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.
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>>> For more options, visit this group at
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>>>
>>>
>>
>>
>> --
>> best wishes!!
>> Vaibhav Shukla
>> DU-MCA
>>
>>
>> --
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>>
>
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