Hi, well i am able to find a loop in a list using two iterators.One iterator runs "two times faster than the other", and if he encounters the first, it means that there is a loop.
Example : 1,2,3,4,5,6,7,8,9,5 the algorithm would generate: start - 1,2 iteration 1- 2, 4 iteration 2- 3, 6 iteration 3- 4, 8 iteration 4- 5, 5 ( match) But how can this help me with counting list elements? :( Thanks, D. On Thu, Jan 8, 2009 at 5:48 PM, Diez B. Roggisch <de...@nospam.web.de> wrote: > > David Hláčik wrote: > > > Hi, > > > > so okay, i will create a helping set, where i will be adding elements > > ID, when element ID will be allready in my helping set i will stop and > > count number of elements in helping set. This is how long my cycled > > linked list is. > > > But what if i have another condition , and that is *i can use only > > helping memory with constant size* ? This means i am not able to > > create any set and adding elements there. I need to have a constant > > size variables . This is complication a complication for me. > > This isn't to hard - think about what you are really interested in - knowing > if *all* other elements are already counted, or a specific one? You can get > away with only one, to detect the cycle and abort. > > Diez > -- > http://mail.python.org/mailman/listinfo/python-list -- http://mail.python.org/mailman/listinfo/python-list
