【leetcode】Fraction to Recurring Decimal

Fraction to Recurring Decimal

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)".

 

    0.16  
6 ) 1.00
    0 
    1 0       <-- Remainder=1, mark 1 as seen at position=0.
    - 6 
      40      <-- Remainder=4, mark 4 as seen at position=1.
    - 36 
       4      <-- Remainder=4 was seen before at position=1, so the fractional part which is 16 starts repeating at position=1 => 1(6).

如果发现余数曾经出现过，则说明开始循环了，利用一个hash表记录余数出现的位置即可

注意INT_MIN转换成正数会溢出，故需要先把数字先转为long long int

注意余数也要去long long int，因为-1,2147483648这种情况下，余数在计算乘以10的过程中会溢出

 

class Solution {
public:
    string fractionToDecimal(int numerator, int denominator) {
       
       
        string ans="";
        if(numerator==0) return "0";
       
        long long int n=numerator;
        long long int d=denominator;
        n=abs(n);
        d=abs(d);
       
        if(numerator<0^denominator<0) ans.push_back('-');
       
        map<long long int,int> hash;
       
        ans+=to_string(n/d);
        long long int rem=n%d;
        if(rem!=0) ans.push_back('.');
       
        while(rem!=0)
        {
            if(hash.find(rem)!=hash.end())
            {
                ans.insert(hash[rem],"(");
                ans.push_back(')');
                break;
            }
           
            hash[rem]=ans.length();
           
            ans+=to_string(rem*10/d);
            rem=(rem*10)%d;
        }
       
        return ans;
    }
};