@Umer: I guess the last node in the LL is _not_ pointing to the head back. Rather, its pointing to one of the intermediate nodes. That way, N2 or N2.next might not ever be _head_. See example: 1->2->3->4->5->3, where the last node points to the 3rd node in LL.
On Mar 29, 9:21 pm, Umer Farooq <the.um...@gmail.com> wrote: > that's why i have a terminating condition. It will keep on iterating until > ((N2 != head)&&(N2->NextPtr != head)) is true. > > head pointer of the linked list will be passed as an argument to the > function. > > On Mon, Mar 29, 2010 at 7:04 AM, Navin Naidu <navinmna...@gmail.com> wrote: > > @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 > > >>> -- > >>> You received this message because you are subscribed to the Google Groups > >>> "Algorithm Geeks" group. > >>> To post to this group, send email to algoge...@googlegroups.com. > >>> To unsubscribe from this group, send email to > >>> algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups.com> > >>> . > >>> For more options, visit this group at > >>>http://groups.google.com/group/algogeeks?hl=en. > > >> -- > >> You received this message because you are subscribed to the Google Groups > >> "Algorithm Geeks" group. > >> To post to this group, send email to algoge...@googlegroups.com. > >> To unsubscribe from this group, send email to > >> algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups.com> > >> . > >> For more options, visit this group at > >>http://groups.google.com/group/algogeeks?hl=en. > > > -- > > Thanks & Regards, > > NMN > > > -- > > You received this message because you are subscribed to the Google Groups > > "Algorithm Geeks" group. > > To post to this group, send email to algoge...@googlegroups.com. > > To unsubscribe from this group, send email to > > algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups.com> > > . > > For more options, visit this group at > >http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
