How to increment version number in a shell script?

by 
$ echo 1.2.3.4 | awk -F. -v OFS=. 'NF==1{print ++$NF}; NF>1{if(length($NF+1)>length($NF))$(NF-1)++; $NF=sprintf("%0*d", length($NF), ($NF+1)%(10^length($NF))); print}'
1.2.3.5

1.2.3.9  => 1.2.4.0
1.2.3.44 => 1.2.3.45
1.2.3.99 => 1.2.4.00
1.2.3.999=> 1.2.4.000
1.2.9    => 1.3.0
999      => 1000

UPDATE:

#!/usr/bin/gawk -f

BEGIN{
    v[1] = "1.2.3.4"
    v[2] = "1.2.3.44"
    v[3] = "1.2.3.99"
    v[4] = "1.2.3"
    v[5] = "9"
    v[6] = "9.9.9.9"
    v[7] = "99.99.99.99"
    v[8] = "99.0.99.99"
    v[9] = ""

    for(i in v)
        printf("#%d: %s => %s\n", i, v[i], inc(v[i])) | "sort | column -t"
}

function inc(s,    a, len1, len2, len3, head, tail)
{
    split(s, a, ".")

    len1 = length(a)
    if(len1==0)
        return -1
    else if(len1==1)
        return s+1

    len2 = length(a[len1])
    len3 = length(a[len1]+1)

    head = join(a, 1, len1-1)
    tail = sprintf("%0*d", len2, (a[len1]+1)%(10^len2))

    if(len2==len3)
        return head "." tail
    else
        return inc(head) "." tail
}

function join(a, x, y,    s)
{
    for(i=x; i<y; i++)
        s = s a[i] "."
    return s a[y]
}

$ chmod +x inc.awk
$ ./inc.awk
#1:  1.2.3.4      =>  1.2.3.5
#2:  1.2.3.44     =>  1.2.3.45
#3:  1.2.3.99     =>  1.2.4.00
#4:  1.2.3        =>  1.2.4
#5:  9            =>  10
#6:  9.9.9.9      =>  10.0.0.0
#7:  99.99.99.99  =>  100.00.00.00
#8:  99.0.99.99   =>  99.1.00.00
#9:  =>           -1

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