Re^2: Determining uniqueness in a string.

Oops, formatting necesse est!

Its only now that I saw the other solutions. I didn't time it, but regex should be to much overhead. Is that correct?

sub uni{
    my ($s) = shift;
    my @a;
    for $i (0..9) {
        $j = int(substr($s,$i,1));
        if ($a[$j]) {return( 0 )};
        $a[$j] = 1;}
    return 1; }

 print uni("1902356784");
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Comment on Re^2: Determining uniqueness in a string. Download Code

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Re^3: Determining uniqueness in a string.
by gam3 (Curate) on Oct 02, 2005 at 05:25 UTC

use Inline C;

use Benchmark qw( cmpthese ) ;

my $digits = '1234567891';
my $bob = {
    u1 => sub {
        return uniq($digits);
    },
    u2 => sub {
        my @l = ();
        for my $x (split '', $digits) {
            if ($l[$x]++) {
                return 1;
            }
        }
        return 0;
    },
    u3 => sub {
        my @l = ();
        my $max = length($digits);
        for (my $i=0; $i < $max; $i++) {
            $x = substr($digits,$i,1);
            if ($l[$x]++) {
                return 1;
            }
        }
        return 0;
    },
    u4 => sub {
        return $digits !~ /(.).*\1/ ? 0:1;
    },
};
print "u1: ", $bob->{u1}->(), "\n";
print "u2: ", $bob->{u2}->(), "\n";
print "u3: ", $bob->{u3}->(), "\n";
print "u4: ", $bob->{u4}->(), "\n";
cmpthese( -5, $bob);
__END__
       Rate    u2    u3    u4    u1
u2  19725/s    --  -10%  -90%  -95%
u3  21847/s   11%    --  -89%  -94%
u4 204817/s  938%  838%    --  -44%
u1 367282/s 1762% 1581%   79%    --
__C__
int uniq(char* name) {
    int bob[10] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
    int max = strlen(name);
    int x;
    for (x = 0; x < max; x++) {
        if (bob[name[x]-'0']++) {
            return 1;
        }
    }
    return 0;
}
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