I this this code will solves it ...
*#include <stdio.h>
#include <stdlib.h>
struct node{
int val;
struct node *next;
};
int main(){
struct node *head = NULL , *next_ptr = head;
int n = 1;
while ( n <= 6 ){
struct node *temp = (struct node*) malloc (sizeof(struct node) );
temp->val = n++;
temp->next = NULL;
if (head == NULL ){
head = temp ;
next_ptr = head;
}else{
next_ptr->next = temp;
next_ptr = temp;
}
}
next_ptr = head;
head = NULL;
struct node *prev = NULL;
while ( next_ptr != NULL && next_ptr->next != NULL ){
struct node *swp = next_ptr->next->next;
if ( head == NULL){
head = next_ptr->next;
}
if ( prev == NULL ){
prev = next_ptr;
}else{
prev->next = next_ptr->next;
prev = next_ptr;
}
next_ptr->next->next = next_ptr;
next_ptr->next = swp;
next_ptr = swp;
}
while ( head != NULL ){
printf("%d \n" , head->val);
head = head->next;
}
return 0;
}
*
On Thu, Jun 2, 2011 at 7:18 AM, Anurag Bhatia <[email protected]> wrote:
> Have a look at this recursive soln in java -
>
> Node convertList(Node head)
> {
> if(head ==null)
> return null;
>
> Node node = head;
> Node next = node.next;
> Node next2next = next.next;
>
> next.next = node;
> node.next = convertList(next2next);
>
> return next;
> }
>
> This soln assumes even number of nodes in the list. The code is quite
> self explanatory.
>
> --Anurag
>
>
> On Wed, Jun 1, 2011 at 6:26 AM, Anand <[email protected]> wrote:
> > Given a linked list of the form, 1->2->3->4->5->6, convert it into the
> form
> > 2->1->4->3->6->5. Note that the nodes need to be altered and not the data
> > contained in them
> >
> > --
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>
--
ROHIT SINDHU
B.Tech (IT) -- 6th sem.
INDIAN INSTITUTE OF INFORMATION TECHNOLOGY, ALLAHABAD.
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