CareerCup

public static int intReverse(int n, int sum) {
		if (n == 0)
			return sum;
		else
			return intReverse(n / 10, sum * 10 + n % 10);
	}
public static void main(String[] args) {
		System.out.println(intReverse(12345, 0));
           }
}

Idea:

let n = 123456

Note that:
- There are ceil(log10 n) = 6 digits in '123456'
- floor(n / pow(10, ceil(log10 n) - 1)) = 1 (i.e. 123456 / 10000 = 1.23456 -> 1)
- n % 10 (i.e. 123456 % 10 = 6)

Putting this all together (Python):

int(( math.pow(10, 6-1) * (n % 10) + n % math.pow(10, 6-1) ) - (n % 10) + math.floor(n / math.pow(10, 6-1) )) = 623451

Rinse and repeat for the rest of the digits, and you should be able to flip it in place.

(I sure hope there's a better way than this!)

hmm without using temp it would definitely have to be recursive way
anyway here the solution with temp variables (in python)
>>> sum = 0
>>> num = 123
>>> while num:
elem = num % 10
sum = sum*10 + elem
num = num // 10

unfortunately even with recursion i had to use a extra variable .. wonder if that s allowed

>>> def reverse(num,sum):
if num == 0 :
return sum
return reverse(num//10,sum*10+num%10)

>>> reverse(123,0)
>>> 321

public class IntegerReverse
{
public static void main(String args[])
{
int n = 123456;
rev(n);
}
static void rev(int n)
{
while( n != 0)
{
System.out.print(n%10);
n = n/10;
}
}
}

public void test() {
		int n1 = 12332;
		int n2 = 22333;
		
		
		n1 ^= n2;
		n2 ^= n1;
		n1 ^= n2;
	}

if int a=10,b=20;

b=a+b;
a=b-a;//a become 20
b=b-a;//b become 10