Equal Rational Numbers - LeetCode solutions

Problem

Given two strings s and t, each of which represents a non-negative rational number, return true if and only if they represent the same number. The strings may use parentheses to denote the repeating part of the rational number.

A rational number can be represented using up to three parts: <IntegerPart>, <NonRepeatingPart>, and a <RepeatingPart>. The number will be represented in one of the following three ways:

<IntegerPart>

For example, 12, 0, and 123.

<IntegerPart>**<.>**<NonRepeatingPart>
For example, 0.5, 1., 2.12, and 123.0001.

<IntegerPart>**<.>**<NonRepeatingPart>**<(>**<RepeatingPart>**<)>**
For example, 0.1(6), 1.(9), 123.00(1212).

The repeating portion of a decimal expansion is conventionally denoted within a pair of round brackets. For example:

1/6 = 0.16666666... = 0.1(6) = 0.1666(6) = 0.166(66).

Example 1:

Input: s = "0.(52)", t = "0.5(25)"
Output: true
Explanation: Because "0.(52)" represents 0.52525252..., and "0.5(25)" represents 0.52525252525..... , the strings represent the same number.

Example 2:

Input: s = "0.1666(6)", t = "0.166(66)"
Output: true

Example 3:

Input: s = "0.9(9)", t = "1."
Output: true
Explanation: "0.9(9)" represents 0.999999999... repeated forever, which equals 1.  [See this link for an explanation.]
"1." represents the number 1, which is formed correctly: (IntegerPart) = "1" and (NonRepeatingPart) = "".

Constraints:

Each part consists only of digits.
The <IntegerPart> does not have leading zeros (except for the zero itself).
1 <= <IntegerPart>.length <= 4
0 <= <NonRepeatingPart>.length <= 4
1 <= <RepeatingPart>.length <= 4

Solution

class Solution {
    public boolean isRationalEqual(String s, String t) {
        int sLeftIndex = s.indexOf("(");
        int tLeftIndex = t.indexOf("(");
        // if they are integer or only with non repeating part
        if (sLeftIndex < 0 && tLeftIndex < 0) {
            return Double.valueOf(s).equals(Double.valueOf(t));
        }
        // get the first 100 (or so) digits of s and t
        double sDouble;
        double tDouble;
        if (sLeftIndex >= 0) {
            s = s.substring(0, sLeftIndex) + repeat(s.substring(sLeftIndex + 1, s.length() - 1));
            sDouble = Double.parseDouble(s.substring(0, 100));
        } else {
            sDouble = Double.parseDouble(s);
        }
        if (tLeftIndex >= 0) {
            t = t.substring(0, tLeftIndex) + repeat(t.substring(tLeftIndex + 1, t.length() - 1));
            tDouble = Double.parseDouble(t.substring(0, 100));
        } else {
            tDouble = Double.parseDouble(t);
        }
        return sDouble == tDouble;
    }

    private String repeat(String a) {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < 100; i++) {
            sb.append(a);
        }
        return sb.toString();
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).