reversing integer numbers?

[NtOrZor]

I just came back from a MS interview and I was asked to reverse integer numbers. Needless to say I completely bombed the question. You can't use toString () methods and use the reverse string function call ( I tried to do that). So how do you reverse integer numbers mathematically?! Say I have an integer 123, how do I mathematically manipulate it so it comes out to be 321?

[Menge]

try this link:

http://forums.belution.com/en/cpp/000/015/37.shtml

didn't know that either

[AndreasV]

Think of the (decimal) number 123 as 1 * 10 ^ 2 + 2 * 10 ^ 1 + 3 * 10 ^ 0, now this should map to 3 * 10 ^ 2 + 2 * 10 ^ 1 + 3 * 10 ^ 0.

In Haskell

reverseInt :: Int -> Int
reverseInt i  = let calculateInt n b
    | null n = 0
  	| otherwise = (head n) * 10  ^ (b) + calculateInt (tail n) (b+1) 
	in
	calculateInt (sepInt i) 0 where
  sepInt i
   | i == 0  = []
   | otherwise = (sepInt (i `div` 10)) ++ [(i `mod` 10)]

[NtOrZor]

I guess that's similar to what I was thinking, but it seemed like too much work and I thought there must be a more simple way to do it (like a magical formula or something). I guess it's just pure brute force then.

[anog]

I'm in the first year of college (Computer Engeneering) and that's the kind of exercises they made us solve in the Programming class.

I had no idea they asked for that kind of thing in interviews!

[NtOrZor]

  Quote

I had no idea they asked for that kind of thing in interviews!

In technical interviews they do. There are also behavioral interviews as well.

[Elagizy]

Hey... Check this >>

Imports vb = Microsoft.VisualBasic

Text1.Text = vb.StrReverse("123") -------- > 321

[MrRogers]

  Elagizy said:

Hey... Check this >>

Imports vb = Microsoft.VisualBasic

Text1.Text = vb.StrReverse("123") -------- > 321

585500504[/snapback]

  Quote

You can't use toString () methods and use the reverse string function call ( I tried to do that). So how do you reverse integer numbers mathematically?!

I never even thought of the string reverse method. That would've been slick.

Here's what I came up with.

private int Reverse(int value)
  {
  	int reverse = 0;
  	while(value > 0)
  	{
    int part = value % 10;
    value /= 10;
    reverse *= 10;
    reverse += part;
  	}
  	return reverse;
  }

[NtOrZor]

That's a great simple elegant way. Thanks for showing us!

[Ianmac45]

for those of you who want this in vb, here it is

Private Shared Function Reverse(ByVal num As Integer) As Integer
	Dim value As Integer

	While num > 0
  Dim part As Integer = num Mod 10
  num \= 10   'must use integer division instead of normal division
  value *= 10
  value += part
	End While

	Return value
End Function

the only weird part is the use of backwards slash, which denotes integer division in vbnet