字符串转为数字，细节题。要考虑空格、正负号，当转化的数字超过最大或最小是怎么办。

int atoi(char *str)      {          int len = strlen(str);          int sign = 1;          int num = 0;          int i = 0;          while (str[i] == ' '&& i < len)i++;          if (str[i] == '+')i++;          else if (str[i] == '-')          {              sign = -1;              i++;          }          for (; i < len; i++)          {              if (str[i]<'0'&&str[i]>'9')break;              if (num>INT_MAX / 10 || (num==INT_MAX / 10 && (str[i] - '0')>INT_MAX % 10))              {                  return sign == -1 ? INT_MIN : INT_MAX;              }              num = num * 10 + str[i] - '0';          }          return num*sign;      }