Leetcode Q7:Reverse Integer

题目7：

Reverse digits of an integer.

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

Have you thought about this?
Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!
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.

#define     MAX_INT32       2147483647
#define     MIN_INT32       (-2147483647)
/***************************************************************************************************
*\Function      MyItoa
*\Description   int -> string，反向存入string
*\Parameter     n
*\Parameter     str
*\Return        void
*\Note
*\Log           2015.07.23    Ver 1.0
*               创建函数。
***************************************************************************************************/
void MyItoa(int n, char* str)
{
int i = 0;
int j = 0;
/* 负数也是可以的，在string --> int时，会减去'\0'，仍然会得到负数 */
while (n)
{
str[i] = n % 10 + '0';
n = n / 10;
i++;
}
str[i] = '\0';
}

/***************************************************************************************************
*\Function      MyAtoi
*\Description   string -> int
*\Parameter     str
*\Return        int
*\Note
*\Log           2015.07.23    Ver 1.0
*               创建函数。
***************************************************************************************************/
long long MyAtoi(char* str)
{
long long temp = 1;
int i = 0;
long long res = 0;
int len = 0;

while (str[len] != '\0')
{
len++;
}

for (i = len - 1; i >= 0; i--)
{
res = res + (str[i] - '0') * temp;
temp *= 10;
}
return res;
}

int reverse(int x)
{
int size = 0;
long long rev = 0;
char* temp = NULL;
char x_str[100] = {0};

if (x == 0)
{
return x;
}

MyItoa(x, x_str);

rev = MyAtoi(x_str);

/* 如果计算结果溢出，直接返回0 */
if (rev > MAX_INT32 || rev < MIN_INT32)
{
return 0;
}

return rev;
}