eetcode 之String to Integer (atoi)(28)

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;
      }
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原文地址:https://www.cnblogs.com/573177885qq/p/5519567.html