[Swift]LeetCode972.相等的有理数 | Equal Rational Numbers

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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.

In general a rational number can be represented using up to three parts: an integer part, a non-repeating part,and a repeating part. The number will be represented in one of the following three ways:

• <IntegerPart> (e.g. 0, 12, 123)
• <IntegerPart><.><NonRepeatingPart>  (e.g. 0.5, 1., 2.12, 2.0001)
• <IntegerPart><.><NonRepeatingPart><(><RepeatingPart><)> (e.g. 0.1(6), 0.9(9), 0.00(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)

Both 0.1(6) or 0.1666(6) or 0.166(66) are correct representations of 1 / 6.

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) = "".

Note:

1. Each part consists only of digits.
2. The <IntegerPart> will not begin with 2 or more zeros.  (There is no other restriction on the digits of each part.)
3. 1 <= <IntegerPart>.length <= 4
4. 0 <= <NonRepeatingPart>.length <= 4
5. 1 <= <RepeatingPart>.length <= 4

---

给定两个字符串 S 和 T，每个字符串代表一个非负有理数，只有当它们表示相同的数字时才返回 true；否则，返回 false。字符串中可以使用括号来表示有理数的重复部分。

通常，有理数最多可以用三个部分来表示：整数部分 <IntegerPart>、小数非重复部分 <NonRepeatingPart> 和小数重复部分 <(><RepeatingPart><)>。数字可以用以下三种方法之一来表示：

• <IntegerPart>（例：0，12，123）
• <IntegerPart><.><NonRepeatingPart> （例：0.5，2.12，2.0001）
• <IntegerPart><.><NonRepeatingPart><(><RepeatingPart><)>（例：0.1(6)，0.9(9)，0.00(1212)）

十进制展开的重复部分通常在一对圆括号内表示。例如：

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

0.1(6) 或 0.1666(6) 或 0.166(66) 都是 1 / 6 的正确表示形式。

示例 1：

输入：S = "0.(52)", T = "0.5(25)"
输出：true
解释：因为 "0.(52)" 代表 0.52525252...，而 "0.5(25)" 代表 0.52525252525.....，则这两个字符串表示相同的数字。

示例 2：

输入：S = "0.1666(6)", T = "0.166(66)"
输出：true

示例 3：

输入：S = "0.9(9)", T = "1."
输出：true
解释：
"0.9(9)" 代表 0.999999999... 永远重复，等于 1 。[有关说明，请参阅此链接]
"1." 表示数字 1，其格式正确：(IntegerPart) = "1" 且 (NonRepeatingPart) = "" 。

提示：

1. 每个部分仅由数字组成。
2. 整数部分 <IntegerPart> 不会以 2 个或更多的零开头。（对每个部分的数字没有其他限制）。
3. 1 <= <IntegerPart>.length <= 4
4. 0 <= <NonRepeatingPart>.length <= 4
5. 1 <= <RepeatingPart>.length <= 4

---

8ms

class Solution {
    func isRationalEqual(_ S: String, _ T: String) -> Bool {
         return abs(convertToDouble(S)  - convertToDouble(T)) < 1e-8
    }
    
    func convertToDouble(_ S: String) -> Double
    {
        var index = S.firstIndex(of: "(") ?? S.endIndex
        if index == S.endIndex
        {
            return Double(S)!
        }
        else
        {
            var sb:String = String(S[..<index])
            let rightIndex = S.firstIndex(of: ")") ?? S.startIndex
            if rightIndex != S.startIndex
            {
                index = S.index(after: index)
                var rep:String = String(S[index..<rightIndex])
                while(sb.count < 50)
                {
                    sb += rep
                }
            }
            return Double(sb)!
        }
    }
}

---

12ms

class Solution {
    func isRationalEqual(_ S: String, _ T: String) -> Bool {
        if abs(parse(str: S) - parse(str: T)) < 0.0000001 {
            return true
        }
        return false
    }
    func parse(str: String) -> Double {
        if let dotIndex = str.firstIndex(of: "(") {
            let ddIndex = str.firstIndex(of: ".")
            var ans: Double = 0
            ans += Double(String(str[str.startIndex..<dotIndex]))!
            var dotStr = String(str[dotIndex..<str.endIndex]).dropFirst().dropLast()
            let nn = dotIndex.encodedOffset -  ddIndex!.encodedOffset + dotStr.count - 1
            for i in 0...10 {
                ans += Double(dotStr)! * pow(10.0, -Double(nn+i*(dotStr.count)))
            }
            return ans
        } else {
            return Double(str)!
        }
    }
}