注意判断是否＞INT_MAX or ＜INT_MIN
时间复杂度：O(N)
C++代码：

class Solution {
public:
	int myAtoi(string str) {
		int flag = 0, num = 0;
		for (int i = 0; i < str.length(); i++)
		{
			if (!flag && str[i] == ' ')
				continue;
			else
			{
				if (!flag)
				{
					if (str[i] == '+')
						flag = 1;
					else if (str[i] == '-')
						flag = -1;
					else if ((str[i] - '0') < 0 || (str[i] - '0') > 9)
						return 0;
					else
					{
						num += str[i] - '0';
						flag = 1;
					}
				}
				else
				{
					if ((str[i] - '0') < 0 || (str[i] - '0') > 9)
					{
						break;
					}
					else
					{
						if (flag == 1 &&
							(num > INT_MAX / 10 ||
							(num == INT_MAX / 10 && (str[i] - '0') > 7)))
							return INT_MAX;
						else if (flag == -1 &&
							(flag*num < INT_MIN / 10 ||
							(flag * num == INT_MIN / 10 && (str[i] - '0') > 8)))
							return INT_MIN;
						num = num * 10 + (str[i] - '0');
					}
				}
			}
		}
		return num * flag;
	}
};