Requirements for atoi:

The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.

The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.

If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.

If no valid conversion could be performed, a zero value is returned. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.

1. 首先需要丢弃字符串前面的空格；

2. 然后可能有正负号（注意只取一个，如果有多个正负号，那么说这个字符串是无法转换的，返回0。比如测试用例里就有个“+-2”）；

3. 字符串可以包含0~9以外的字符，如果遇到非数字字符，那么只取该字符之前的部分，如“-00123a66”返回为“-123”；

4. 如果超出int的范围，返回边界值（2147483647或-2147483648）。

因为可能超出int的界限，所以result可以是double（8个字节）型的变量 C++：

int Solution::myAtoi(string str) {

    //判断空
    if(str.empty() || str.length() == 0)
        return 0;

    //去首尾空格
    str.erase(0,str.find_first_not_of(" "));
    str.erase(str.find_last_not_of(" ") + 1);

    //判断正负
    bool positive = true;
    int i = 0;
    if(str[i] == '-'){
        positive = false;
        i++;
    }else if(str[i] == '+'){
        i++;
    }

    //读数
    double result = 0;
    for( ; i < str.length(); i++ ){
        int temp = str[i] - '0';

        if(temp < 0 || temp >9 )
            break;

        if (positive){
            result = result * 10 + temp;
//            if(result > INT_MAX/10)
//                return INT_MAX;
        }else {
            result = result * 10 - temp;
//            if(result < INT_MIN/10)
//                return INT_MIN;
        }
    };

    //判断是否超界限
    if(result > INT_MAX || result < INT_MIN){
        if(positive)
            return INT_MAX;
        else
            return INT_MIN;
    }

    return (int)result;
}