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
>>>>
>>>> --
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>>>>
>>>>
>>>
>>>
>>> --
>>> best wishes!!
>>> Vaibhav Shukla
>>> DU-MCA
>>>
>>>
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>>
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>
>
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