潜在问题：（1）随着求和可能精度会溢出int 范围，需要使用long 来辅助判断是否溢出，此时返回 0
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
（2）去前缀0
eg: reverse(1534236469); 会丢精度，如果不校验 所以WA了一次
int reverse(int x) {
long sumLong = 0;
int sum = 0;
int num = 0;
while (x!= 0) { //支持正负数
num = x % 10; //末尾数字
sum = sum * 10;//进位
sum += num;
x = x / 10;
//校验精度
sumLong = sumLong * 10;
sumLong += num;
if (sumLong != sum) {
sum = 0;
break;
}
}
return sum;
}