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)".
It is not hard in the sense that the algorithm is not hard: basic divide algorithm.
It is hard because
1. Dealing with the remainder part. In the case of denominator >> numerator, we may have lots of decimal numbers before we find the repeating part. One possible approach is to use a hash map. Since we need to add parentheses for the repeating part. We can store the position in the result string of the reminder. Thus when a repeating part shows, we can add parentheses at that position.
2. Overflow. In the case of denominator = Integer.MAX_VALUE or Integer.MIN_VALUE, we will face an overflow problem. To deal with that, we may use long in our method.
public class FractiontoDecimal { public String fractionToDecimal(int numerator, int denominator) { // return 0 condition if (denominator == 0) return null; else if (numerator == 0) return "0"; StringBuilder rst = new StringBuilder(); if (numerator < 0 && denominator > 0 || (denominator < 0 && numerator > 0)) rst.append("-"); long numer = Math.abs((long)numerator); long denom = Math.abs((long)denominator); rst.append (String.valueOf(numer / denom)); long remainder = numer % denom; if (remainder == 0) return rst.toString(); rst.append("."); HashMaphm = new HashMap (); //ArrayList list = new ArrayList (); while (remainder != 0) { if (hm.containsKey(remainder)) { rst.insert(hm.get(remainder),"("); rst.append(")"); break; } hm.put(remainder, rst.length()); long decimal = remainder * 10 / denom; rst.append(decimal); remainder = (remainder * 10) % denom; } return rst.toString(); } }
gr8.....thanks for the solution
ReplyDeleteYou are welcome! :)
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