# Reverse even elements in a Linked List

Given a linked list, the task is to reverse the contiguous even elements and print the updated linked list.

Input: 1 -> 2 -> 3 -> 3 -> 4 -> 6 -> 8 -> 5 -> NULL
Output: 1 2 3 3 8 6 4 5
Initial list: 1 -> 2 -> 3 -> 3 -> 4 -> 6 -> 8 -> 5 -> NULL
Reversed list: 1 -> 2 -> 3 -> 3 -> 8 -> 6 -> 4 -> 5 -> NULL

Input: 11 -> 32 -> 7 -> NULL
Output: 11 32 7

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Reversing the contiguous even elements will not take place when:

1. The node’s value is odd.
2. The node’s value is even but adjacent values are odd.

In rest of the cases, the continuous block of even nodes can be reversed.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Structure of a node of the linked list ` `struct` `node { ` `    ``int` `data; ` `    ``struct` `node* next; ` `}; ` ` `  `// Function to create a new node ` `struct` `node* newNode(``int` `d) ` `{ ` `    ``struct` `node* newnode = ``new` `node(); ` `    ``newnode->data = d; ` `    ``newnode->next = NULL; ` `    ``return` `newnode; ` `} ` ` `  `// Recursive function to reverse the consecutive ` `// even nodes of the linked list ` `struct` `node* reverse(``struct` `node* head, ``struct` `node* prev) ` `{ ` ` `  `    ``// Base case ` `    ``if` `(head == NULL) ` `        ``return` `NULL; ` ` `  `    ``struct` `node* temp; ` `    ``struct` `node* curr; ` `    ``curr = head; ` ` `  `    ``// Reversing nodes until curr node's value ` `    ``// turn odd or Linked list is fully traversed ` `    ``while` `(curr != NULL && curr->data % 2 == 0) { ` `        ``temp = curr->next; ` `        ``curr->next = prev; ` `        ``prev = curr; ` `        ``curr = temp; ` `    ``} ` ` `  `    ``// If elements were reversed then head ` `    ``// pointer needs to be changed ` `    ``if` `(curr != head) { ` ` `  `        ``head->next = curr; ` ` `  `        ``// Recur for the remaining linked list ` `        ``curr = reverse(curr, NULL); ` `        ``return` `prev; ` `    ``} ` ` `  `    ``// Simply iterate over the odd value nodes ` `    ``else` `{ ` `        ``head->next = reverse(head->next, head); ` `        ``return` `head; ` `    ``} ` `} ` ` `  `// Utility function to print the ` `// contents of the linked list ` `void` `printLinkedList(``struct` `node* head) ` `{ ` `    ``while` `(head != NULL) { ` `        ``cout << head->data << ``" "``; ` `        ``head = head->next; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 3, 3, 4, 6, 8, 5 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` `  `    ``struct` `node* head = NULL; ` `    ``struct` `node* p; ` ` `  `    ``// Constructing linked list ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``if` `(head == NULL) { ` `            ``p = newNode(arr[i]); ` `            ``head = p; ` `            ``continue``; ` `        ``} ` `        ``p->next = newNode(arr[i]); ` `        ``p = p->next; ` `    ``} ` ` `  `    ``// Head of the updated linked list ` `    ``head = reverse(head, NULL); ` ` `  `    ``// Printing the reversed linked list ` `    ``printLinkedList(head); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Structure of a node of the linked list ` `static` `class` `node ` `{ ` `    ``int` `data; ` `    ``node next; ` `}; ` ` `  `// Function to create a new node ` `static` `node newNode(``int` `d) ` `{ ` `    ``node newnode = ``new` `node(); ` `    ``newnode.data = d; ` `    ``newnode.next = ``null``; ` `    ``return` `newnode; ` `} ` ` `  `// Recursive function to reverse the consecutive ` `// even nodes of the linked list ` `static` `node reverse(node head, node prev) ` `{ ` ` `  `    ``// Base case ` `    ``if` `(head == ``null``) ` `        ``return` `null``; ` ` `  `    ``node temp; ` `    ``node curr; ` `    ``curr = head; ` ` `  `    ``// Reversing nodes until curr node's value ` `    ``// turn odd or Linked list is fully traversed ` `    ``while` `(curr != ``null` `&& curr.data % ``2` `== ``0``)  ` `    ``{ ` `        ``temp = curr.next; ` `        ``curr.next = prev; ` `        ``prev = curr; ` `        ``curr = temp; ` `    ``} ` ` `  `    ``// If elements were reversed then head ` `    ``// pointer needs to be changed ` `    ``if` `(curr != head)  ` `    ``{ ` `        ``head.next = curr; ` ` `  `        ``// Recur for the remaining linked list ` `        ``curr = reverse(curr, ``null``); ` `        ``return` `prev; ` `    ``} ` ` `  `    ``// Simply iterate over the odd value nodes ` `    ``else` `    ``{ ` `        ``head.next = reverse(head.next, head); ` `        ``return` `head; ` `    ``} ` `} ` ` `  `// Utility function to print the ` `// contents of the linked list ` `static` `void` `printLinkedList(node head) ` `{ ` `    ``while` `(head != ``null``) ` `    ``{ ` `        ``System.out.print(head.data + ``" "``); ` `        ``head = head.next; ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``1``, ``2``, ``3``, ``3``, ``4``, ``6``, ``8``, ``5` `}; ` `    ``int` `n = arr.length; ` ` `  `    ``node head = ``null``; ` `    ``node p = ``new` `node(); ` ` `  `    ``// Constructing linked list ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `        ``if` `(head == ``null``)  ` `        ``{ ` `            ``p = newNode(arr[i]); ` `            ``head = p; ` `            ``continue``; ` `        ``} ` `        ``p.next = newNode(arr[i]); ` `        ``p = p.next; ` `    ``} ` ` `  `    ``// Head of the updated linked list ` `    ``head = reverse(head, ``null``); ` ` `  `    ``// Printing the reversed linked list ` `    ``printLinkedList(head); ` `} ` `}  ` ` `  `// This code is contributed by PrinciRaj1992 `

