题目

Given a 32-bit signed integer, reverse digits of an integer.
给定32位有符号整数，翻转数字
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: [−2^31,  2^31 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
注意：假设我们正在处理一个只能在32位有符号整数范围内存储整数的环境：[ - 2^31,2^31 - 1]。 出于此问题的目的，假设当反向整数溢出时，函数返回0。

代码

注意范围

class Solution {
    public int reverse(int x) {
        int rev = 0;
        while (x != 0) {
            int pop = x % 10;
            x /= 10;
            if (rev > Integer.MAX_VALUE/10 || (rev == Integer.MAX_VALUE / 10 && pop > 7)) return 0;
            if (rev < Integer.MIN_VALUE/10 || (rev == Integer.MIN_VALUE / 10 && pop < -8)) return 0;
            rev = rev * 10 + pop;
        }
        return rev;
    }
}