LeetCode in Kotlin
166. Fraction to Recurring Decimal
Medium
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 - 1denominator != 0
Solution
class Solution {
fun fractionToDecimal(numerator: Int, denominator: Int): String {
if (numerator == 0) {
return "0"
}
val sb = StringBuilder()
// negative case
if (numerator > 0 && denominator < 0 || numerator < 0 && denominator > 0) {
sb.append("-")
}
val x = Math.abs(java.lang.Long.valueOf(numerator.toLong()))
val y = Math.abs(java.lang.Long.valueOf(denominator.toLong()))
sb.append(x / y)
var remainder = x % y
if (remainder == 0L) {
return sb.toString()
}
// decimal case
sb.append(".")
// store the remainder in a Hashmap because in the case of recurring decimal, the remainder
// repeats as dividend.
val map: MutableMap<Long, Int> = HashMap()
while (remainder != 0L) {
if (map.containsKey(remainder)) {
sb.insert(map.getValue(remainder), "(")
sb.append(")")
break
}
// store the remainder and the index of it's occurence in the String
map[remainder] = sb.length
remainder *= 10
sb.append(remainder / y)
remainder %= y
}
return sb.toString()
}
}