Note:
Assume we are dealing with an environment which could only hold integers within the 32-bit signed integer range.
重点来了：
For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
所以我们应该考虑溢出，我们用一个long的数据类型来容纳int类型的运输，以保证运算不会在long数据类型上发生溢出。
上代码：
class Solution {
public:
int reverse(int x) {
int flag = 1;
if(x < 0){
flag = -1;
x = -x;
}
long long sum = 0; //用long来得到结果
while(x){
sum = (sum * 10 + (x % 10));
x /= 10;
}
sum = sum * flag;
return (sum > INT_MAX || sum < INT_MIN)? 0 : sum;
}
};
