Reverse Integer(leetcode)

题目：整数反转

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

题目很简单，但是有些问题要注意：

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?

For
the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

以下代码是自己写的通过了leetcode的测试，思路从个位数开始依次取得每个数便入队列，然后再从队列中取出合并成整数。这部分代码现在看来其实挺冗余的。队列完全不需要。其中INT_MAX，
INT_MIN是C++编译器预定义的宏量。不知道有这两个宏存在程序员就可能跟我开始一样自定义，但VS编译器，总报错。

class Solution {
public:
int reverse(int x) {
queue<int> temp;
long long ans = 0, n;

n = abs(x);
while (n % 10 || n/10) //n/10, 为了防止x为10的整数倍也能进入循环
{
temp.push(n%10);
n /= 10;
}

n = 0;
while (!temp.empty())
{
n = temp.front();
temp.pop();
ans = ans * 10 + n;
}

if (x < 0) //符号处理
ans = -ans;
if (ans > INT_MAX || ans < INT_MIN)
ans = 0;
return (int)ans;
}
};

下面看大神们的最简洁的代码（作者不知）：

class Solution {
public:
int reverse(int x) {
long long res = 0;
while(x) {
res = res*10 + x%10;
x /= 10;
}
return (res<INT_MIN || res>INT_MAX) ? 0 : res;
}
};