166. Fraction to Recurring Decimal 未完成

Given two integers representing the numerator and denominator of a fraction, return the fraction in string format.

If the fractional part is repeating, enclose the repeating part in parentheses.

For example,

• Given numerator = 1, denominator = 2, return "0.5".
• Given numerator = 2, denominator = 1, return "2".
• Given numerator = 2, denominator = 3, return "0.(6)".

此题的思路并不难，但是程序较复杂，各种情况都容易出错

我的代码：当处理1/90时，答案为0.0(11)，应该是因子是10的倍数导致的

class Solution {public:    string fractionToDecimal(int numerator, int denominator) {        string left;        string right;        bool re=false;        if(numerator==0) return left+"0";        if(denominator==0) return left;        unordered_map<int,int> hash;        hash[numerator]=0;        int first=0,last=0;        if(numerator>=denominator)        {            left+=to_string(numerator/denominator);            if(numerator%denominator)            {                fraction(numerator%denominator,denominator,right,numerator,re,hash,last,first);            }                    }        else            {                fraction(numerator,denominator,right,numerator,re,hash,last,first);            }        string res;        if(!left.empty()) res+=left;        else res+="0";        if(!right.empty())        {            if(re) {                res.push_back('.');                for(int i=0;i<first;i++)                {                    res.push_back(right[i]);                }                res+="("+right.substr(first)+")";            }            else res+="."+right;        }        return res;            }    void fraction(int num,const int& de,string& right,const int& nu,bool& re,unordered_map<int,int>& hash,int& last,int& first){        int zero=0;        while(num<de)        {            zero++;            num*=10;        }        int temp=num/de;        int ba=zero;        while(ba-->1)        {            right+="0";        }        right+=to_string(temp);        last++;        while(zero>nu&&zero%10==0)        {            zero/=10;                    }        if(hash.find(num%de)!=hash.end()) {            re=true;            first=hash[num%de];            return ;        }        if(num%de==0) return ;        hash[num%de]=last;        fraction(num%de,de,right,nu,re,hash,last,first);    }};