Fraction to Recurring Decimal
<p>Given two integers representing the <code>numerator</code> and <code>denominator</code> of a fraction, return <em>the fraction in string format</em>.</p>

<p>If the fractional part is repeating, enclose the repeating part in parentheses.</p>

<p>If multiple answers are possible, return <strong>any of them</strong>.</p>

<p>It is <strong>guaranteed</strong> that the length of the answer string is less than <code>10<sup>4</sup></code> for all the given inputs.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> numerator = 1, denominator = 2
<strong>Output:</strong> &quot;0.5&quot;
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> numerator = 2, denominator = 1
<strong>Output:</strong> &quot;2&quot;
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> numerator = 4, denominator = 333
<strong>Output:</strong> &quot;0.(012)&quot;
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>-2<sup>31</sup> &lt;=&nbsp;numerator, denominator &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>denominator != 0</code></li>
</ul>
Hash Table; Math; String
class Solution:
    def fractionToDecimal(self, numerator: int, denominator: int) -> str:
        if numerator == 0:
            return "0"
        ans = []
        neg = (numerator > 0) ^ (denominator > 0)
        if neg:
            ans.append("-")
        a, b = abs(numerator), abs(denominator)
        ans.append(str(a // b))
        a %= b
        if a == 0:
            return "".join(ans)
        ans.append(".")
        d = {}
        while a:
            d[a] = len(ans)
            a *= 10
            ans.append(str(a // b))
            a %= b
            if a in d:
                ans.insert(d[a], "(")
                ans.append(")")
                break
        return "".join(ans)
