【LeetCode 7： Reverse Integer】

Description:

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

Have you thought about this?
Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!
If the integer’s last digit is 0, what should the output be? ie, cases such as 10, 100.
Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?
For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

参考代码：

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

代码已测试通过

思考与小结：

这道题不要尝试将输入的数先转换位整数再进行运算，因为可能负数在转换为整数的时候就可能发生了溢出，但实际上应得到的结果不应该是溢出是的结果0；因此直接对原数进行操作即可；