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;
}
};