The solution is essentially the same as Anurag's; Only that it would be
simple to reverse both the lists
So now the lists look like

Num 1 = 123456
Num 2= 1234
Link-1<-Link-2<-Link-3<-Link-4<-Link5<-Link6 head_num1
Link-1<-Link-2<-Link-3<-Link-4 <-head_num2

Now basically loop, and add the keys, remembering to take the carry ... and
if there is carry in the last stage, then make an extra node.

Regards



On Wed, Jan 27, 2010 at 11:07 AM, Anurag Bhatia <abhati...@gmail.com> wrote:

> Let us take an example -
>
> Num 1 = 123456
> Num 2= 1234
> Link-1->Link-2->Link-3->Link-4->Link5->Link6
> Link-1->Link-2->Link-3->Link-4
>
> Add nodes into linkedlist 1 till either one of the list is not null.
> Make sure you process the carry in each iteration.
>
>
> --AB
>
>
> On Tue, Jan 26, 2010 at 9:47 PM, Algoose Chase <harishp...@gmail.com>
> wrote:
> > conditions:
> > NO extra memory (@ stack or Heap) at all. No recursion.
> >
> > Any body has got any hint about how to get this done ?
> >
> >
> >
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-- 
We are here on earth to do good for others. What the others are here for, I
don't know.

Afroz Mohiuddin
Final Year Masters Student
Dept Computer Science and Engineering
Indian Institute of Technology Kanpur
Kanpur - 208016
INDIA

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