*node *reverse(node *head)
{
    if(head==NULL)
          return head;
    if(head->next==NULL)
          return head;
    node *temp = reverse(head->next);
    head->next->next = head;
    head->next = NULL;
     return (temp);
}*

On Sun, Jul 17, 2011 at 4:30 PM, Anika Jain <anika.jai...@gmail.com> wrote:

> initial call to this will be rev(head);
>
>
> On Sun, Jul 17, 2011 at 4:28 PM, Anika Jain <anika.jai...@gmail.com>wrote:
>
>> node *listing::rev(node *p)
>> {
>>     if(p->next==NULL)
>>     {
>>         head=p;
>>         return p;
>>     }
>>     else
>>     {
>>         node *t=rev(p->next);
>>         t->next=p;
>>         p->next=NULL;
>>         tail=p;
>>         return p;
>>
>>     }
>> }
>>
>> On Sun, Jul 17, 2011 at 3:21 PM, Nishant Mittal <
>> mittal.nishan...@gmail.com> wrote:
>>
>>> 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.
>>>>> To unsubscribe from this group, send email to
>>>>> algogeeks+unsubscr...@googlegroups.com.
>>>>> For more options, visit this group at
>>>>> http://groups.google.com/group/algogeeks?hl=en.
>>>>>
>>>>>
>>>>
>>>>
>>>> --
>>>>   best wishes!!
>>>> Vaibhav Shukla
>>>>     DU-MCA
>>>>
>>>>
>>>>  --
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>>>>
>>>
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>>>
>>
>>
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>



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*Piyush Sinha*
*IIIT, Allahabad*
*+91-7483122727*
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