Because the golf isn't complete without an insanely short regex solution: 
sub japhy {  # 74
our$P="";pop=~m{(.+).?(??{reverse$1})
(?{length$P<length$&and$P=$&})^}xs;$P
}

[download]
Update: inspired by ccn's array approach: 
sub japhy {  # 68
our@P="";pop=~m{(.+).?(??{reverse$
1})(?{$P[length$&]=$&})^}xs;$P[-1]
}

[download]
_____________________________________________________
Jeff japhy Pinyan, P.L., P.M., P.O.D, X.S.: Perl, regex, and perl hacker
How can we ever be the sold short or the cheated, we who for every service have long ago been overpaid? ~~ Meister Eckhart

In reply to Re: Finding longest palindrome from a string by japhy
in thread Finding longest palindrome from a string by BUU

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