String to Integer (atoi)

標籤：

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.

//沒有時間做最佳化，不過個人估計測試集中缺少例如1.2e、.123e4、123..1e3這些數位判斷

class Solution {public:    int myAtoi(string str) {                //判斷字串是否未定義或者為空白        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) //對字串初始部分進行處理            {                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; //初始部分處理完畢標記sign            if(isdigit(str[i]))            {                if(!x) //無小數部分                {                    num=num*10+str[i]-‘0‘;                }else  //含小數部分                {                    num+=(str[i]-‘0‘)/pow(10,x);                    x++;                }            }else if(‘.‘==str[i])            {                if(x) return _sign*num; //遇到小數點重複情況                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; //跳出字串迴圈            }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;        }    }};

此處學習到str[i]是char，傳遞進入函數的時候需注意。

""是字串,‘‘是字元，亦需要注意；

最後num=str[i]-‘0‘，很巧妙啊，isdigit亦可以用此代替。

String to Integer (atoi)