# 166. Fraction to Recurring Decimal

## 刷题内容

• https://leetcode.com/problems/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.

Example 1:

Input: numerator = 1, denominator = 2
Output: "0.5"
Example 2:

Input: numerator = 2, denominator = 1
Output: "2"
Example 3:

Input: numerator = 2, denominator = 3
Output: "0.(6)"

˼· **- ʱ�临�Ӷ�: O(nlgn)*- �ռ临�Ӷ�: O(n)***

class Solution {
public:
void join_str(string& str,long long num,int t)
{
if(!num)
{
if(t < 0)
str.push_back('-');
str.push_back('0');
return;
}
while(num)
{
str.push_back(num % 10 + '0');
num /= 10;
}
if(t < 0)
str.push_back('-');
reverse(str.begin(),str.end());
}
string fractionToDecimal(int numerator, int denominator) {
map<long long,int> m;
if(!numerator)
return "0";
int t = 1;
long long n1 = numerator,d1 = denominator;
if(numerator < 0)
{
t *= -1;
n1 *= -1;
}
if(denominator < 0)
{
t *= -1;
d1 *= -1;
}
long long num = n1 / d1;
string ans;
join_str(ans,num,t);
long long quotient = n1 % d1;
if(quotient)
ans.push_back('.');
string temp;
int count1 = 0;
while(quotient)
{
long long new_num = quotient * 10;
if(m[new_num])
{
temp.push_back(')');
count1 = m[new_num];
break;
}
m[new_num] = temp.length() + 1;
temp.push_back(new_num / d1 + '0');
quotient = new_num % d1;
}
if(count1)
{
ans.append(temp.begin(),temp.begin() + count1 - 1);
ans.push_back('(');
ans.append(temp.begin() + count1 - 1,temp.end());
}
else
ans.append(temp.begin(),temp.end());
return ans;
}
};