leetcode 1096. Brace Expansion II

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}") = {"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: ["ac","ad","ae","bc","bd","be"]

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.

解题思路：

一对花括号便是完整的表达式 ，重点在于如何处理逗号之间表达式的并集和非逗号间隔表达式的连接集。

从题意中可以了解到只要有逗号出现，逗号之前的表达式便与之后的表达式不存在连接的关系了，可以直接将逗号之前组成的字符串的集合加到结果中，但是如果是非逗号间隔的表达式就需要知道之前的表达式组成了什么样的字符串，也就是需要一个前缀，所以定义一个前缀所有字符串的集合unordered_set prefixes,但遇到逗号时，将prefixes加入到res中，并清空prefixes ;

class Solution {
public:
    vector braceExpansionII(string expression) 
    {
        int pos = 0 ;
        unordered_set strs = brace(expression , pos) ;
        vector res(strs.begin() , strs.end()) ;
        sort(res.begin() , res.end()) ;
        return res ;
    }
    
    unordered_set brace(string &expression , int &pos)
    {
        unordered_set res , prefixes ;
        int len = expression.size() ;
        
        while(pos < len)
        {
            if(expression[pos] == '{')
            {
                pos++ ;
                unordered_set words = brace(expression , pos) ;
                if(!prefixes.empty())
                {
                    str_multi(prefixes , words) ;
                }
                else
                    prefixes = words ;
            }
            else if(expression[pos] == ',')
            {
                str_add(res , prefixes) ;
                prefixes.clear() ;
                pos++ ;
            }
            else if(expression[pos] >='a' && expression[pos] <= 'z')
            {
                string s = "" ;
                while(expression[pos] >='a' && expression[pos] <= 'z')
                {
                    s += expression[pos] ;
                    pos++ ;
                }
                
                if(!prefixes.empty())
                {
                    unordered_set new_prefixes ;
                    for(auto str : prefixes)
                        new_prefixes.insert(str + s) ;
                    
                    prefixes = new_prefixes ;
                }
                else
                    prefixes.insert(s) ;
            }
            else 
            {
                pos++ ;
                break ;    
            }
        }
        
        str_add(res , prefixes) ;
        
        return res ;
    }
    
    void str_multi(unordered_set &prefixes , unordered_set &words)
    {
        unordered_set res ;
        
        if(words.empty()) return ;
        
        for(auto prefix : prefixes)
        {
            for(auto word : words)
                res.insert(prefix + word) ;
        }
        
        prefixes = res ;
    }
    
    void str_add(unordered_set &res , unordered_set &prefixes)
    {
        for(auto prefix : prefixes)
            res.insert(prefix) ;
    }
};