Palindrome Number

Easy Numbers

Check whether a number reads the same forwards and backwards.

Sample input

121

Sample output

true

Solution

n = 121
original = n
rev = 0
while n > 0:
    rev = rev * 10 + n % 10
    n //= 10
print(rev == original)

let n = 121;
const original = n;
let rev = 0;
while (n > 0) {
  rev = rev * 10 + (n % 10);
  n = Math.floor(n / 10);
}
console.log(rev === original);

public class Main {
    public static void main(String[] args) {
        int n = 121, original = n, rev = 0;
        while (n > 0) {
            rev = rev * 10 + n % 10;
            n /= 10;
        }
        System.out.println(rev == original);
    }
}

fun main() {
    var n = 121
    val original = n
    var rev = 0
    while (n > 0) {
        rev = rev * 10 + n % 10
        n /= 10
    }
    println(rev == original)
}

var n = 121
let original = n
var rev = 0
while n > 0 {
    rev = rev * 10 + n % 10
    n /= 10
}
print(rev == original)

void main() {
  int n = 121;
  int original = n;
  int rev = 0;
  while (n > 0) {
    rev = rev * 10 + n % 10;
    n ~/= 10;
  }
  print(rev == original);
}

#include <iostream>
using namespace std;

int main() {
    int n = 121, original = n, rev = 0;
    while (n > 0) {
        rev = rev * 10 + n % 10;
        n /= 10;
    }
    cout << (rev == original ? "true" : "false") << endl;
    return 0;
}

#include <stdio.h>

int main() {
    int n = 121, original = n, rev = 0;
    while (n > 0) {
        rev = rev * 10 + n % 10;
        n /= 10;
    }
    printf("%s\n", rev == original ? "true" : "false");
    return 0;
}