## Python

 `# Python implementation of the approach  ` ` `  `# Node of a linked list  ` `class` `Node:  ` `    ``def` `__init__(``self``, ``next` `=` `None``, data ``=` `None``):  ` `        ``self``.``next` `=` `next` `        ``self``.data ``=` `data  ` ` `  `# Function to create a new node ` `def` `newNode( d): ` ` `  `    ``newnode ``=` `Node() ` `    ``newnode.data ``=` `d ` `    ``newnode.``next` `=` `None` `    ``return` `newnode ` ` `  `# Recursive function to reverse the consecutive ` `# even nodes of the linked list ` `def` `reverse(head, prev): ` ` `  `    ``# Base case ` `    ``if` `(head ``=``=` `None``): ` `        ``return` `None` ` `  `    ``temp ``=` `None` `    ``curr ``=` `None` `    ``curr ``=` `head ` ` `  `    ``# Reversing nodes until curr node's value ` `    ``# turn odd or Linked list is fully traversed ` `    ``while` `(curr !``=` `None` `and` `curr.data ``%` `2` `=``=` `0``) : ` `        ``temp ``=` `curr.``next` `        ``curr.``next` `=` `prev ` `        ``prev ``=` `curr ` `        ``curr ``=` `temp ` ` `  `    ``# If elements were reversed then head ` `    ``# pointer needs to be changed ` `    ``if` `(curr !``=` `head) : ` `     `  `        ``head.``next` `=` `curr ` ` `  `        ``# Recur for the remaining linked list ` `        ``curr ``=` `reverse(curr, ``None``) ` `        ``return` `prev ` `     `  `    ``# Simply iterate over the odd value nodes ` `    ``else``: ` `     `  `        ``head.``next` `=` `reverse(head.``next``, head) ` `        ``return` `head ` `     `  `# Utility function to print the ` `# contents of the linked list ` `def` `printLinkedList(head): ` ` `  `    ``while` `(head !``=` `None``): ` `     `  `        ``print``(head.data ,end``=` `" "``) ` `        ``head ``=` `head.``next` `     `  `# Driver code ` `arr ``=` `[ ``1``, ``2``, ``3``, ``3``, ``4``, ``6``, ``8``, ``5` `] ` `n ``=` `len``(arr) ` ` `  `head ``=` `None` `p ``=` `Node() ` ` `  `i ``=` `0` ` `  `# Constructing linked list ` `while` `( i < n ): ` `     `  `    ``if` `(head ``=``=` `None``): ` `         `  `        ``p ``=` `newNode(arr[i]) ` `        ``head ``=` `p ` `    ``else``: ` `        ``p.``next` `=` `newNode(arr[i]) ` `        ``p ``=` `p.``next` `    ``i ``=` `i ``+` `1``; ` ` `  `# Head of the updated linked list ` `head ``=` `reverse(head, ``None``) ` ` `  `# Printing the reversed linked list ` `printLinkedList(head) ` ` `  `# This code is contributed by Arnab Kundu `

