Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
解题：设定一个变量sum存放反转后的答案，每次取输入x的最后一位n，并用sum = sum*10+n更新sum。
例如题设的数字123，初始sum为0，过程如下：
取末尾的3，sum=3，x=12
取末尾的2，sum=32，x=1
取末尾的1，sum=321，x=0
特别注意x一开始就是0的处理，直接返回0就可以了。
代码：
1 class Solution { 2 public: 3 int reverse(int x) { 4 int sum = 0; 5 if(x == 0) 6 return 0; 7 while(x != 0){ 8 int temp = x%10; 9 sum = sum*10+temp; 10 x /= 10; 11 } 12 return sum; 13 } 14 };
题目还给出了一些思考：
1.If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.
上述的方法避免了这个问题，10和100的输出都是1
2.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).