void rev_recursion(NODE **head)
{
if(*head==NULL)
return;
NODE *first, *rest;
first=*head;
rest=first->next;
if(!rest)
return;
rev_recursion(&rest);
first->next->next=first;
first->next=NULL;
*head=rest;
}
On Sun, Jul 17, 2011 at 2:53 PM, vaibhav shukla <vaibhav200...@gmail.com>wrote:
> struct node *reverse_recurse(struct node *start)
> {
> if(start->next)
> {
> reverse_recurse(start->next);
> start->next->next=start;
> return(start);
> }
> else
> {
> head=start;
> }
> }
>
>
> in main
>
> if(head)
> {
> temp = reverse_recurse(head);
> temp->next =NULL;
> }
>
> "head and temp are global"
>
>
>
>
> On Sun, Jul 17, 2011 at 2:42 PM, Navneet Gupta <navneetn...@gmail.com>wrote:
>
>> Hi,
>>
>> I was trying to accomplish this task with the following call , header
>> = ReverseList(header)
>>
>> I don't want to pass tail pointer or anything and just want that i get
>> a reversed list with new header properly assigned after this call. I
>> am getting issues in corner conditions like returning the correct node
>> to be assigned to header.
>>
>> Can anyone give an elegant solution with above requirement? Since it
>> is with recursion, please test for multiple scenarios (empty list, one
>> node list, twe nodes
>
>
>
>> list etc) before posting your solution. In case
>> of empty list, the procedure should report that.
>>
>> --
>> Regards,
>> Navneet
>>
>> --
>> You received this message because you are subscribed to the Google Groups
>> "Algorithm Geeks" group.
>> To post to this group, send email to algogeeks@googlegroups.com.
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>>
>>
>
>
> --
> best wishes!!
> Vaibhav Shukla
> DU-MCA
>
>
> --
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