Reverse a Number

Easy Numbers

Reverse the digits of an integer.

Sample input

12345

Sample output

54321

Solution

n = 12345
rev = 0
while n > 0:
    rev = rev * 10 + n % 10
    n //= 10
print(rev)

let n = 12345, rev = 0;
while (n > 0) {
  rev = rev * 10 + (n % 10);
  n = Math.floor(n / 10);
}
console.log(rev);

public class Main {
    public static void main(String[] args) {
        int n = 12345, rev = 0;
        while (n > 0) {
            rev = rev * 10 + n % 10;
            n /= 10;
        }
        System.out.println(rev);
    }
}

fun main() {
    var n = 12345
    var rev = 0
    while (n > 0) {
        rev = rev * 10 + n % 10
        n /= 10
    }
    println(rev)
}

var n = 12345
var rev = 0
while n > 0 {
    rev = rev * 10 + n % 10
    n /= 10
}
print(rev)

void main() {
  int n = 12345, rev = 0;
  while (n > 0) {
    rev = rev * 10 + n % 10;
    n ~/= 10;
  }
  print(rev);
}

#include <iostream>
using namespace std;

int main() {
    int n = 12345, rev = 0;
    while (n > 0) {
        rev = rev * 10 + n % 10;
        n /= 10;
    }
    cout << rev << endl;
    return 0;
}

#include <stdio.h>

int main() {
    int n = 12345, rev = 0;
    while (n > 0) {
        rev = rev * 10 + n % 10;
        n /= 10;
    }
    printf("%d\n", rev);
    return 0;
}