LeetCode OJ----Reverse Integer

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.

Update (2014-11-10):
Test cases had been added to test the overflow behavior.

这道简单的题目暗藏玄机！

我提交了数次才AC，都是在溢出的问题上，我查了一下网上答案，全都没考虑到溢出，根本不可能AC！

贴一下我最后的代码：

class Solution {public:    int reverse(int x) {                const int Int_MAX = 0x7fffffff;          const int Int_MIN = 0x80000000;          long long ret = 0;                 while(x != 0)        {            ret = ret * 10 + x % 10;            if (ret > Int_MAX || ret < Int_MIN)   //关键：每一步都要检查溢出            {                return 0;            }            x = x / 10;        }        return ret;    }};

这里我用比较大小来检查溢出，其实还有一种方法，就是先取绝对值，再每一步检查ret是否小于0，就可以了。