When the fast pointer points to head again, the slow pointer would point to
the middle of the linked list.
On Sat, Mar 27, 2010 at 11:51 PM, vikrant singh vikrantsing...@gmail.comwrote:
Well gaurav, i think by this method you can only check for a cycle in the
list.
If u have any idea how can
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.comwrote:
Well gaurav, i think by this method you can only check for a cycle
@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.comwrote:
vikrant you havent read properly the post by Gaurav when the second pointer
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
if(faster==head||faster-next=
=head)
return slower
else
faster=faster-next-next
slower=slower-next
HERE we hv three poniter faster,headr,slower.
as u hv mentioned that linked list has cycle list so better is that we hv
to use for loop
as :-
if we take headr,link,temp;
if(temp==headr)
{
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
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.comwrote:
Hi,
Keep two pointers both initially pointing to Head.
