[Swift]LeetCode1096. 花括号展开 II | Brace Expansion II

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Under a grammar given below, strings can represent a set of lowercase words.  Let's use R(expr) to denote the set of words the expression represents.

Grammar can best be understood through simple examples:

• Single letters represent a singleton set containing that word.
  • R("a") = {"a"}
  • R("w") = {"w"}
• When we take a comma delimited list of 2 or more expressions, we take the union of possibilities.
  • R("{a,b,c}") = {"a","b","c"}
  • R("{{a,b},{b,c}}") = {"a","b","c"} (notice the final set only contains each word at most once)
• When we concatenate two expressions, we take the set of possible concatenations between two words where the first word comes from the first expression and the second word comes from the second expression.
  • R("{a,b}{c,d}") = {"ac","ad","bc","bd"}
  • R("{a{b,c}}{{d,e},f{g,h}}") = R("{ab,ac}{dfg,dfh,efg,efh}") = {"abdfg", "abdfh", "abefg", "abefh", "acdfg", "acdfh", "acefg", "acefh"}

Formally, the 3 rules for our grammar:

• For every lowercase letter x, we have R(x) = {x}
• For expressions e_1, e_2, ... , e_k with k >= 2, we have R({e_1,e_2,...}) = R(e_1) ∪ R(e_2) ∪ ...
• For expressions e_1 and e_2, we have R(e_1 + e_2) = {a + b for (a, b) in R(e_1) × R(e_2)}, where + denotes concatenation, and × denotes the cartesian product.

Given an expression representing a set of words under the given grammar, return the sorted list of words that the expression represents. 

Example 1:

Input: "{a,b}{c{d,e}}"
Output: ["acd","ace","bcd","bce"] 

Example 2:

Input: "{{a,z},a{b,c},{ab,z}}"
Output: ["a","ab","ac","z"]
Explanation: Each distinct word is written only once in the final answer.

Constraints:

1. 1 <= expression.length <= 50
2. expression[i] consists of '{', '}', ','or lowercase English letters.
3. The given expression represents a set of words based on the grammar given in the description.

---

若是你熟悉 Shell 编程，那么必定了解过花括号展开，它能够用来生成任意字符串。

花括号展开的表达式能够看做一个由 花括号、逗号 和 小写英文字母 组成的字符串，定义下面几条语法规则：

• 若是只给出单一的元素 x，那么表达式表示的字符串就只有 "x"。 
  • 例如，表达式 {a} 表示字符串 "a"。
  • 而表达式 {ab} 就表示字符串 "ab"。
• 当两个或多个表达式并列，以逗号分隔时，咱们取这些表达式中元素的并集。
  • 例如，表达式 {a,b,c} 表示字符串 "a","b","c"。
  • 而表达式 {a,b},{b,c} 也能够表示字符串 "a","b","c"。
• 要是两个或多个表达式相接，中间没有隔开时，咱们从这些表达式中各取一个元素依次链接造成字符串。
  • 例如，表达式 {a,b}{c,d} 表示字符串 "ac","ad","bc","bd"。
• 表达式之间容许嵌套，单一元素与表达式的链接也是容许的。
  • 例如，表达式 a{b,c,d} 表示字符串 "ab","ac","ad"​​​​​​。
  • 例如，表达式 {a{b,c}}{{d,e}f{g,h}} 能够代换为 {ab,ac}{dfg,dfh,efg,efh}，表示字符串 "abdfg", "abdfh", "abefg", "abefh", "acdfg", "acdfh", "acefg", "acefh"。

给出表示基于给定语法规则的表达式 expression，返回它所表示的全部字符串组成的有序列表。

假如你但愿以「集合」的概念了解此题，也能够经过点击 “显示英文描述” 获取详情。 

示例 1：

输入："{a,b}{c{d,e}}"
输出：["acd","ace","bcd","bce"]

示例 2：

输入："{{a,z}, a{b,c}, {ab,z}}"
输出：["a","ab","ac","z"]
解释：输出中 不该 出现重复的组合结果。 

提示：

1. 1 <= expression.length <= 50
2. expression[i] 由 '{'，'}'，',' 或小写英文字母组成
3. 给出的表达式 expression 用以表示一组基于题目描述中语法构造的字符串

---

Runtime: 16 ms

Memory Usage: 13.2 MB

 1 class Solution {
 2     func braceExpansionII(_ expression: String) -> [String] {
 3         var str:String = expression
 4         var pos:Int = 0
 5         return Array(parseRule2(&str, &pos).sorted())
 6     }
 7     
 8     func merge(_ a:Set<String>,_ b:Set<String>) -> Set<String>
 9     {
10         if a.isEmpty {return b}
11         if b.isEmpty {return a}
12         var ans:Set<String> = Set<String>()
13         for v1 in a
14         {
15             for v2 in b
16             {
17                 ans.insert(v1 + v2)
18             }
19         }
20         return ans
21     }
22     
23     //{a,b,c}
24     func parseRule1(_ str:inout String,_ i:inout Int) -> Set<String>
25     {
26         var ans:Set<String> = Set<String>()
27         i += 1
28         ans = parseRule2(&str, &i)
29         i += 1
30         return ans
31     }
32     
33     //{a,b},{c,d}
34     func parseRule2(_ str:inout String,_ i:inout Int) -> Set<String>
35     {
36         var ans:Set<String> = Set<String>()
37         ans = parseRule3(&str, &i)
38         let arrStr:[Character] = Array(str)
39         while(i < str.count)
40         {
41             if arrStr[i] != "," {break}
42             i += 1
43             let temp:Set<String> =  parseRule3(&str, &i)
44             ans = ans.union(temp)
45         }
46         return ans
47     }
48     
49     //a{c,d}b{e,f}
50     func parseRule3(_ str:inout String,_ i:inout Int) -> Set<String>
51     {
52         var ans:Set<String> = Set<String>()
53         let arrStr:[Character] = Array(str)
54         while(i < str.count)
55         {
56             if arrStr[i] == "}" || arrStr[i] == "," {break}
57             if arrStr[i] == "{"
58             {
59                 let temp:Set<String> =  parseRule1(&str, &i)
60                 ans = merge(ans, temp)
61             }
62             else
63             {
64                 var temp:Set<String> = Set<String>()
65                 var tmpStr:String = String()
66                 while(i < str.count && arrStr[i] <= "z" && arrStr[i] >= "a")
67                 {
68                     tmpStr.append(arrStr[i])
69                     i += 1
70                 }
71                 temp.insert(tmpStr)
72                 ans = merge(ans,temp)
73             }
74         }
75         return ans
76     }
77 }