题目：

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output:  321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note:
Assume we are dealing with an environment which could only hold integers within the 32-bit signed integer range. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

这个题目是一个easy的题目，但是细节要注意，第一点要注意记得处理负数的情况，

第二点就是记得看Note,Assume we are dealing with an environment which could only hold integers within the 32-bit signed integer range. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

就是超过int的范围了就需要返回0。这一点非常重要，因为你要判断的是可能反转后会溢出，那么需要注意的就是

需要用一个long long来存储答案ans。判断使用的语句就是：

if(ans>INT_MAX||ans<INT_MIN)
            return 0;

啦啦啦！

代码如下：

 class Solution {

 public:

     int reverse(int x) {

         long long ans=;

         bool isLittle=false;

         if(x<)

         {

             x=-x;

             isLittle=true;

         }

         while(x>)

         {

             int num=x%;

             ans=ans*+num;

             x=x/;

         }

         if(ans>INT_MAX||ans<INT_MIN)

             return ;

         if(isLittle)

             return -ans;

         else return ans;

     }

 };