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)
{
link->temp=first;
}
else
return temp;
for(temp->first=null;temp>headr;temp->link++)
{
we can print the middle element;
}
or inplace of this we can use the pointer concept to assign the middle and
faster element..........
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