sub fizbin { return $_[0] unless ($_[0] and length($_[0]) > 1); my @string = (300, unpack("U*", $_[0]), 301); my $palstart, $palend; my ($bestlen, $beststart, $bestend) = (-1,-1,-1); for ($palmid = 1; $palmid < $#string; $palmid++) { if ($string[$palmid] == $string[$palmid+1]) { # try even-length palindrome ($palstart, $palend) = ($palmid, $palmid+1); while ($string[$palend+1] == $string[$palstart-1]) { $palend++; $palstart--; } if ($bestlen < $palend - $palstart) { ($bestlen, $bestend, $beststart) = ($palend - $palstart, $palend, $palstart); } } # try odd-length palindrome ($palstart, $palend) = ($palmid, $palmid); while ($string[$palend+1] == $string[$palstart-1]) { $palend++; $palstart--; } if ($bestlen < $palend - $palstart) { ($bestlen, $bestend, $beststart) = ($palend - $palstart, $palend, $palstart); } } pack("U*", @string[$beststart..$bestend]); }

It's also unfortunately an O(n^2) algorithm, but my initial O(n) idea turned out to be badly flawed. (Actually, I guess it's O(n*m), where "n" is the length of the input and "m" is the length of the longest palindrome - in the worst case, a string of all the same letter, it'd be O(n^2))

Note that it'll also work on unicode strings, assuming that perl knows that its argument is a unicode string.

--
@/=map{[/./g]}qw/.h_nJ Xapou cets krht ele_ r_ra/;
map{y/X_/\n /;print}map{pop@$_}@/for@/
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