## C#

 `// C# implementation of the above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Structure of a node of the linked list ` `public` `class` `node ` `{ ` `    ``public` `int` `data; ` `    ``public` `node next; ` `}; ` ` `  `// Function to create a new node ` `static` `node newNode(``int` `d) ` `{ ` `    ``node newnode = ``new` `node(); ` `    ``newnode.data = d; ` `    ``newnode.next = ``null``; ` `    ``return` `newnode; ` `} ` ` `  `// Recursive function to reverse the consecutive ` `// even nodes of the linked list ` `static` `node reverse(node head, node prev) ` `{ ` ` `  `    ``// Base case ` `    ``if` `(head == ``null``) ` `        ``return` `null``; ` ` `  `    ``node temp; ` `    ``node curr; ` `    ``curr = head; ` ` `  `    ``// Reversing nodes until curr node's value ` `    ``// turn odd or Linked list is fully traversed ` `    ``while` `(curr != ``null` `&& curr.data % 2 == 0)  ` `    ``{ ` `        ``temp = curr.next; ` `        ``curr.next = prev; ` `        ``prev = curr; ` `        ``curr = temp; ` `    ``} ` ` `  `    ``// If elements were reversed then head ` `    ``// pointer needs to be changed ` `    ``if` `(curr != head)  ` `    ``{ ` `        ``head.next = curr; ` ` `  `        ``// Recur for the remaining linked list ` `        ``curr = reverse(curr, ``null``); ` `        ``return` `prev; ` `    ``} ` ` `  `    ``// Simply iterate over the odd value nodes ` `    ``else` `    ``{ ` `        ``head.next = reverse(head.next, head); ` `        ``return` `head; ` `    ``} ` `} ` ` `  `// Utility function to print the ` `// contents of the linked list ` `static` `void` `printLinkedList(node head) ` `{ ` `    ``while` `(head != ``null``) ` `    ``{ ` `        ``Console.Write(head.data + ``" "``); ` `        ``head = head.next; ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]arr = { 1, 2, 3, 3, 4, 6, 8, 5 }; ` `    ``int` `n = arr.Length; ` ` `  `    ``node head = ``null``; ` `    ``node p = ``new` `node(); ` ` `  `    ``// Constructing linked list ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``if` `(head == ``null``)  ` `        ``{ ` `            ``p = newNode(arr[i]); ` `            ``head = p; ` `            ``continue``; ` `        ``} ` `        ``p.next = newNode(arr[i]); ` `        ``p = p.next; ` `    ``} ` ` `  `    ``// Head of the updated linked list ` `    ``head = reverse(head, ``null``); ` ` `  `    ``// Printing the reversed linked list ` `    ``printLinkedList(head); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```1 2 3 3 8 6 4 5
```

Time Complexity: O(N)
Space Complexity: O(1)

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Improved By : princiraj1992, andrew1234