Leetcode 0007: Reverse Integer

题目描述：

Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [-2^31, 2^31 - 1], then return 0.
Assume the environment does not allow you to store 64-bit integers (signed or unsigned).

Example 1:

Input: x = 123
Output: 321

Example 2:

Input: x = -123
Output: -321

Example 3:

Input: x = 120
Output: 21

Example 4:

Input: x = 0
Output: 0

Constraints:

-2^31 <= x <= 2^31 -1

Time complexity: O(logn)
只使用int 记录：

1. 依次从右往左计算出每位数字，然后逆序累加在一个整数中
2. 在每次累加操作之前，需要判断累加后是否会越出int范围。当res > 0时，若 res * 10 + x % 10 > INT_MAX，等价于res > (INT_MAX - x % 10) / 10，若成立，则超出范围（注意，这里x是正数）
   当res < 0时，若 res * 10 + x % 10 < INT_MIN，等价于res < (INT_MIN - x % 10) / 10，若成立，则超出范围（注意，这里x是负数）

class Solution {
    
    
    public int reverse(int x) {
    
    
        int res = 0;
        while(x != 0){
    
    
            if(res > 0 && res > (Integer.MAX_VALUE - x % 10)/10) return 0;
            if(res < 0 && res < (Integer.MIN_VALUE - x % 10)/10) return 0;
            res = res * 10 + x % 10;
            x /= 10;
        }
        return res;
    }
}