@Umer: Its a singly LL and it has cycle. Both N1 and N2 will keep traversing
within the cycle.


On Sun, Mar 28, 2010 at 9:56 PM, Umer Farooq <the.um...@gmail.com> wrote:

> I'll appreciate comments on the solution proposed by me. It works the
> following way:
>
> take two pointers, N1 and N2 pointing on the head of the list.
>
> Move N2 by two nodes, and N1 by a single node.
>
> When N2 reaches head again (or N2->Next is a head)
>
> then return N1 which will be pointing to the middle element of the list.
>
> Regards
> Umer Farooq
>
>
> On Sun, Mar 28, 2010 at 2:17 PM, Mukesh Kumar thakur <
> mukeshraj8...@gmail.com> wrote:
>
>> hi! in my opinion we can find the middle element in the singly linked
>> which hv the cycle........
>> as we know the list doesnt support the concept of cycle coz it has only
>> one direction traversal........
>>
>>> but if we take the case when the list hvng  the no of element equal
>>>
>> as we hv :
>>                      n element in the list
>> we hv to find the middle one
>>     in genral;;;;;simply we divide it in .........
>> n/2; or
>> if consider middle elment as a key ;;;;
>>          temp->link=null;
>>           temp->first=middle
>>
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Thanks & Regards,
NMN

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