How to reverse a number as an integer and not as a string?

Question

I came across a question \"How can one reverse a number as an integer and not as a string?\" Could anyone please help me to find out the answer? Reversal should reverse the

Answer 1

int sum = 0; int remainder = 0;
        int n = 123;
        for (int i = n; i > 0; i = i / 10)
        {
            remainder = i % 10;
            sum = (sum * 10) + remainder;
        }
Console.WriteLine(sum);
Console.ReadLine();

Answer 2

I looked at the following solution. But how about when n is negative?

Lets say n = -123456

Here is a tweak I added to it to handle negative numbers, with that assumption it will threat negative numbs the same way.

        int reverse = 0;
        bool isNegative = false;

        if (n < 0)
            isNegative = true;

        n = Math.Abs(n);
        while (n > 0)
        {
            int rem = n % 10;
            reverse = (reverse * 10) + rem;
            n = n / 10;
        }

        if (isNegative)
            reverse = reverse * (-1);

        return reverse;

Answer 3

    static void Main(string[] args)
    {
        Console.WriteLine("Please enter number");
        int number = Convert.ToInt32(Console.ReadLine());
        Console.WriteLine(ReverseLoop(number));
        Console.WriteLine(ReverseRecursive(number));
        Console.ReadLine();
    }

    public static int ReverseRecursive(int number)
    {
        int remainder = number % 10;
        number = number / 10;
        if (number < 1)
            return remainder;
        return ReverseRecursive(number) + remainder * Convert.ToInt32(Math.Pow(10, number.ToString().Length));
    }

    public static int ReverseLoop(int number)
    {
        int reversed = 0;
        while (number > 0)
        {
            reversed = reversed * 10 + (number % 10);
            number = number / 10;
        }
        return reversed;
    }

Answer 4

Something like this?

public int ReverseInt(int num)
{
    int result=0;
    while (num>0) 
    {
       result = result*10 + num%10;
       num /= 10;
    }
    return result;
}

As a hackish one-liner (update: used Benjamin's comment to shorten it):

num.ToString().Reverse().Aggregate(0, (b, x) => 10 * b + x - '0');

A speedier one-and-a-quarter-liner:

public static int ReverseOneLiner(int num)
{
    for (int result=0;; result = result * 10 + num % 10, num /= 10) if(num==0) return result;
    return 42;
}

It's not a one-liner because I had to include return 42;. The C# compiler wouldn't let me compile because it thought that no code path returned a value.

P.S. If you write code like this and a co-worker catches it, you deserve everything he/she does to you. Be warned!

EDIT: I wondered about how much slower the LINQ one-liner is, so I used the following benchmark code:

public static void Bench(Func<int,int> myFunc, int repeat)
{
    var R = new System.Random();
    var sw = System.Diagnostics.Stopwatch.StartNew();
    for (int i = 0; i < repeat; i++)
    {
        var ignore = myFunc(R.Next());
    }
    sw.Stop();
    Console.WriteLine("Operation took {0}ms", sw.ElapsedMilliseconds);
}

Result (10^6 random numbers in positive int32 range):

While loop version:
Operation took 279ms

Linq aggregate:
Operation took 984ms

Answer 5

int x = 9876543;
char[] arr = x.ToString().ToCharArray();
Array.Reverse(arr);
Console.WriteLine(int.Parse(new string(arr)));

Answer 6

the shortest solution:

        int reverse=0;
        int number = 21;

        while (number > 0)
        {
            reverse = number % 10;
            Console.Write(reverse);
            number = number / 10;  
        }