1
1 1
2 1
1 2 1 1
1 1 1 2 2 1
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If the 5th line was 3 1 1 2 passing an argument gives that solution-

my $anchor = @ARGV ? '' : '^';
printf "%d\n", $_ = '1';
do {
    my $new;
    my ( $n ) = /^(\d)/o;
    while ( $_ ) {
        my $c = 0;
        $c++ while s/$anchor\s?$n\s?//;
        $new .= "$c $n " if $c;
        ( $n ) = /^(\d)/o;
    }
    print $_ = $new;
} until ( <STDIN> =~ /\S/ );
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