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/**

* Reverse digits of an integer.

* <p>

* Example1: x = 123, return 321

* Example2: x = -123, return -321

* <p>

* 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!

* <p>

* 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?

* For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

* <p>

* Note:

* The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer

* overflows.

* <p>

* Created by drfish on 25/06/2017.

*/

public class _007ReverseInteger {

public int reverse(int x) {

int result = 0;

while (x != 0) {

int tail = x % 10;

int temp = result * 10 + tail;

if ((temp - tail) / 10 != result) {

return 0;

}

result = temp;

x /= 10;

}

return result;

}

public static void main(String[] args) {

_007ReverseInteger solution = new _007ReverseInteger();

assert 321 == solution.reverse(123);

assert -321 == solution.reverse(-123);

assert 0 == solution.reverse(2147483647);

}

}