[LeetCode] Reverse Integer

Problem Statement (link):

Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
Have you thought about this?
Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!
If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.
Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?
Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).

Analysis:
There are several ways to approach this problem, we could convert the integer to string, inverse the string, then convert the string to int. Or, we could read each digit into a stack, pop out each digit in sequence and form a new integer.

Here, I simple extract each digit from the end of int x by taking the reminder. The time complexity is O(n) where n is the length of the integer.

Code:

class Solution {
public:
    int reverse(int x) {
        int sign=x<0?-1:1;
        x=abs(x);
        int rev=0;
        while (x>0) {
            rev=rev*10+x%10;
            x/=10;
            if (rev<0) return 0; // overflow
        }
        return rev*sign;
    }
};