will print**(n-2)**blank spaces and the 2^{nd}element and then again carriage return "r" and so on… until we reach the end of the list.

**Note:** What we are doing above is, after printing **(n-1)** blank spaces and then the element, we are commanding the cursor to jump to the start of the line. So that, when the next printing portion happens, the **(n-2)** spaces are given from the start of the line and then the next element is printed and hence in this way the printing will happen in a reverse fashion.

### Algorithm

- Find the length of the given linked list, and store it in
**n**. - Initialize a variable
**j**with 0. - Now initialize a pointer
**curr**with head and start traversing the list using**curr**. - While traversing the list:
- Run a for loop to print the proper number of blank spaces before printing the current node’s data. Actually, print
**(n-j)**spaces, because that is the proper spacing which will give the perfect look of reverse printing. - After printing the blank spaces, print the current’s nodes data and use carriage return (
**printf(“%d\r”, curr→data)**). - Increment the current node by
**curr = curr→next**. - Increment
**j**by 1.

- Run a for loop to print the proper number of blank spaces before printing the current node’s data. Actually, print
- This way, our list will get printed in reverse order.

### Dry Run

### Code Implementation

#include <stdio.h> #include <stdlib.h> /* Link list node */ struct Node { int data; struct Node* next; }; /* Function to reverse the linked list */ static void reverse(struct Node** head_ref) { struct Node* prev = NULL; struct Node* current = *head_ref; struct Node* next = NULL; while (current != NULL) { // Store next next = current->next; // Reverse current node's pointer current->next = prev; // Move pointers one position ahead. prev = current; current = next; } *head_ref = prev; } /* 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 */ void printList(struct Node* head) { struct Node* temp = head; while (temp != NULL) { printf("%d ", temp->data); temp = temp->next; } } /* Driver code*/ int main() { /* Start with the empty list */ struct Node* head = NULL; push(&head, 20); push(&head, 4); push(&head, 15); push(&head, 85); printf("Given linked list\n"); printList(head); reverse(&head); printf("\nReversed Linked list \n"); printList(head); getchar(); }

#include <bits/stdc++.h> using namespace std; typedef struct node { int val; struct node* next; } node; node* head = NULL; // Function to return the No of nodes present in the linked list int count(node* head) { node* p = head; int k = 1; while (p != NULL) { p = p->next; k++; } return k; } node* ll_reverse(node* head) // to reverse the linked list { node* p = head; long int i = count(head), j = 1; long int arr[i]; while (i && p != NULL) { arr[j++] = p->val; p = p->next; i--; } j--; while (j) // loop will break as soon as j=0 cout << arr[j--] << " "; return head; } // Function to insert node at the end of linked list node* insert_end(node* head, int data) { node *q = head, *p = (node*)malloc(sizeof(node)); p->val = data; while (q->next != NULL) q = q->next; q->next = p; p->next = NULL; return head; } node* create_ll(node* head, int data) // create ll { node* p = (node*)malloc(sizeof(node)); p->val = data; if (head == NULL) { head = p; p->next = NULL; return head; } else { head = insert_end(head, data); return head; } } // Driver code int main() { int i = 5, j = 1; while (i--) head = create_ll(head, j++); head = ll_reverse(head); return 0; }

import java.io.*; import java.util.*; /* Structure of a linked list node */ class Node { int data; Node next; Node(int val) { data = val; next = null; } } public class PrepBytes { /* Using this function we will be printing the reverse of the given linked list */ static void printReverse(Node head, int n) { int j = 0; Node curr = head; while (curr != null) { for (int i = 0; i < (n-j); i++) System.out.print(" "); System.out.print(curr.data +"\r" ); curr = curr.next; j++; } } /* Using this function we will be adding a new node to the head of the given linked list */ static Node push(Node head, int data) { Node new_node = new Node(data); new_node.next = head; head = new_node; return head; } /* Using this function we will be printing the elements of the given linked list */ 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[]) { Node head = null; head = push(head, 10); head = push(head, 2); head = push(head, 12); head = push(head, 23); System.out.println("Printing original given linked list: "); int n = printList(head); System.out.println(); System.out.println("Printing reversed linked list: "); printReverse(head, n); System.out.println(); } }

# Structure of a linked list node class Node: def __init__(self): self.data= 0 self.next=None # Using this function we will be printing the reverse of the given linked list def printReverse( head_ref, n): j = 0 current = head_ref while (current != None): i = 0 while ( i < (n - j) ): print(end=" ") i = i + 1 print( current.data, end = "\r") current = current.next j = j + 1 # Using this function we will be adding a new node to the head of the given linked list 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 # Using this function we will be printing the elements of the given linked list def printList( head): i = 0 temp = head while (temp != None): print( temp.data,end = " ") temp = temp.next i = i + 1 return i head = None head = push(head, 10) head = push(head, 2) head = push(head, 12) head = push(head, 23) print("Printing original given linked list:") n = printList(head) print("\nPrinting reversed linked list: ") printReverse(head, n) print()

#### Output

Printing original given linked list:

23 12 2 10

Printing reversed Linked list:

10 2 12 23

**Time Complexity:** O(n), as list traversal is needed

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So, in this article, we have tried to explain an interesting method to print the reverse of a linked list. The unique constraints of this problem are what makes this problem an important one for coding interviews. If you want to solve more questions on Linked List, which are curated by our expert mentors at PrepBytes, you can follow this link Linked List.