leetcode——Reverse Integer 反转整数数字（AC）

Reverse digits of an integer.

Example1: x = 123, return 321

Example2: x = -123, return -321

这个题比较简单，考虑特殊情况如12000，注意检查反转后数字是否会越界溢出。代码如下：

class Solution {
public:
    int reverse(int x) {
         bool minus = false;
		short int splitNum[10];
		int i = 0, j = 0;
		unsigned int num, result = 0;
		unsigned int MAX = 2147483647;
		unsigned int MIN = 2147483648;
        if(x < 0)
		{
			num = 0 - x;
			minus = true;
		}
		else
		{
			num = x;
		}
		while(num != 0)
		{
			splitNum[i++] = num%10;
			num = num/10;
		}
		while(splitNum[j] == 0)
			j++;
		while(j < i)
		{
			if(!minus && (MAX-splitNum[j])/10 >= result)
			{
				result = result*10 + splitNum[j];
				j++;
			}
			else if(minus && (MIN-splitNum[j])/10 >= result)
			{
				result = result*10 + splitNum[j];
				j++;
			}
			else
			{
				return 0;
			}
		}
		return minus ? (0-result) : result;
    }
};

leetcode——Reverse Integer 反转整数数字（AC）