how about this one?

Node* reverseBy2(Node* head){
    Node* p1 = head;
    if(p1 == NULL)
        return NULL;
    Node* p2 = p1->next;
    if(p2 == NULL)
        return head;
    Node* nextHead = p2->next;
    p2->next = p1;
    p1->next = reverseBy2(nextHead);
    return p2;
}
[?]
2011/6/1 Shivaji Varma <shivaji...@gmail.com>

> Hi,
>
> Please take a look at this link.
>
> http://mycsinterviewsexperiences.blogspot.com/
>
>
> --
> Shivaji
>
>
> On Wed, Jun 1, 2011 at 6:26 AM, Anand <anandut2...@gmail.com> 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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