[LeetCode84]Reverse Integer

Reverse digits of an integer.

Example1: x = 123, return 321

Example2: x = -123, return -321

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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:
we don't need to use special logic to handle '+' and '-', because the sign can be kept during calculating.

Idea is to keep x/10 for the original int and *10 for the result int + x%10.

Java

public int reverse(int x) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int reverseNum = 0;
		while(x != 0){
			reverseNum = reverseNum*10 + x%10;
			x /= 10;
		}
        return reverseNum;
    }

c++

int reverse(int x) {
        int reverseNum = 0;
		while(x != 0){
			reverseNum = reverseNum*10 + x%10;
			x /= 10;
		}
        return reverseNum;
    }