in starting we keep track of head pointer..
we are having three pointers
head,slower,faster,
know we start traversing the list
if(faster==head||faster->next==head)
  return slower
else
   faster=faster->next->next
   slower=slower->next

On Sun, Mar 28, 2010 at 12:15 PM, vikrant singh <vikrantsing...@gmail.com>wrote:

> @diamond
> sorry but how do we know the fast pointer is pointing to head again?
> u see, in case u dont have O(n) space to record visited nodes
>
> On Sun, Mar 28, 2010 at 10:28 AM, blackDiamond <patidarc...@gmail.com>wrote:
>
>> vikrant you havent read properly the post by Gaurav when the second
>> pointer will come back to the head first will be pointing the middle.(Think
>> it)!!
>>
>>
>> On Sun, Mar 28, 2010 at 10:21 AM, vikrant singh <vikrantsing...@gmail.com
>> > wrote:
>>
>>> Well gaurav, i think by this method you can only check for a cycle in the
>>> list.
>>> If u have any idea how can you implement this to solve originally posted
>>> problem?
>>>
>>> On Sat, Mar 27, 2010 at 9:03 AM, gaurav kishan 
>>> <gauravkis...@gmail.com>wrote:
>>>
>>>> Hi,
>>>>
>>>> Keep two pointers both initially pointing to Head.
>>>> Move first pointer one by one and the second pointer by two nodes in
>>>> each iteration.
>>>> When second pointer next link points to head again,return first pointer.
>>>>
>>>> Please let me know if this can be further imporved upon or there is some
>>>> fallacy in the approach.
>>>>
>>>> Regards,
>>>> Gaurav.
>>>>
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>>>
>>>
>>>
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