Thanks for the link Ritesh, if (isMatch(s, p+2)) return true; isnt this line incorrect in the code, as it can lead to segmentation fault... how can we directly access p+2 element, we know for sure that p is not '\0', but p+1 element can be '\0' , therefore leading to p+2 to be undefined.
On Thu, Dec 27, 2012 at 6:23 AM, Ritesh Mishra <rforr...@gmail.com> wrote: > try to solve it by recursion .. > http://www.leetcode.com/2011/09/regular-expression-matching.html > > > Regards, > > Ritesh Kumar Mishra > Information Technology > Third Year Undergraduate > MNNIT Allahabad > > > On Sun, Dec 23, 2012 at 11:14 PM, Prem Krishna Chettri <hprem...@gmail.com > > wrote: > >> Well I can tell you Something about design pattern to solve this case.. >> >> What I mean is by using The State Machine Design Pattern, Anyone >> can solve this. but Ofcourse it is complicated. >> >> >> >> >> On Sun, Dec 23, 2012 at 11:01 PM, shady <sinv...@gmail.com> wrote: >> >>> that's the point, Have to implement it from scratch... otherwise java >>> has regex and matcher, pattern to solve it........... >>> >>> >>> On Sun, Dec 23, 2012 at 10:28 PM, saurabh singh <saurab...@gmail.com>wrote: >>> >>>> If you need to implement this for some project then python and java >>>> have a very nice library >>>> >>>> >>>> Saurabh Singh >>>> B.Tech (Computer Science) >>>> MNNIT >>>> blog:geekinessthecoolway.blogspot.com >>>> >>>> >>>> On Sun, Dec 23, 2012 at 7:48 PM, shady <sinv...@gmail.com> wrote: >>>> >>>>> >>>>> http://stackoverflow.com/questions/13144590/to-check-if-two-strings-match-with-alphabets-digits-and-special-characters >>>>> >>>>> any solution for this......... we need to implement such regex >>>>> tester................ >>>>> >>>>> some complex cases : >>>>> *string* * regex * -> * status* >>>>> * >>>>> * >>>>> reesd re*.d -> match >>>>> re*eed reeed -> match >>>>> >>>>> can some one help with this ? >>>>> >>>>> -- >>>>> >>>>> >>>>> >>>> >>>> -- >>>> >>>> >>>> >>> >>> -- >>> >>> >>> >> >> -- >> >> >> > > -- > > > --