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 <[email protected]> wrote:
> @Karen: It is more complicated than scanning character by character.
> E.g., "1.10.3" is older than "1.9.7". I think you need to parse the
> numbers between the dots and compare them 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, "[email protected]"
> <[email protected]> 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
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