How do I check if a number is a palindrome? - algorithm

Question

How do I check if a number is a palindrome?
Any language. Any algorithm. (except the algorithm of making the number a string and then reversing the string).

Answer 1

For any given number:
n = num;
rev = 0;
while (num > 0)
{
dig = num % 10;
rev = rev * 10 + dig;
num = num / 10;
}
If n == rev then num is a palindrome:
cout << "Number " << (n == rev ? "IS" : "IS NOT") << " a palindrome" << endl;

Answer 2

This is one of the Project Euler problems. When I solved it in Haskell I did exactly what you suggest, convert the number to a String. It's then trivial to check that the string is a pallindrome. If it performs well enough, then why bother making it more complex? Being a pallindrome is a lexical property rather than a mathematical one.

Answer 3

def ReverseNumber(n, partial=0):
if n == 0:
return partial
return ReverseNumber(n // 10, partial * 10 + n % 10)
trial = 123454321
if ReverseNumber(trial) == trial:
print("It's a Palindrome!")
Works for integers only. It's unclear from the problem statement if floating point numbers or leading zeros need to be considered.

Answer 4

Above most of the answers having a trivial problem is that the int variable possibly might overflow.
Refer to http://articles.leetcode.com/palindrome-number/
boolean isPalindrome(int x) {
if (x < 0)
return false;
int div = 1;
while (x / div >= 10) {
div *= 10;
}
while (x != 0) {
int l = x / div;
int r = x % 10;
if (l != r)
return false;
x = (x % div) / 10;
div /= 100;
}
return true;
}

Answer 5

int is_palindrome(unsigned long orig)
{
unsigned long reversed = 0, n = orig;
while (n > 0)
{
reversed = reversed * 10 + n % 10;
n /= 10;
}
return orig == reversed;
}

Answer 6

Push each individual digit onto a stack, then pop them off. If it's the same forwards and back, it's a palindrome.

Answer 7

I didn't notice any answers that solved this problem using no extra space, i.e., all solutions I saw either used a string, or another integer to reverse the number, or some other data structures.
Although languages like Java wrap around on integer overflow, this behavior is undefined in languages like C. (Try reversing 2147483647 (Integer.MAX_VALUE) in Java)
Workaround could to be to use a long or something but, stylistically, I don't quite like that approach.
Now, the concept of a palindromic number is that the number should read the same forwards and backwards. Great. Using this information, we can compare the first digit and the last digit. Trick is, for the first digit, we need the order of the number. Say, 12321. Dividing this by 10000 would get us the leading 1. The trailing 1 can be retrieved by taking the mod with 10. Now, to reduce this to 232. (12321 % 10000)/10 = (2321)/10 = 232. And now, the 10000 would need to be reduced by a factor of 2. So, now on to the Java code...
private static boolean isPalindrome(int n) {
if (n < 0)
return false;
int div = 1;
// find the divisor
while (n / div >= 10)
div *= 10;
// any number less than 10 is a palindrome
while (n != 0) {
int leading = n / div;
int trailing = n % 10;
if (leading != trailing)
return false;
// % with div gets rid of leading digit
// dividing result by 10 gets rid of trailing digit
n = (n % div) / 10;
// got rid of 2 numbers, update div accordingly
div /= 100;
}
return true;
}
Edited as per Hardik's suggestion to cover the cases where there are zeroes in the number.

Answer 8

Fastest way I know:
bool is_pal(int n) {
if (n % 10 == 0) return 0;
int r = 0;
while (r < n) {
r = 10 * r + n % 10;
n /= 10;
}
return n == r || n == r / 10;
}

Answer 9

In Python, there is a fast, iterative way.
def reverse(n):
newnum=0
while n>0:
newnum = newnum*10 + n % 10
n//=10
return newnum
def palindrome(n):
return n == reverse(n)
This also prevents memory issues with recursion (like StackOverflow error in Java)

Answer 10

Just for fun, this one also works.
a = num;
b = 0;
if (a % 10 == 0)
return a == 0;
do {
b = 10 * b + a % 10;
if (a == b)
return true;
a = a / 10;
} while (a > b);
return a == b;

