The goal: Any language. The smallest function which will return whether a string is a palindrome. Here is mine in Python:
R=lambda s:all(a==b for a,b in zip(s,reversed(s))) 50 characters.
The accepted answer will be the current smallest one - this will change as smaller ones are found. Please specify the language your code is in.
7 characters in J: Not sure if this is the best way, I'm somewhat new to J :)
p=:-:|. explanation: |. reverses the input. -: compares. the operands are implicit.
p 'radar' 1 p 'moose' 0 Here's mine; it's written in a domain-specific language I invented, called 'palindrome'.
p Edit: Less flippant version (i386 asm, AT&T syntax)
xor %eax, %eax mov %esi, %edi #cld not necessary, assume DF=0 as per x86 ABI repne scasb scan: dec %edi cmpsb .byte 0x75, 6 #jnz (short) done dec %edi cmp %esi, %edi .byte 0x72, -9 #jb (short) scan inc %eax done: 16 bytes, string pointer goes in ESI, result is in EAX.
Sadly, I'm unable to get under a thousand words...
(LabVIEW. Yeah, they'll let just about any hobo post here ;)
Haskell, 15 chars:
p=ap(==)reverse More readable version, 16 chars:
p x=x==reverse x Another python version that is rather shorter (21 chars):
R=lambda s:s==s[::-1] At the risk of getting down votes, most all of these just call a command reverse of some sort that hides all the real programming logic.
I wonder what the shortest manual way to do this is in each of these languages.
With C# and LINQ operators:
public bool IsPalindrome(string s) { return s.Reverse().SequenceEqual(s); } If you consider Reverse as cheating, you can do the entire thing with a reduction:
public bool IsPalindrome(string s) { return s.Aggregate(new StringBuilder(), (sb, c) => sb.Insert(0, c), (sb) => sb.ToString() == s); } Perl (27 chars):
sub p{$_[0]eq reverse$_[0]} Ruby (24 chars):
def p(a)a==a.reverse end 73 clean, readable, chars written in java
boolean p(String s){return s.equals(""+new StringBuffer(s).reverse());} peace :)
Pointless Haskell version (15 chars, though doesn't really work unless you include Control.Arrow and Control.Monad and ignore the monomorphism restriction):
p=ap(==)reverse Lua aims more at readability than conciseness, yet does an honest 37 chars:
function p(s)return s==s:reverse()end variant, just for fun (same size):
p=function(s)return s==s:reverse''end The JavaScript version is more verbose (55 chars), because it doesn't has a string reverse function:
function p(s){return s==s.split('').reverse().join('')} (equal p (reverse p)) lisp. 18 characters.
ok, this is a special case. This would work if typed directly into a lisp interpreter and p was already defined.
otherwise, this would be necessary:
(defun g () (equal p (reverse p))) 28 characters.
I'll take it a little bit further: full c code, compile and go.
90 characters
main(int n,char**v){char*b,*e;b=e=v[1];while(*++e);for(e--;*b==*e&&b++<e--;);return b>e;} F# (a lot like the C# example)
let p s=let i=0;let l=s.Length;while(++i<l)if(s[i]!=[l-i-1]) 0; 1;; PHP:
function p($s){return $s==strrev($s);} // 38 chars or, just
$s==strrev($s); // 15 chars Isn't using the reverse function in your language kind of cheating a bit? I mean, looking at the Ruby solution give as
def p(a)a==a.reverse end you could easily rewrite that as
def p(a)a==a.r end and just say that you made an extension method in your code so that "r" called reverse. I'd like to see people post solutions that don't contain calls to other functions. Of course, the string length function should be permitted.
