*
*regards
- Sumit Kumar Pathak
(Sumit/ Pathak/ SKP ...)
*Smile is only good contagious thing.*
*Spread it*!!!!!
On Tue, Oct 11, 2011 at 11:22 AM, DIPANKAR DUTTA <dutta.dipanka...@gmail.com
> wrote:
> what's happen if the versions are not in same length..
>
> For example
>
> v1: 1.1.1.133.2
> v2: 1.2
> v3: 1.2.3.4.5555.333
> v4: 1.2.3.4.5554.222
> v5: 1.3.2.2.2.2.2.2.2.2.2.2.2.2
>
> It implies we must be scan from left side one by one ...
>
> In general the we make a some sort of lexical comparison where each
> character is a number and separated by number ..
>
> the problem definition is as :
>
> let V1,V2,V3 ...Vn be the n version
>
> let S={1,2,3...n} ,index=0
>
> Latest[S,index]= return Vi if { Max(ALL Vi[k] where i belongs to S )}
> is a singleton, else
> = return Latest[ set of all index belongs to { Max(ALL
> Vi[k] where i belongs to S )}, index+1 ]
>
>
>
>
>
> On 10/11/11, Dave <dave_and_da...@juno.com> wrote:
> > @Karen: It is more complicated than scanning character by character.
> > E.g., "1.10.3" is older than "1.9.7". I think
>
<snip>
> you need to parse the
> > numbers between the dots and compare them
>
</snip>
simple, easier and correct way.
points:
- compare left to right (ofcourse) each version being vector of numbers
- for diffrent size, assume extra ones to be zero while comapring (no need
to store trailing zeros)
> one by one. Thus, in the
> > above example, 1 compares equal to 1, so you keep scanning. Then 10
> > compares greater than 9 so the first string is number of the newer
> > version. I did this many years ago in a csh install script for a unix
> > product.
> >
> > Dave
> >
> > On Oct 10, 9:52 pm, "bagaria.ka...@gmail.com"
> > <bagaria.ka...@gmail.com> wrote:
> >> Given two strings describing the version of a particular software need
> to
> >> find the later version.
> >>
> >> For eg.
> >> 1st string = "1.2.4.5"
> >> 2nd string="1.2.3.5"
> >>
> >> 1st string is the later one.
> >>
> >> Can be done using traversing the string and comparing each character one
> >> after the another. Looking for a better solution with lesser complexity.
> >>
> >> --
> >> Thanks and Regards
> >>
> >> *Karan Bagaria*
> >> *MCA Final Year*
> >> Training and Placement Representative
> >> *NIT Durgapur*
> >
> > --
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> >
> >
>
>
> --
> Thanks and Regards,
> ------------------------------
> **DIPANKAR DUTTA
> Software Development Engineer
> Xen Server - OpenStack Development Team (DataCenter and Cloud)
>
> Citrix R&D India Pvt Ltd
> 69/3, Millers Road, Bangalore – 560052
> Phone: +91 8147830733
> Office: Extn: 16429
> Email: dipankar.du...@citrix.com
>
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