LeetCode 7. 整数反转 Reverse Integer

Posted 2020-12-05 zsy-blog

 

在程序循环内终止更好。

class Solution {
public:
    int reverse(int x) {
        int num = 0;
        while (x != 0)
        {
            int n = x % 10;
            x /= 10;
            //-2,147,483,648 ~ 2,147,483,647
            if (num > INT_MAX/10 || (num == INT_MAX/10 && n > INT_MAX%10)) return 0;
            if (num < INT_MIN/10 || (num == INT_MIN/10 && n < INT_MIN%10)) return 0;
            num = num * 10 + n;
        }
        return num;
    }
};