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).

解答：

如果为负数，变为正数后处理，处理完后再变为负数。

处理过程：

采用队列，不断取余，将各位存入队列中，低位先入队列，高位后入队列。

从队列中先取低位，不断地*10 + 下一位， 直到队列为空。

如果结果为负数，说明溢出。

正负数判断输出。

代码：

class Solution {public:    int reverse(int x) {		if(x == 0)			return x;        int flag;		(x >0 ? flag = 1 : (flag = -1, x = -x));		int num = 0;		queue
    
      st;		while(x)		{			st.push(x % 10);			x /= 10;		}		while(!st.empty())		{			num = num*10 + st.front();			st.pop();		}		if(num < 0)		{			exit(0);		}		num = num * flag;		return num;    }};