Under the grammar given below, strings can represent a set of lowercase words. Let R(expr) denote the set of words the expression represents.
The 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 two 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}") = {"abdfg", "abdfh", "abefg", "abefh", "acdfg", "acdfh", "acefg", "acefh"}
Formally, the three rules for our grammar:
For every lowercase letter x, we have R(x) = {x}.
For expressions e1, e2, ... , ek with k >= 2, we have R({e1, e2, ...}) = R(e1) ∪ R(e2) ∪ ...
For expressions e1 and e2, we have R(e1 + e2) = {a + b for (a, b) in R(e1) × R(e2)}, 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: expression = "{a,b}{c,{d,e}}"
Output: ["ac","ad","ae","bc","bd","be"]
Example 2:
Input: expression = "{{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 <= expression.length <= 60
expression[i] consists of '{', '}', ','or lowercase English letters.
The given expression represents a set of words based on the grammar given in the description.
這題給我們一個字串 expression，
希望我們從中利用三種模式來拆分可能字串。
模式：
- R("a") = {"a"}
- R("a, b, c") = {"a", "b", "c"}
- R("{a,b}{c,d}") = {"ac","ad","bc","bd"}
簡單來說就是中括號內是同一組，
不同括號要相互組合。
然後這種題目靠北的地方在括號狀態很亂，
第一直覺是找前後括號配對，用雙指標之類的。
但是看到括號狀態，瓶頸就是要保留左右側狀態計算，
通常都是用 stack。
這題我有偷看一下解答，
因為組字串的時候要保留狀態，用 getline 比較好寫，
可以嘗試一下用雙指標(#。
思路就是找到最裡面那層括號，
把這組括號的組合歷遍（無論 prev, post 都會先存回 stack），
待無括號時就存回答案陣列。
最後要記得 sort，題目有要求。
今天程式碼太長了就先不放。
假設每個子陣列中的最大選項有 K 個，
這樣的子陣列有 M 組， expression 的長度是 N。
======晚上更新複雜度後======
計算複雜度：O(K^M * M * N * log(K))
-> 展開總共有 K^M 個計算
-> 展開每組 K^M 最多歷遍 N 個字元
-> 展開部分 O(K^M * N)
-> 排序部分總共 O(K^M * log(K*M) * N) == O(K^M * M * N *log(K))
空間複雜度：O(K^M * N)
Github程式碼：
