# Print reverse of a Linked List without extra space and modifications

• Difficulty Level : Easy

Given a Linked List, display the linked list in reverse without using recursion, stack or modifications to given list.

Examples:

Input: 1->2->3->4->5->NULL
Output: 5 4 3 2 1

Input: 10->5->15->20->24->NULL
Output: 24 20 15 5 10

Below are different solutions that are now allowed here as we cannot use extra space and modify list.
1) Recursive solution to print reverse a linked list. Requires extra space.
2) Reverse linked list and then print. This requires modifications to original list.
3) Stack based solution to print linked list reverse. Push all nodes one by one to a stack. Then one by one pop elements from stack and print. This also requires extra space.
Algorithms:

```1) Find n = count nodes in linked list.
2) For i = n to 1, do following.
Print i-th node using get n-th node function```

## C++

 `// C/C++ program to print reverse of linked list``// without extra space and without modifications.``#include``#include` `/* Link list node */``struct` `Node``{``    ``int` `data;``    ``struct` `Node* next;``};` `/* Given a reference (pointer to pointer) to the head``  ``of a list and an int, push a new node on the front``  ``of the list. */``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;``}` `/* Counts no. of nodes in linked list */``int` `getCount(``struct` `Node* head)``{``    ``int` `count = 0;  ``// Initialize count``    ``struct` `Node* current = head;  ``// Initialize current``    ``while` `(current != NULL)``    ``{``        ``count++;``        ``current = current->next;``    ``}``    ``return` `count;``}` `/* Takes head pointer of the linked list and index``    ``as arguments and return data at index*/``int` `getNth(``struct` `Node* head, ``int` `n)``{``    ``struct` `Node* curr = head;``    ``for` `(``int` `i=0; inext;``    ``return` `curr->data;``}` `void` `printReverse(Node *head)``{``    ``// Count nodes``    ``int` `n = getCount(head);` `    ``for` `(``int` `i=n; i>=1; i--)``        ``printf``(``"%d "``, getNth(head, i));``}` `/* Driver program to test count function*/``int` `main()``{``    ``/* Start with the empty list */``    ``struct` `Node* head = NULL;` `    ``/* Use push() to construct below list``     ``1->2->3->4->5 */``    ``push(&head, 5);``    ``push(&head, 4);``    ``push(&head, 3);``    ``push(&head, 2);``    ``push(&head, 1);` `    ``printReverse(head);` `    ``return` `0;``}`

## Java

 `// Java program to print reverse of linked list``// without extra space and without modifications.``class` `GFG``{``    ` `/* Link list node */``static` `class` `Node``{``    ``int` `data;``    ``Node next;``};``static` `Node head;` `/* Given a reference (pointer to pointer) to``the head of a list and an int, push a new node``on the front of the list. */``static` `void` `push(Node head_ref, ``int` `new_data)``{``    ``Node new_node = ``new` `Node();``    ``new_node.data = new_data;``    ``new_node.next = (head_ref);``    ``(head_ref) = new_node;``    ``head = head_ref;``}` `/* Counts no. of nodes in linked list */``static` `int` `getCount(Node head)``{``    ``int` `count = ``0``; ``// Initialize count``    ``Node current = head; ``// Initialize current``    ``while` `(current != ``null``)``    ``{``        ``count++;``        ``current = current.next;``    ``}``    ``return` `count;``}` `/* Takes head pointer of the linked list and``index as arguments and return data at index*/``static` `int` `getNth(Node head, ``int` `n)``{``    ``Node curr = head;``    ``for` `(``int` `i = ``0``; i < n - ``1` `&& curr != ``null``; i++)``    ``curr = curr.next;``    ``return` `curr.data;``}` `static` `void` `printReverse()``{``    ``// Count nodes``    ``int` `n = getCount(head);` `    ``for` `(``int` `i = n; i >= ``1``; i--)``        ``System.out.printf(``"%d "``, getNth(head, i));``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``/* Start with the empty list */``    ``head = ``null``;` `    ``/* Use push() to construct below list``    ``1->2->3->4->5 */``    ``push(head, ``5``);``    ``push(head, ``4``);``    ``push(head, ``3``);``    ``push(head, ``2``);``    ``push(head, ``1``);` `    ``printReverse();``}``}` `// This code is contributed by PrinciRaj1992`

