LeetCode 7. Reverse Integer

problem

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: [−2^31, 2^31 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

题目大意：

给你一个数字，在int可以表示范围内的数，将数逆置，不在范围内的数返回0，逆置之后不在范围内的数，也输出0；

class Solution {
public:
    int reverse(int x) {
        int rev = 0;
        while(x != 0){
            int y = x%10;
            x /= 10;
            if(rev > INT_MAX/10 || (rev == INT_MAX/10 && y > 7)) return 0;
            if(rev < INT_MIN/10 || (rev == INT_MIN/10 && y < -8)) return 0;
            rev = rev*10 + y;
        }
        return rev;
    }
};

边进行每一位的取值，边判断是否溢出，注意：INT_MAX 和 INT_MIN 是C++中limits.h头文件中所包含的值，表示的是带符号的int型变量的最大和最小可表示的范围