The Solution is pretty straight forward when you long number is represented
in reverse order in linked list.
If the number is not in reverse order, We need an Explicit stack or we must
Use Recursion .
Other way around this is to construct another parallel linked list along
with Sum(linked list) fo
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(
@Manjunath: A small snippet of my explanation of what? What do you not
understand?
Dave
On Aug 15, 1:49 am, Manjunath Manohar
wrote:
> @Dave..Can u provide a small snippet for ur explanation pls..
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@Dave..Can u provide a small snippet for ur explanation pls..
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Don't reverse the list. Just store it from low to high order digits.
I.e., the head points to the ones digit, which points to the tens
digit, etc. That is the assumption I made with my algorithm presented
in the second post of this thread, because almost all operations use
the number beginning with
i think we can use recursion method to reverse the list
-- Prashant Kulkarni
On Sat, Aug 14, 2010 at 11:40 PM, Lokesh Agarwal wrote:
> how can you traverse from last without reversing it.
>
> and there is no need fof using extra stack space.
>
> --
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Reversing and then again reversing the answer will not be an efficient
algorithm...
on the fly computation of sum must be done...any ideas
On 8/14/10, Rahul Singhal wrote:
> I men to say ki just traverse from last instead of reversing it and storing
> result in a stack in linked list form so that
I men to say ki just traverse from last instead of reversing it and storing
result in a stack in linked list form so that we dont need to reverse
again.Hope,i made myself clear.
On Sat, Aug 14, 2010 at 11:40 PM, Lokesh Agarwal wrote:
> how can you traverse from last without reversing it.
>
> and
how can you traverse from last without reversing it.
and there is no need fof using extra stack space.
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i think we can access numbers from last so no need to reverse it and also we
can store it in linke list in stack way so again no need to reverse the
linked list.
On Sat, Aug 14, 2010 at 7:03 PM, Gaurav Singh wrote:
> Reversing the lists and then adding and then reversing the final list
> is the
Reversing the lists and then adding and then reversing the final list
is the most appropriate method. Bcoz the lists may contain
arbitarily large numbers, so forming integers then adding is not
logical here.
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Set carry = 0
Walk the linked lists of the addends simultaneoulsy
At each pair of list nodes, add the digits and carry
Create a linked list node for the sum
If the sum is less than 10:
Store the sum
Set carry = 0
Otherwise
Store sum - 10
Set carry = 1
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