# [LeetCode 166] Fraction to Recurring Decimal

### Question

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)".

Credits:
Special thanks to @Shangrila for adding this problem and creating all test cases.

### Analysis

Wow, this is just another incredible question on Leetcode. Quite a few difficult corner cases in the OJ.

Current AC rate is first lowest at 12.4%.

### Solution

There’re 3 things that we must take note:

1. Handle positive/nagetive cases well. Note that both numerator and denominator can be negative number. And there’re also case like:

1 / -3 = -0.(3)

1. Note that we have to match the repetation of the actual numerator. Not the quotient. What do I mean by that?

1 / 6 = 0.1(6)

1 / 333 = 0.(003)

1. Note below is how to override the equals() method (first one is wrong and second is right):

public boolean equals(Pair p) {

}

public boolean equals(Object obj) {

}

One more thing: Most other guys' solutions like this, this and this are using Hashing. It’s fine and definitely good. I used linearly search though, which is a small time compromise. I am presenting my code below and please just consider it as something different. Keep in mind it’s not the optimized solution.

### Code

``````public class Solution {
public String fractionToDecimal(int numerator, int denominator) {
long quotient = (long) numerator / denominator;
long reminder = (long) numerator % denominator;
if (reminder == 0) {
return String.valueOf(quotient);
}

// The result has 3 parts: sign, integer part, and fraction part
// eg. -4 / 3 = -1 and the result of (1/3) = (3)
String sign = ((long) numerator * denominator >= 0) ? "" : "-";
long integer = Math.abs(quotient);
String fraction = fraction(Math.abs((long)reminder), Math.abs((long)denominator));

// why do we have to seperate sign from integer?
// cuz 1 / -3 = -0.(3), while quotient is 0.
// So, we can't simply concatenate quotient with fraction
return sign + integer + "." + fraction;
}

String fraction(long num, long denum) {
// eg. num = 1, denum = 4, should return "25"

List<Pair> list = new ArrayList<Pair>();
String result = "";
while (num != 0) {
num *= 10;
long digit = num / denum;

// eg. 1 / 333 = (003), so the pairs would be like this:
// {10, 0}, {100, 0}, {1000, 3}, {10, 0}...
Pair cur = new Pair(num, digit);
num %= denum;

// now add cur Pair to the list
if (list.indexOf(cur) == -1) {
} else {
// found a recurring dicimal in the previous output stream
int pos = list.indexOf(cur);
for (int i = 0; i < pos; i++) {
result += list.get(i).digit;
}
result += "(";
for (int i = pos; i < list.size(); i++) {
result += list.get(i).digit;
}
result += ")";
break;
}
}

// if there is recurring digit, the result should have already been generated
if (result.length() == 0) {
for (Pair p: list) {
result += p.digit;
}
}
return result;
}

class Pair {
long num;
long digit;

public Pair(long a, long b) {
num = a;
digit = b;
}

public boolean equals(Object obj) {
// note the equals interface passes in (Object obj)
// instead of a Pair object
Pair p = (Pair) obj;
return this.num == p.num && this.digit == p.digit;
}
}
}
``````