int add(struct node* pL1, struct node * pl2,int gap/*no of digits in l1 -no
of digits in l2*/)
{ //assumption, # of nodes in L1 is > # of nodes in L2, make sure this
before calling this, counting nodes is less costlier than reversal
if (!(pl1) || !(pl2)) return 0;
if (gap>0)
{
carry = add(pL1->next, pL2, gap-1);
carry = (pl1->data+carry)/10;
pl1->data =(pl1->data+carry) %10;
return carry;
}
else
{
carry = add(pL1->next, pl2->next, gap -1);
carry = (pl1->data+pl2->data+carry)/10;
pl1->data =(pl1->data+pl2->data+carry) %10;
return carry;
}
}
Best Regards
Ashish Goel
"Think positive and find fuel in failure"
+919985813081
+919966006652
On Sun, Aug 15, 2010 at 12:19 PM, Manjunath Manohar <
manjunath.n...@gmail.com> wrote:
> @Dave..Can u provide a small snippet for ur explanation pls..
>
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