C# LeetCode 9: Palindrome Number - (Easy)

Problem Description:

Given an integer x, return true if x is a palindrome, and false otherwise

Code:

public class Solution
{
    public bool IsPalindrome(int x)
    {
        // we know it can't be a palindrome
        if (x < 0) return false;

        var oNum = x;
        var rNum = 0;

        while (x > 0)
        {
            var digit = x % 10;
            rNum = rNum * 10 + digit;
            x = (int)Math.Floor((decimal)(x / 10));
        }

        return rNum == oNum;
    }
}

Approach

The approach is to reverse the integer and compare it to the original value.

If the reversed number equals the original number, it is a palindrome; otherwise, it is not. This avoids extra memory for strings and checks each digit only once.

Step Through

var digit = x % 10;
rNum = rNum * 10 + digit;
x = (int)Math.Floor((decimal)(x / 10));

% 10 gives the remainder when dividing by 10, which is the last digit of the number.

Example: 123 % 10 = 3

then

rNum = rNum * 10 + digit;

• Multiply rNum by 10 to shift its digits left.
• Add the extracted digit to append it to the end.

Example:
rNum = 12;
digit = 3;
rNum * 10 + digit = 123

x = (int)Math.Floor((decimal)(x / 10));

Divide by 10 to drop the last digit.

Floor ensures it truncates toward zero (though integer division in C# already does this).

Example: 123 / 10 = 12

Iteration	Current x	digit = x % 10	rNum = rNum * 10 + digit	New x = x / 10
1	1234	4	0*10 + 4 = 4	123
2	123	3	4*10 + 3 = 43	12
3	12	2	43*10 + 2 = 432	1
4	1	1	432*10 + 1 = 4321	0

And there you have a simple solution, of reversing the number, and then comparing.

Another option is the two‑pointer approach, where you convert the number to a string and use a left and right pointer to compare digits from the outside in. While easier to visualise, it requires extra space for the string conversion.

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