that will result into an infinite loop,
On Mon, Mar 29, 2010 at 9:51 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<algogeeks%2bunsubscr...@googlegroups.com> > . > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- Thanks & Regards, - Navin Mohan Naidu Great Minds Discuss Ideas Average Minds Discuss Events Small Minds Discuss People -- 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.
