Reverse a doubly linked list

Problem Statement

You’re given the pointer to the head node of a doubly linked list. Reverse the order of the nodes in the list. The head node might be NULL to indicate that the list is empty.

Input Format 
You have to complete the Node* Reverse(Node* head) method which takes one argument - the head of the doubly linked list. You should NOT read any input from stdin/console.

Output Format 
Change the next and prev pointers of all the nodes so that the direction of the list is reversed. Then return the head node of the reversed list. Do NOT print anything to stdout/console.

Sample Input

NULL 
NULL <-- 2 <--> 4 <--> 6 --> NULL

Sample Output

NULL
NULL <-- 6 <--> 4 <--> 2 --> NULL

Explanation 
1. Empty list, so nothing to do. 
2. 2,4,6 become 6,4,2 o reversing in the given doubly linked list.

/*

   Reverse a doubly linked list, input list may also be empty

   Node is defined as

   struct Node

   {

     int data;

     Node *next;

     Node *prev;

   }

*/

Node* Reverse(Node* head)

{

    // Complete this function

    // Do not write the main method.

    Node* tail = head;

    Node* forward = head;

    if(head == NULL){

        return NULL;

    }

    if(head->next == NULL){

        return head;

    }

    while(tail->next != NULL){

        tail = tail->next;

    }

    while(forward != tail && forward->prev != tail){

        int temp = forward->data;

        forward->data = tail->data;

        tail->data = temp;

        forward = forward->next;

        tail = tail->prev;

    }

    return head;

}