folks ..
for second problem .. the mentioned algo doesn't work for last node.. I mean
copying data of next node and aalso making nxt->nxt as nxt node.. and
deleting the next node..

in case of last node.. how can we delete the last node being on that node?
even though if we use free (node) ... the prev node->next value doesn't
become NULL .. It will contains some garbage..

On Sat, Aug 13, 2011 at 11:34 AM, aditya kumar <aditya.kumar130...@gmail.com
> wrote:

> @mohit : +1
>
>
> On Sat, Aug 13, 2011 at 9:20 AM, Raghavan <its...@gmail.com> wrote:
>
>> First question:
>>
>>    - .Read the data from the first list and put it in a stack
>>    - Traverse the next list and compare by reading elements from the
>>    stack
>>    - This would solve it
>>
>>
>>  Second question:
>>
>>    - Take an list like 1->2->-3->4
>>    - If you are given with 2, juz copy the value and reference of next
>>    node(i.e : 3) into 2 and delete the node 3 this would solve it
>>
>> Thanks.
>>
>>
>> On Thu, Aug 11, 2011 at 10:21 AM, Abhishek gupta <
>> mailatabhishekgu...@gmail.com> wrote:
>>
>>> Q2). i think for second question it will be enough to just swap the data
>>> of current node to next node,
>>> and delete the next node.
>>> it will be like,
>>> //for swap
>>> int temp=current->data;
>>> current->data=current->next->data;
>>> current->next->data=temp;
>>>
>>> //for delete
>>> struct node *temp;
>>> temp=current->next;
>>> current->next=current->next->next;
>>> free(temp);
>>>
>>> i think it will be enough even in case of last node.
>>> correct me if i am wrong.
>>>
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>>
>>
>>
>> --
>> Thanks and regards,
>> Raghavan.K.L
>> <http://in.linkedin.com/in/raghavankl>
>>
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-- 
Thx,
--Gopi

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