Ruby without reverse - 41 characters
def m(a)a==a.split('').inject{|r,l|l+r}end VB.Net - 173 Chars
Function P(ByVal S As String) As Boolean For i As Integer = 0 To S.Length - 1 If S(i) <> S(S.Length - i - 1) Then Return False End If Next Return True End Function Golfscript, 5 char
.-1%= $ echo -n abacaba | ruby golfscript.rb palindrome.gs 1 $ echo -n deadbeef | ruby golfscript.rb palindrome.gs 0 Common Lisp, short-and-cheating version (23 chars):
#L(equal !1(reverse !1)) #L is a reader macro character implemented by SHARPL-READER in the iterate package. It's basically equivalent to (lambda (!1) ...).
Common Lisp, long version using only primitives (137 including whitespace, compressible down to 108):
(defun p (s) (let ((l (1- (length s)))) (iter (for i from l downto (/ l 2)) (always (equal (elt s i) (elt s (- l i))))))) Again, it uses iterate, which is basically a cleaner version of the builtin LOOP facility, so I tend to treat it as being in the core language.
Not the shortest, and very after-the-fact, but I couldn't help giving it a try in MATLAB:
R=@(s)all(s==fliplr(s)); 24 chars.
C# Without Reverse Function 84 chars
int p(char[]s){int i=0,l=s.Length,t=1;while(++i<l)if(s[i]!=s[l-i-1])t&=0;return t;} C# Without Reverse Function 86 chars
int p(char[]s){int i=0;int l=s.Length;while(++i<l)if(s[i]!=s[l-i-1])return 0;return 1;} VBScript 41 chars
function p:p=s=strreverse(s):end function 18 character perl regex
/^(.?|(.)(?1)\2)$/ 52 characters in C, with the caveat that up to half the string will be overwritten:
p(char*s){return!*s||!(s[strlen(s)-1]-=*s)&&p(++s);}
Without library calls it's 64 characters:
p(char*s){char*e=s;while(*e)++e;return!*s||!(*--e-=*s)&&p(++s);}
Inspired by previous post, 69 characters
p(char*a){char*b=a,q=0;while(*++b);while(*a)q|=*a++!=*--b;return!q;} EDIT: Down one char:
p(char*a){char*b=a,q=0;while(*++b);while(*a)q|=*a++%*--b;return!q;} EDIT2: 65 chars:
p(char*a){char*b=a;while(*b)b++;while(*a&&*a++==*--b);return!*a;} Haskell, 28 chars, needs Control.Arrow imported.
p=uncurry(==).(id&&&reverse) Straightforward implementation in C using standard library functions, inspired by the strlen in the other C answer.
Number of characters: 57
p(char*s){char*r=strdup(s);strrev(r);return strcmp(r,s);} Confession: I'm being the bad guy by not freeing r here. My current attempt at being good:
p(char*s){char*r=strdup(s);s[0]=strcmp(strrev(r),s);free(r);return s[0];} brings it to 73 characters; I'm thinking of any ways to do it shorter.
Clojure using 37 characters:
user=> (defn p[s](=(seq s)(reverse(seq s)))) #'user/p user=> (p "radar") true user=> (p "moose") false 24 characters in Perl.
sub p{$_[0]eq+reverse@_} Groovy 17B:
p={it==it[-1..0]}
Downside is that it doesn't work with emptry string.
On second thought, throwing exception for empty string is reasonable since you can't tell if nothing is palindrome or not.
Without using any library functions (because you should really add in the #include cost as well), here's a C++ version in 96:
int p(char*a,char*b=0,char*c=0){return c?b<a||p(a+1,--b,c)&&*a==*b:b&&*b?p(a,b+1):p(a,b?b:a,b);} My attempt in C (70 chars):
P(char*s){char*e=s+strlen(s)-1;while(s<e&&*s==*e)s++,e--;return s>=e;} [Edit] Now actually working
[Edit 2] Reduced from 74 to 70 by using default int return
In response to some of the comments: I'm not sure if that preprocessor abuse counts - you could just define the whole thing on the command line and make the function one character.
来源：https://stackoverflow.com/questions/228518/palindrome-golf