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.

while (x != 0) {
            res = res * 10 + x % 10;
            x /= 10;
        }

是翻轉整數的方法，同樣適用於負數，不受影響，但是要考慮到過程中res越界的問題，所以加入if (Math.abs(res) > Integer.MAX_VALUE/10) return 0;提前考慮是否大於max的10分之1 

while (x != 0) {
            if (Math.abs(res) > Integer.MAX_VALUE/10) return 0;
            res = res * 10 + x % 10;
            x /= 10;
        }
 

也考慮一下這種寫法，但是如果題目啊直接要求就是long long 型別的話就處理不了了 

class Solution {
public:
    int reverse(int x) {
        long long res = 0;
        while (x != 0) {
            res = 10 * res + x % 10;
            x /= 10;
        }
        return (res > INT_MAX || res < INT_MIN) ? 0 : res;
    }
};