Perl : print the sequence number without missing number

Shell Programming and Scripting
1

Dear Perl users,
I need your help to solve my problem below.
I want to print the sequence number without missing number within the range.
E.g. my sequence number :

1 2 3 4 5 6 7 8  11 12 13 14

my desired output:

1 -8 , 11-14

my code below but still problem with the result:

1 - 14
1 - 14
1 - 14
1 - 14
1 - 14
1 - 14
1 - 14
1 - 8
11 - 14
1 - 14
1 - 14

######Perl Code ######

#!/usr/bin/perl -w
use strict;
my $i;
my $start;
my @lst_array = qw /1 2 3 4 5 6 7  8 11 12 13 14/;
   $start = $lst_array[0];

      for ($i=1; $i<$#lst_array; $i++){
                 if (($lst_array[$i] - $start) == 1){
                             print "$lst_array[0] - $lst_array[$#lst_array]\n";
                 }
                   elsif (($lst_array[$i] - $start) > 1){
                              print "$lst_array[0] - $start\n";
                              print "$lst_array[$i] - $lst_array[$#lst_array]\n";
                  }
       $start = $lst_array[$i];
}

Thank alot
Regards
Mandai

2

Can you be more clear about what your expected output?`

3

Try this:

#!/usr/bin/perl 

use warnings;
use strict; 

my @lst_array = qw (1 2 3 4 5 6 7  8 11 12 13 14);
my $start = $lst_array[0];

for (my $i=1; $i<@lst_array; $i++){
  if (($lst_array[$i] - $lst_array[$i -1]) != 1 ){
    print "$start - $lst_array[$i-1], ";
    $start = $lst_array[$i];
  }
}
print "$start -$lst_array[-1]\n"'