leetcode——string to integer

class Solution
{
public:
    int myAtoi(string str)
    {
        if (str.empty())
        {
            return 0;
        }
        int result = 0;
        int current_position = 0;   //current position 永遠指向下一個要讀取的字符
        bool sig_set = false;
        bool number_set = false;
        bool is_negative = false;

        //stage 1
        while (sig_set == false && number_set == false && current_position < str.size())
        {
            if (str[current_position] == ' ')
            {
                current_position++;
            }
            else if (str[current_position] == '+' || str[current_position] == '-')
            {
                sig_set = true;
                if (str[current_position] == '-')
                {
                    is_negative = true;
                }
                current_position++;
            }
            else if (str[current_position] >= '0' && str[current_position] <= '9')
            {
                number_set = true;
                result = result * 10 + (str[current_position] - '0');   //should not be overflowed here
                current_position++;
            }
            else
            {
                return result;
            }
        }
        //stage 2 這個階段有可能不執行
        while (number_set == false && current_position < str.size())
        {
            if (str[current_position] == ' ')
            {
                current_position++;
            }
            else if (str[current_position] >= '0' && str[current_position] <= '9')
            {
                number_set = true;
                result = result * 10 + (str[current_position] - '0');   //should not be overflowed here
                current_position++;
            }
            else
            {
                return result;
            }
        }

        while (current_position < str.size())
        {
            if (str[current_position] >= '0' && str[current_position] <= '9')
            {
                number_set = true;
                int a = str[current_position] - '0';
                if ((INT_MAX - a) / 10 < result)
                {
                    if (is_negative)
                    {
                        return INT_MIN;
                    }else{
                        return INT_MAX;
                    }
                }
                result = result * 10 + str[current_position] - '0';
                current_position++;
            }
            else
            {
                break;
            }
        }

        if (is_negative)
        {
            return -result;
        }else{
            return result;
        }
    }
};