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.
If multiple answers are possible, return any of them.
It is guaranteed that the length of the answer string is less than 104
for all the given inputs.
Example 1:
Input: numerator = 1, denominator = 2 Output: "0.5"
Example 2:
Input: numerator = 2, denominator = 1 Output: "2"
Example 3:
Input: numerator = 4, denominator = 333 Output: "0.(012)"
Constraints:
-231 <= numerator, denominator <= 231 - 1
denominator != 0
Fraction to Recurring Decimal Solutions
✅Time: O(n)
✅Space: O(1)
C++
class Solution {
public:
string fractionToDecimal(int numerator, int denominator) {
if (numerator == 0)
return "0";
string ans;
if (numerator < 0 ^ denominator < 0)
ans += "-";
long n = labs(numerator);
long d = labs(denominator);
ans += to_string(n / d);
if (n % d == 0)
return ans;
ans += '.';
unordered_map<int, int> seen;
for (long r = n % d; r; r %= d) {
if (seen.count(r)) {
ans.insert(seen[r], 1, '(');
ans += ')';
break;
}
seen[r] = ans.size();
r *= 10;
ans += to_string(r / d);
}
return ans;
}
};
Java
class Solution {
public String fractionToDecimal(int numerator, int denominator) {
if (numerator == 0)
return "0";
StringBuilder sb = new StringBuilder();
if (numerator < 0 ^ denominator < 0)
sb.append("-");
long n = Math.abs((long) numerator);
long d = Math.abs((long) denominator);
sb.append(n / d);
if (n % d == 0)
return sb.toString();
sb.append(".");
Map<Long, Integer> seen = new HashMap<>();
for (long r = n % d; r > 0; r %= d) {
if (seen.containsKey(r)) {
sb.insert(seen.get(r), "(");
sb.append(")");
break;
}
seen.put(r, sb.length());
r *= 10;
sb.append(r / d);
}
return sb.toString();
}
}
Python
class Solution:
def fractionToDecimal(self, numerator: int, denominator: int) -> str:
if numerator == 0:
return '0'
ans = ''
if (numerator < 0) ^ (denominator < 0):
ans += '-'
numerator = abs(numerator)
denominator = abs(denominator)
ans += str(numerator // denominator)
if numerator % denominator == 0:
return ans
ans += '.'
dict = {}
remainder = numerator % denominator
while remainder:
if remainder in dict:
ans = ans[:dict[remainder]] + '(' + ans[dict[remainder]:] + ')'
break
dict[remainder] = len(ans)
remainder *= 10
ans += str(remainder // denominator)
remainder %= denominator
return ans
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