leetcode 虐我篇之（八）Reverse Integer

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Reverse Integer 这道题的AC率也很高，在打击到没信心的时候可以来做做。先来看看题目：

Reverse digits of an integer.Example1: x = 123, return 321Example2: x = -123, return -321click to show spoilers.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?Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).

其实这道题目讲得也很清楚了。就是将一个整数翻转过来，一开始就想到用整除和取余来做。如果不考虑整型溢出，那么这样子做就已经可以Accepted了。代码如下

int reverse(int x) {int result = 0;if (x < 10 && x > (-10)){return x;}while(x){result = result*10 + x%10;x /= 10;}return result;}

题目最后面抛出了一个问题，如果反转的整数溢出怎么办？建议是抛出异常，或者是传递另外一个参数。那就是说加一个引用类型的形参，用来标志是否溢出。但是关键的问题时如何判断还有在哪里判断结果已经溢出了。不知道谁有什么好方法呢？

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