String to Integer (atoi)

Implement atoi to convert a string to an integer.

Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.

Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button  to reset your code definition.

spoilers alert... click to show requirements for atoi.

//没有时间做优化，不过个人估计测试集中缺少例如1.2e、.123e4、123..1e3这些数字的判断

class Solution {
public:
    int myAtoi(string str) {
        
        //判断字符串是否未定义或者为空
        if(""==str) return 0;
        
        bool sign=false,esign=false; 
        int i=0,j=0,_sign=1,x=0,e=0;
        double num=0.0,tmp_num;
        
        for(i=0;i<str.length();i++)
        {
            if(!sign) //对字符串初始部分进行处理
            {
                if(' '==str[i]) continue;
                else if('+'==str[i])
                {
                    sign=true; continue;
                }else if('-'==str[i])
                {
                    sign=true; _sign=-1; continue;
                }else if('e'==str[i]||'E'==str[i])
                {
                    return 0;
                }
            }
            
            sign=true; //初始部分处理完毕标记sign
            if(isdigit(str[i]))
            {
                if(!x) //无小数部分
                {
                    num=num*10+str[i]-'0';
                }else  //含小数部分
                {
                    num+=(str[i]-'0')/pow(10,x);
                    x++;
                }
            }else if('.'==str[i])
            {
                if(x) return _sign*num; //遇到小数点重复情况
                else x++;
            }else if('e'==str[i]||'E'==str[i])
            {
                for(j=i+1;j<str.length();j++)
                {
                    if(isdigit(str[j]))
                    {
                        e=e*10+str[j]-'0';
                    }else
                    {
                        break;
                    }
                }
                break; //跳出字符串循环
            }else
            {
                num*=_sign; tmp_num=num*pow(10,e);
                if(tmp_num>INT_MAX||tmp_num<INT_MIN)
                {
                    if(num>INT_MAX) return INT_MAX;
                    else if(num<INT_MIN) return INT_MIN;
                    else return (int)num;
                }else
                {
                    return (int)tmp_num;
                }
            }
        }
        
        num*=_sign; tmp_num=num*pow(10,e);
        if(tmp_num>INT_MAX||tmp_num<INT_MIN)
        {
            if(num>INT_MAX) return INT_MAX;
            else if(num<INT_MIN) return INT_MIN;
            else return (int)num;
        }else
        {
            return (int)tmp_num;
        }
    }
};

此处学习到str[i]是char，传递进入函数的时候需注意。

""是字符串,''是字符，亦需要注意；

最后num=str[i]-'0'，很巧妙啊，isdigit亦可以用此代替。