Reversing a Number using bitwise shift

Question

I am trying to find a way to reverse a number without

1. Converting it to a string to find the length
2. Reversing the string and parsing

Answer 1

How about:

int revnum = 0;
while (num != 0) {
  revnum = revnum * 10 + (num % 10);
  num /= 10;
}
return revnum;

The code expects a non-negative input.

This may or may not matter to you, but it's worth noting that getReverse(getReverse(x)) does not necessarily equal x as it won't preserve trailing zeroes.

Answer 2

How about this? It handles negative numbers as well.

public int getReverse(int num){
   int rst=0;
   int sign;
   sign=num>0?1:-1;

   num*=sign;
   while(num>0){
      int lastNum = num%10;
      rst=rst*10+lastNum
      num=num/10;
   }
   return rst*sign;
}