166. Fraction to Recurring Decimal 未完成

最新推荐文章于 2020-10-15 11:22:00 发布

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本文介绍了一个将任意两个整数作为分数的分子和分母转换成字符串格式的方法，特别关注了循环小数部分的处理。

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);
    }
};