DFA 简易正则表达式匹配

　　一个只能匹配非常简单的（字母 . + *）共 4 种状态的正则表达式语法的自动机（注意，仅限 DFA，没考虑 NFA）：

　　好久之前写的了，记得有个 bug 一直没解决...

#include <iostream>
//#include <fstream>
#include <vector>
#include <string>

class DFA
{
	void construction(std::string regex)
	{
		std::vector<AM*> worker;
		Match = std::make_unique<AM>();
		h = std::make_unique<AM>(toRange('H'), nullptr);
		AM* h_ptr = h.get();
		for (auto iter = regex.begin(); iter != regex.end(); ++iter)
		{
			AM * temp = new AM(toRange(*iter), h_ptr);
			switch (*iter)
			{
			case '.':
				{
					h_ptr->next[temp->ch] = temp;
					h_ptr = temp;
					if (iter + 1 != regex.end() && *(iter + 1) != '*')
					{
						while (!worker.empty())
						{
							AM*c_ptr = worker.front();
							worker.erase(worker.begin());
							c_ptr->next[h_ptr->ch] = h_ptr;
						}
					}
				}
				break;
			case '*':
				{
					h_ptr->next[h_ptr->ch] = h_ptr;
					for (std::vector<AM*>::iterator i = worker.begin(); i != worker.end(); i++)
						(*i)->next[h_ptr->ch] = h_ptr;
					if (h_ptr->prev != nullptr)
						worker.push_back(h_ptr->prev);
					worker.push_back(h_ptr);
					delete temp;
					temp = nullptr;
				}
				break;
			case '+':
				{
					h_ptr->next[h_ptr->ch] = h_ptr;
					while (!worker.empty())
					{
						AM*c_ptr = worker.front();
						worker.erase(worker.begin());
						c_ptr->next[h_ptr->ch] = h_ptr;
					}
					delete temp;
					temp = nullptr;
				}
				break;
			default:
				{
					h_ptr->next[temp->ch] = temp;
					h_ptr = temp;
					if (iter + 1 != regex.end() && *(iter + 1) != '*')
					{
						while (!worker.empty())
						{
							AM*c_ptr = worker.front();
							worker.erase(worker.begin());
							c_ptr->next[h_ptr->ch] = h_ptr;
						}
					}
				}
				break;
			}
		}
		while (!worker.empty())
		{
			AM*c_ptr = worker.front();
			worker.erase(worker.begin());
			if (h_ptr->next[h_ptr->ch] == h_ptr)
				c_ptr->next[0] = Match.get();
			else
				c_ptr->next[h_ptr->ch] = h_ptr;
		}
		h_ptr->next[0] = Match.get();
	}

	char toRange(char c) const
	{
		if (c == '.')
			return 27;
		return c - 'a' + 1;
	}

public:
	bool isMatch(std::string s, std::string regex)
	{
		construction(regex);
		AM * am = h.release();
		for (auto i:s)
		{
			char c = toRange(i);
			if (am == nullptr)
				return false;
			if (am->next[c] != nullptr)
				am = am->next[c];
			else if (am->next[27] != nullptr)
				am = am->next[27];
			else
				am = am->next[c];
		}
		return am != nullptr && am->next[0] == Match.get();
	}

private:
	struct AM {
		char ch;
		AM *prev, *next[28];
		AM() : ch(), prev(), next() {}
		AM(char v, AM * prev) : ch(v), prev(prev), next() {}
	};

	std::unique_ptr<AM> Match, h;
};

int main(int argc, char const *argv[])
{
	DFA s;
	
	std::cout << (s.isMatch("abc", "aa*b*c+p*") ? "true":"false");
	
	return 0;
}

　　示例1：a*b*c+d*

　　该正则表达式的DFA如下图

 

　　示例2：(a|b)*a

 

　　这是一个NFA，我的代码并没有实现NFA转DFA，因而会导致匹配失败。（2020-04-05 21:00:39 补充：所以 leetcode 上 a*a 过不了，因为它也是NFA。虽然可以转换为正则表达式 a+ 来匹配，它的构造图：

 

 　　但 a+ 的 DFA 构造图也可以是这样的：

 　　实际上我们写出状态转移表，然后直接查状态表效率会更高，不过我个人觉得模拟匹配的过程更有意思。