CareerCup

The program is pretty straightforward, i assume we are talking about positive number, for negative numbers you need more clarification from the interviewer. Also number like '2340' will return '432' - this is should be ok.

public static int reverseInteger(int number){
		int num = number;
		int temp;
		int result = 0;
		
		while(num>0){
			temp = num%10;
			num = num/10;
			result = result*10+temp;
		}
		return result;
	}

it has traps, you don't take care of overflow and negative numbers

int reverse( int a)
{
bool negate = false;
if (a < 0)
{
a = a*(-1);
negate = true;
}
num=0;
while(a>0)
{
num = num*10 + a%10;
a= a/10;
}

if(negate)
return num* (-1);
else
return num;
}

I think in a recursive way you can solve this problem.

public class InverseInteger {
    public static void main(String[] args) {
        int n = 1234;
        System.out.println(inverse(n));
        n = -12345;
        System.out.println(inverse(n));
    }

    private static String inverse(int n) {
        if (n != 0)
            return "" + (n % 10) + inverse((n < 0 ? n * -1 : n) / 10);
        return "";
    }

}

Result:
4321
-54321

int InverseInteger(const int& n){
int temp=n,result=0;
while(temp){
result=result*10+temp%10;
temp/=10;
}
return result;
}
output:
-123456
-654321

void reverse(int&);

int main()
{

int x = 12345;
cout << x << endl;
reverse(x);
cout << x << endl;

return 0;
}


void reverse(int &a){
int b = 0;
while(a>0){
b = b*10 + a%10;
a = a /10;
}
a = b;
}