[Swift]LeetCode1096. 花括號展開 II | Brace Expansion II

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➤微信公眾號：山青詠芝（shanqingyongzhi）
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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.

---

如果你熟悉 Shell 編程，那么一定了解過花括號展開，它可以用來生成任意字符串。

花括號展開的表達式可以看作一個由 花括號、逗號 和 小寫英文字母 組成的字符串，定義下面幾條語法規則：

• 如果只給出單一的元素 x，那么表達式表示的字符串就只有 "x"。 
  • 例如，表達式 {a} 表示字符串 "a"。
  • 而表達式 {ab} 就表示字符串 "ab"。
• 當兩個或多個表達式並列，以逗號分隔時，我們取這些表達式中元素的並集。
  • 例如，表達式 {a,b,c} 表示字符串 "a","b","c"。
  • 而表達式 {a,b},{b,c} 也可以表示字符串 "a","b","c"。
• 要是兩個或多個表達式相接，中間沒有隔開時，我們從這些表達式中各取一個元素依次連接形成字符串。
  • 例如，表達式 {a,b}{c,d} 表示字符串 "ac","ad","bc","bd"。
• 表達式之間允許嵌套，單一元素與表達式的連接也是允許的。
  • 例如，表達式 a{b,c,d} 表示字符串 "ab","ac","ad"​​​​​​。
  • 例如，表達式 {a{b,c}}{{d,e}f{g,h}} 可以代換為 {ab,ac}{dfg,dfh,efg,efh}，表示字符串 "abdfg", "abdfh", "abefg", "abefh", "acdfg", "acdfh", "acefg", "acefh"。

給出表示基於給定語法規則的表達式 expression，返回它所表示的所有字符串組成的有序列表。

假如你希望以「集合」的概念了解此題，也可以通過點擊 “顯示英文描述” 獲取詳情。 

示例 1：

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

示例 2：

輸入："{{a,z}, a{b,c}, {ab,z}}"
輸出：["a","ab","ac","z"]
解釋：輸出中 不應 出現重復的組合結果。 

提示：

1. 1 <= expression.length <= 50
2. expression[i] 由 '{'，'}'，',' 或小寫英文字母組成
3. 給出的表達式 expression 用以表示一組基於題目描述中語法構造的字符串

---

Runtime: 16 ms

Memory Usage: 13.2 MB

class Solution {
    func braceExpansionII(_ expression: String) -> [String] {
        var str:String = expression
        var pos:Int = 0
        return Array(parseRule2(&str, &pos).sorted())
    }
    
    func merge(_ a:Set<String>,_ b:Set<String>) -> Set<String>
    {
        if a.isEmpty {return b}
        if b.isEmpty {return a}
        var ans:Set<String> = Set<String>()
        for v1 in a
        {
            for v2 in b
            {
                ans.insert(v1 + v2)
            }
        }
        return ans
    }
    
    //{a,b,c}
    func parseRule1(_ str:inout String,_ i:inout Int) -> Set<String>
    {
        var ans:Set<String> = Set<String>()
        i += 1
        ans = parseRule2(&str, &i)
        i += 1
        return ans
    }
    
    //{a,b},{c,d}
    func parseRule2(_ str:inout String,_ i:inout Int) -> Set<String>
    {
        var ans:Set<String> = Set<String>()
        ans = parseRule3(&str, &i)
        let arrStr:[Character] = Array(str)
        while(i < str.count)
        {
            if arrStr[i] != "," {break}
            i += 1
            let temp:Set<String> =  parseRule3(&str, &i)
            ans = ans.union(temp)
        }
        return ans
    }
    
    //a{c,d}b{e,f}
    func parseRule3(_ str:inout String,_ i:inout Int) -> Set<String>
    {
        var ans:Set<String> = Set<String>()
        let arrStr:[Character] = Array(str)
        while(i < str.count)
        {
            if arrStr[i] == "}" || arrStr[i] == "," {break}
            if arrStr[i] == "{"
            {
                let temp:Set<String> =  parseRule1(&str, &i)
                ans = merge(ans, temp)
            }
            else
            {
                var temp:Set<String> = Set<String>()
                var tmpStr:String = String()
                while(i < str.count && arrStr[i] <= "z" && arrStr[i] >= "a")
                {
                    tmpStr.append(arrStr[i])
                    i += 1
                }
                temp.insert(tmpStr)
                ans = merge(ans,temp)
            }
        }
        return ans
    }
}