Answer 11

except making the number a string and then reversing the string.
Why dismiss that solution? It's easy to implement and readable. If you were asked with no computer at hand whether 2**10-23 is a decimal palindrome, you'd surely test it by writing it out in decimal.
In Python at least, the slogan 'string operations are slower than arithmetic' is actually false. I compared Smink's arithmetical algorithm to simple string reversal int(str(i)[::-1]). There was no significant difference in speed - it happened string reversal was marginally faster.
In compiled languages (C/C++) the slogan might hold, but one risks overflow errors with large numbers.
def reverse(n):
rev = 0
while n > 0:
rev = rev * 10 + n % 10
n = n // 10
return rev
upper = 10**6
def strung():
for i in range(upper):
int(str(i)[::-1])
def arithmetic():
for i in range(upper):
reverse(i)
import timeit
print "strung", timeit.timeit("strung()", setup="from __main__ import strung", number=1)
print "arithmetic", timeit.timeit("arithmetic()", setup="from __main__ import arithmetic", number=1)
Results in seconds (lower is better):
strung 1.50960231881
arithmetic 1.69729960569

Answer 12

I answered the Euler problem using a very brute-forcy way. Naturally, there was a much smarter algorithm at display when I got to the new unlocked associated forum thread. Namely, a member who went by the handle Begoner had such a novel approach, that I decided to reimplement my solution using his algorithm. His version was in Python (using nested loops) and I reimplemented it in Clojure (using a single loop/recur).
Here for your amusement:
(defn palindrome? [n]
(let [len (count n)]
(and
(= (first n) (last n))
(or (>= 1 (count n))
(palindrome? (. n (substring 1 (dec len))))))))
(defn begoners-palindrome []
(loop [mx 0
mxI 0
mxJ 0
i 999
j 990]
(if (> i 100)
(let [product (* i j)]
(if (and (> product mx) (palindrome? (str product)))
(recur product i j
(if (> j 100) i (dec i))
(if (> j 100) (- j 11) 990))
(recur mx mxI mxJ
(if (> j 100) i (dec i))
(if (> j 100) (- j 11) 990))))
mx)))
(time (prn (begoners-palindrome)))
There were Common Lisp answers as well, but they were ungrokable to me.

Answer 13

Here is an Scheme version that constructs a function that will work against any base. It has a redundancy check: return false quickly if the number is a multiple of the base (ends in 0).
And it doesn't rebuild the entire reversed number, only half.
That's all we need.
(define make-palindrome-tester
(lambda (base)
(lambda (n)
(cond
((= 0 (modulo n base)) #f)
(else
(letrec
((Q (lambda (h t)
(cond
((< h t) #f)
((= h t) #t)
(else
(let*
((h2 (quotient h base))
(m (- h (* h2 base))))
(cond
((= h2 t) #t)
(else
(Q h2 (+ (* base t) m))))))))))
(Q n 0)))))))

Answer 14

Recursive solution in ruby, without converting the number to string.
def palindrome?(x, a=x, b=0)
return x==b if a<1
palindrome?(x, a/10, b*10 + a%10)
end
palindrome?(55655)

Answer 15

Golang version:
package main
import "fmt"
func main() {
n := 123454321
r := reverse(n)
fmt.Println(r == n)
}
func reverse(n int) int {
r := 0
for {
if n > 0 {
r = r*10 + n%10
n = n / 10
} else {
break
}
}
return r
}

Answer 16

Pop off the first and last digits and compare them until you run out. There may be a digit left, or not, but either way, if all the popped off digits match, it is a palindrome.

Answer 17

Here is one more solution in c++ using templates . This solution will work for case insensitive palindrome string comparison .
template <typename bidirection_iter>
bool palindrome(bidirection_iter first, bidirection_iter last)
{
while(first != last && first != --last)
{
if(::toupper(*first) != ::toupper(*last))
return false;
else
first++;
}
return true;
}

Answer 18

a method with a little better constant factor than #sminks method:
num=n
lastDigit=0;
rev=0;
while (num>rev) {
lastDigit=num%10;
rev=rev*10+lastDigit;
num /=2;
}
if (num==rev) print PALINDROME; exit(0);
num=num*10+lastDigit; // This line is required as a number with odd number of bits will necessary end up being smaller even if it is a palindrome
if (num==rev) print PALINDROME

Answer 19

here's a f# version:
let reverseNumber n =
let rec loop acc = function
|0 -> acc
|x -> loop (acc * 10 + x % 10) (x/10)
loop 0 n
let isPalindrome = function
| x when x = reverseNumber x -> true
| _ -> false

Answer 20

A number is palindromic if its string representation is palindromic:
def is_palindrome(s):
return all(s[i] == s[-(i + 1)] for i in range(len(s)//2))
def number_palindrome(n):
return is_palindrome(str(n))

Answer 21

def palindrome(n):
d = []
while (n > 0):
d.append(n % 10)
n //= 10
for i in range(len(d)/2):
if (d[i] != d[-(i+1)]):
return "Fail."
return "Pass."

