CareerCup

struct node
{
void * data;
struct Node * Prev;
struct Node * Next;
};


struct node * Prev = NULL;
struct node * Next = NULL;
struct node * Cur = Head;

while(Cur != NULL)
{
Next = Cur -> Next;
Cur -> Next = Prev;
Cur -> Prev = Next;
Prev = Cur;
Cur = Next;
}

Head = Prev;

why would you want to reverse a doubly lined list?

No Duh!..
1. Recurse till end of list, and return pointer.

Done!

Change the name of the pointer "next" to "prev" and vice-versa and return the last node

struct node
{
void * data;
struct Node * Prev;
struct Node * Next;
};
struct node * Cur = Head;
struct node * Temp = NULL;

while(Cur!=NULL){
if(Cur->Prev==NULL){//First Node
Cur->Prev=Cur->Next;
Cur->Next=NULL;
}
Temp=Cur->Next;
Cur->Next=Cur->Prev;
Cur->Prev=Temp;
Cur=Cur->Prev;
}
return Cur;

//@CyberPhoenix
//my dear i think when u had returened Cur its NULL
//is that of any significance
//well have a look on this!!

struct node
{
void * data;
struct Node * prev;
struct Node * next;
};
struct node * tra = Head;

while(tra){
head= tra->prev;
tra->prev = tra->next;
tra->next = head;
tra = tra->prev;
}
head = head->prev;

Sorry for using the SMS Language.

//Iterative
void reverdll(node *root){
	node *cur=root;
	node *res=NULL;
	while(cur){
		temp=cur->next;
		cur->next=cur->prev;
		cur->prev=temp;
		cur=temp;
	}
	*root=cur;
}
//Recursion
node* reversedll(node *root){
	if(root==NULL) return NULL;
	else{
		root->prev=reversedll(root->next);
		root->next=root->prev;
		return root;
	}
}

Treat this like a singly linked list, and swap the pointers. thats all

/* Ignoring the sentinel cases */
while(root)
{
    SWAP(root->next, root->prev);
    root = root->prev; /* This is next actually */
}