Given a 32-bit signed integer, reverse digits of an integer.

Example 1:
Input: 123
Output: 321

Example 2:
Input: -123
Output: -321

Example 3:
Input: 120
Output: 21

Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

难度：easy

题目：
给定32位带符号整数，反转该整数。

注意：
假定处理的环境只能存储32位带符号整数范轩为[负的2的32次方， 2的32次方－1]。 这个问题的目的在于溢出的整数返回0.

思路：两次反转，如果不变则没有溢出，如果不同分为两种情况一种是变之前尾数有0的，另一种则是溢出的。

Runtime: 16 ms, faster than 97.34% of Java online submissions for Reverse Integer.

public class Solution {
    public int reverse(int x) {
        int result = justReverse(x);
        int reverseResult = justReverse(result);
        // 120 -> 21 -> 12, so x % reverseResult == 0 is also not overflow
        return reverseResult == x || x % reverseResult == 0 ? result : 0;
        
    }
    
    private int justReverse(int x) {
        int result = 0;
        while(x != 0) {
            int m = x % 10;
            x = x / 10;
            
            result = result * 10 + m;
        } 
        
        return result;
    }
}