## Python3

 `# Python3 program to print reverse of linked list``# without extra space and without modifications.`` ` `''' Link list node '''``class` `Node:``    ` `    ``def` `__init__(``self``, data):``        ``self``.data ``=` `data``        ``self``.``next` `=` `None``     ` `''' Given a reference (pointer to pointer) to the head``  ``of a list and an int, push a new node on the front``  ``of the list. '''``def` `push( head_ref, new_data):   ``    ``new_node ``=` `Node(new_data)``    ``new_node.``next` `=` `head_ref;``    ``head_ref ``=` `new_node;``    ``return` `head_ref` `''' Counts no. of nodes in linked list '''``def` `getCount(head):``    ``count ``=` `0``;  ``# Initialize count``    ``current ``=` `head;  ``# Initialize current   ``    ``while` `(current !``=` `None``):   ``        ``count ``+``=` `1``        ``current ``=` `current.``next``;   ``    ``return` `count;`` ` `''' Takes head pointer of the linked list and index``    ``as arguments and return data at index'''``def` `getNth(head, n):``    ``curr ``=` `head;  ``    ``i ``=` `0`   `    ``while` `i < n ``-` `1` `and` `curr !``=` `None``:``        ``curr ``=` `curr.``next``;``        ``i ``+``=` `1`   `    ``return` `curr.data;`` ` `def` `printReverse(head):` `    ``# Count nodes``    ``n ``=` `getCount(head);``    ` `    ``for` `i ``in` `range``(n, ``0``, ``-``1``):``        ``print``(getNth(head, i), end ``=` `' '``);``     ` `# Driver code``if` `__name__``=``=``'__main__'``:``    ` `    ``''' Start with the empty list '''``    ``head ``=` `None``;`` ` `    ``''' Use push() to construct below list``     ``1.2.3.4.5 '''``    ``head ``=` `push(head, ``5``);``    ``head ``=` `push(head, ``4``);``    ``head ``=` `push(head, ``3``);``    ``head ``=` `push(head, ``2``);``    ``head ``=` `push(head, ``1``);`` ` `    ``printReverse(head);``     ` `# This code is contributed by rutvik_56`

## C#

 `// C# program to print reverse of``// linked list without extra space``// and without modifications.``using` `System;``    ` `class` `GFG``{``    ` `/* Link list node */``public` `class` `Node``{``    ``public` `int` `data;``    ``public` `Node next;``};``static` `Node head;` `/* Given a reference (pointer to pointer)``to the head of a list and an int, push a``new node on the front of the list. */``static` `void` `push(Node head_ref,``                 ``int` `new_data)``{``    ``Node new_node = ``new` `Node();``    ``new_node.data = new_data;``    ``new_node.next = (head_ref);``    ``(head_ref) = new_node;``    ``head = head_ref;``}` `/* Counts no. of nodes in linked list */``static` `int` `getCount(Node head)``{``    ``int` `count = 0; ``// Initialize count``    ``Node current = head; ``// Initialize current``    ``while` `(current != ``null``)``    ``{``        ``count++;``        ``current = current.next;``    ``}``    ``return` `count;``}` `/* Takes head pointer of the linked list and``index as arguments and return data at index*/``static` `int` `getNth(Node head, ``int` `n)``{``    ``Node curr = head;``    ``for` `(``int` `i = 0;``             ``i < n - 1 && curr != ``null``; i++)``    ``curr = curr.next;``    ``return` `curr.data;``}` `static` `void` `printReverse()``{``    ``// Count nodes``    ``int` `n = getCount(head);` `    ``for` `(``int` `i = n; i >= 1; i--)``        ``Console.Write(``"{0} "``, getNth(head, i));``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``/* Start with the empty list */``    ``head = ``null``;` `    ``/* Use push() to construct below list``    ``1->2->3->4->5 */``    ``push(head, 5);``    ``push(head, 4);``    ``push(head, 3);``    ``push(head, 2);``    ``push(head, 1);` `    ``printReverse();``}``}` `// This code is contributed by Princi Singh`

## Javascript

 ``

Output:

`5 4 3 2 1`

Time complexity: O(n2) where n is the number of nodes in the given linked list
Auxiliary Space: O(1), as constant extra space is required.

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