Problem: Write a Java program that reverses a given integer number.

This challenge got me to start slowing down when I want to think well through a problem. That’s always a good thing!

Here is my solution:

public class Main {
    public static void main(String[] args) {
        int num = 1234;
        int reversed = 0;

        while (num != 0) {
            int digit = num % 10;
            reversed = reversed * 10 + digit;
            num /= 10;
        }
        System.out.println("Reversed: " + reversed);
    }


}

Output: Reversed: 4321 Reverse Number

Explanation

The above code looks pretty straightforward, but let’s break it down in simple terms like I always like it!

You have a number: 1234 You want to turn it into:4321 But you’re not allowed to convert it to a string. You must do it mathematically.

Think of the number like a stack of digits

Number = 1234

  • Last digit = 4
  • Then 3
  • Then 2
  • Then 1

We remove digits from the end one by one.

  1. Extract the Last Digit: Use num % 10 to get the last digit.
  2. Build the Reversed Number: Multiply the current reversed number by 10 and add the extracted digit. Perhaps you are wondering why we are multiplying by 10? Continue reading! It is explained below this solution.
  3. Remove the Last Digit: Use integer division num /= 10 to drop the last digit from the original number.
  4. Repeat: Continue until there are no digits left (num becomes 0). Here is where we use the while loop.

Why Multiply by 10?

When building the reversed number, multiplying by 10 shifts the digits to the left, making space for the new digit at the end.

For example:

  • Start with reversed = 0
  • Add digit 4: reversed = 0 * 10 + 4 = 4
  • Next digit 3: reversed = 4 * 10 + 3 = 43
  • Next digit 2: reversed = 43 * 10 + 2 = 432
  • Last digit 1: reversed = 432 * 10 + 1 = 4321
  • Final Output: 4321

In case we did it this way it would be wrong:

  • Start with reversed = 0
  • Add digit 4: reversed = 0 + 4 = 4
  • Next digit 3: reversed = 4 + 3 = 7
  • Next digit 2: reversed = 7 + 2 = 9
  • Last digit 1: reversed = 9 + 1 = 10
  • Final Output: 10 (which is incorrect)

General Rules for Reversing a Number

Every time you want to add a new digit at the end of a number, you must multiply the old number by 10.

Old Add Digit Multiply by 10 Final
4 3 4*10 = 40 43
43 2 43*10 = 430 432
432 1 432*10 = 4320 4321

This is exactly what the code is doing.

This code can be found in my github.

Happy hacking!