Answer 22

To check the given number is Palindrome or not (Java Code)
class CheckPalindrome{
public static void main(String str[]){
int a=242, n=a, b=a, rev=0;
while(n>0){
a=n%10; n=n/10;rev=rev*10+a;
System.out.println(a+" "+n+" "+rev); // to see the logic
}
if(rev==b) System.out.println("Palindrome");
else System.out.println("Not Palindrome");
}
}

Answer 23

A lot of the solutions posted here reverses the integer and stores it in a variable which uses extra space which is O(n), but here is a solution with O(1) space.
def isPalindrome(num):
if num < 0:
return False
if num == 0:
return True
from math import log10
length = int(log10(num))
while length > 0:
right = num % 10
left = num / 10**length
if right != left:
return False
num %= 10**length
num /= 10
length -= 2
return True

Answer 24

I always use this python solution due to its compactness.
def isPalindrome(number):
return int(str(number)[::-1])==number

Answer 25

int reverse(int num)
{
assert(num >= 0); // for non-negative integers only.
int rev = 0;
while (num != 0)
{
rev = rev * 10 + num % 10;
num /= 10;
}
return rev;
}
This seemed to work too, but did you consider the possibility that the reversed number might overflow? If it overflows, the behavior is language specific (For Java the number wraps around on overflow, but in C/C++ its behavior is undefined). Yuck.
It turns out that comparing from the two ends is easier. First, compare the first and last digit. If they are not the same, it must not be a palindrome. If they are the same, chop off one digit from both ends and continue until you have no digits left, which you conclude that it must be a palindrome.
Now, getting and chopping the last digit is easy. However, getting and chopping the first digit in a generic way requires some thought. The solution below takes care of it.
int isIntPalindrome(int x)
{
if (x < 0)
return 0;
int div = 1;
while (x / div >= 10)
{
div *= 10;
}
while (x != 0)
{
int l = x / div;
int r = x % 10;
if (l != r)
return 0;
x = (x % div) / 10;
div /= 100;
}
return 1;
}

Answer 26

Try this:
reverse = 0;
remainder = 0;
count = 0;
while (number > reverse)
{
remainder = number % 10;
reverse = reverse * 10 + remainder;
number = number / 10;
count++;
}
Console.WriteLine(count);
if (reverse == number)
{
Console.WriteLine("Your number is a palindrome");
}
else
{
number = number * 10 + remainder;
if (reverse == number)
Console.WriteLine("your number is a palindrome");
else
Console.WriteLine("your number is not a palindrome");
}
Console.ReadLine();
}
}

Answer 27

Here is a solution usings lists as stacks in python :
def isPalindromicNum(n):
"""
is 'n' a palindromic number?
"""
ns = list(str(n))
for n in ns:
if n != ns.pop():
return False
return True
popping the stack only considers the rightmost side of the number for comparison and it fails fast to reduce checks

Answer 28

public class Numbers
{
public static void main(int givenNum)
{
int n= givenNum
int rev=0;
while(n>0)
{
//To extract the last digit
int digit=n%10;
//To store it in reverse
rev=(rev*10)+digit;
//To throw the last digit
n=n/10;
}
//To check if a number is palindrome or not
if(rev==givenNum)
{
System.out.println(givenNum+"is a palindrome ");
}
else
{
System.out.pritnln(givenNum+"is not a palindrome");
}
}
}

Answer 29

let isPalindrome (n:int) =
let l1 = n.ToString() |> List.ofSeq |> List.rev
let rec isPalindromeInt l1 l2 =
match (l1,l2) with
| (h1::rest1,h2::rest2) -> if (h1 = h2) then isPalindromeInt rest1 rest2 else false
| _ -> true
isPalindromeInt l1 (n.ToString() |> List.ofSeq)

Answer 30

checkPalindrome(int number)
{
int lsd, msd,len;
len = log10(number);
while(number)
{
msd = (number/pow(10,len)); // "most significant digit"
lsd = number%10; // "least significant digit"
if(lsd==msd)
{
number/=10; // change of LSD
number-=msd*pow(10,--len); // change of MSD, due to change of MSD
len-=1; // due to change in LSD
} else {return 1;}
}
return 0;
}