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 <[email protected]> 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 <[email protected]
> > 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 <[email protected]> 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 <[email protected]>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 <[email protected]> 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 ?
>>>>>
>>>>> --
>>>>>
>>>>>
>>>>>
>>>>
>>>> --
>>>>
>>>>
>>>>
>>>
>>> --
>>>
>>>
>>>
>>
>> --
>>
>>
>>
>
> --
>
>
>
--
