7. Reverse Integer

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

Note:

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

Thought

x will drop one digit (ones) in each iteration which divided by in 10
+-----------+
| 1 | 2 | 3 |  rev = 0 + 123 % 10 = 0 + 3 = 3, x = 123 / 10 = 12
+-----------+

          ↓ iterate
+-----------+
|   |   | 3 |
+-----------+
rev will increase one digit in each iteration which multiplied by 10, and keep the remainder

Getting the wrong answer for the first submission because I did not take the overflow into account. For example, given 1534236469 will raise the integer overflows, and we should return 0. Since I change the variable rev type from int to long to avoid reversed integer overflows. Lastly, check the valid range values of integer before return the reversed result.

Solution

public class Solution {
    public int reverse(int x) {
        long rev = 0;
        while (x != 0) {
            rev = rev * 10 + x % 10;
            x = x / 10;
        }

        if (rev < Integer.MIN_VALUE || rev > Integer.MAX_VALUE) return 0;
        return (int) rev;
    }
}