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

Input: S = true

Input: S = 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

 1 class Solution {
 2     func isRationalEqual(_ S: String, _ T: String) -> Bool {
 3          return abs(convertToDouble(S)  - convertToDouble(T)) < 1e-8
 4     }
 5     
 6     func convertToDouble(_ S: String) -> Double
 7     {
 8         var index = S.firstIndex(of: "(") ?? S.endIndex
 9         if index == S.endIndex
10         {
11             return Double(S)!
12         }
13         else
14         {
15             var sb:String = String(S[..<index])
16             let rightIndex = S.firstIndex(of: ")") ?? S.startIndex
17             if rightIndex != S.startIndex
18             {
19                 index = S.index(after: index)
20                 var rep:String = String(S[index..<rightIndex])
21                 while(sb.count < 50)
22                 {
23                     sb += rep
24                 }
25             }
26             return Double(sb)!
27         }
28     }
29 }

12ms

 1 class Solution {
 2     func isRationalEqual(_ S: String, _ T: String) -> Bool {
 3         if abs(parse(str: S) - parse(str: T)) < 0.0000001 {
 4             return true
 5         }
 6         return false
 7     }
 8     func parse(str: String) -> Double {
 9         if let dotIndex = str.firstIndex(of: "(") {
10             let ddIndex = str.firstIndex(of: ".")
11             var ans: Double = 0
12             ans += Double(String(str[str.startIndex..<dotIndex]))!
13             var dotStr = String(str[dotIndex..<str.endIndex]).dropFirst().dropLast()
14             let nn = dotIndex.encodedOffset -  ddIndex!.encodedOffset + dotStr.count - 1
15             for i in 0...10 {
16                 ans += Double(dotStr)! * pow(10.0, -Double(nn+i*(dotStr.count)))
17             }
18             return ans
19         } else {
20             return Double(str)!
21         }
22     }
23 }