LeetCode--Regular Expression Matching

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Implement regular expression matching with support for'.'and'*'.

'.' Matches any single character.
'*' Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

The function prototype should be:
bool isMatch(const char *s,const char *p)

Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa","a*") → true
isMatch("aa",".*") → true
isMatch("ab",".*") → true
isMatch("aab","c*a*b") → true
本题使用了正则表达式匹配的方法,用非确定状态自动机模拟字符串p,从而进行字符串的匹配,算法的时间复杂度为O(N);

解题代码

class graph
{
private:
	struct node
	{
		int wei;
		node* next;
	};
	node* g;
	int v;
	bool* marked;
	void dfs(const int s)
	{
		marked[s] = true;
		vector<int> temp = adj(s);
		for(int i=0; i<temp.size(); i++)
		{
			if(!marked[temp[i]])
				dfs(temp[i]);
		}
		return ;
	}
public:
	graph(const int v)
	{
		this->v = v;
		marked = new bool[v];
		g = new node[v];
		for(int i=0; i<v; i++)
		{
			marked[i] = false;
			g[i].wei = 0;
			g[i].next = nullptr;
		}	
	}
	~graph()
	{
		delete[] marked;
		for(int i=0; i<v; i++)
		{
			node* temp1 = g[i].next;
			while(temp1!=nullptr)
			{
				node* temp2 = temp1;
				temp1 = temp1->next;
				delete temp2;
			}	
		}
		delete[] g;
	}
	bool is_marked(const int v){ return marked[v];}
	void restore_marked()
	{
		for(int i=0; i<v; i++)
			marked[i] = false;
	}
	void add_edge(const int v,const int w)
	{
		node* temp = new node;
		temp->wei = w;
		temp->next = g[v].next;
		g[v].next = temp;
	}
	void dfs(vector<int>& start)
	{
		int n = start.size();
		for(int i=0; i<n; i++)
			if(!marked[start[i]])
				dfs(start[i]);
	}
	vector<int>& adj(const int v)
	{
		vector<int>* res = new vector<int>;
		node* temp = g[v].next;
		while(temp!=nullptr)
		{
			res->push_back(temp->wei);
			temp = temp->next;
		}
		return *res;
	}
};
class Solution 
{
public:
	bool isMatch(const char *s,const char *p) 
	{
		if(has_star(p))
			return match(s,p);
		int i=0;
		while(s[i]!='\0' && p[i] != '\0')
		{
			if(s[i]=='.' || p[i]=='.')
			{
				i++;
				continue;
			}
			if(s[i] != p[i])
				return false;
			i++;
		}
		if(s[i]!='\0' || p[i]!='\0')
		    return false;
		return true;
	}
	bool match(const char* s,const char* p)
	{
		string temp_s = char_to_string(s);
		string temp_p = char_to_string(p);
		int n = temp_p.length();
		graph g(n+1);
		for(int i=0; i<n; i++)
		{
			if(p[i] == '*')
			{
				g.add_edge(i,i-1);
				g.add_edge(i-1,i);
				g.add_edge(i,i+1);
			}
		}
		vector<int> pc;
		vector<int> marked;
		marked.push_back(0);
		g.dfs(marked);
		for(int i=0; i<n+1; i++)
		{
			if(g.is_marked(i))
				pc.push_back(i);
		}
		for(int i=0; i<temp_s.length(); i++)
		{	
			marked.clear();
			for(int v=0; v<pc.size(); v++)
			{
				if(pc[v]<n)
				{
					if(p[pc[v]] == s[i] || p[pc[v]] == '.')
						marked.push_back(pc[v]+1);
				}
			}
			pc.clear();
			g.restore_marked();
			g.dfs(marked);
			for(int i=0; i<n+1; i++)
			{
				if(g.is_marked(i))
					pc.push_back(i);
			}
		}
		for(int i=0; i<pc.size(); i++)
		{
			if(pc[i] == n)
				return true;
		}
		return false;
	}
	bool has_star(const char* s)
	{
		int i=0;
		while(s[i] != '\0')
		{
			if(s[i] == '*')
				return true;
			i++;
		}
		return false;
	}
	string char_to_string(const char* s)
	{
		string temp = "";
		int i=0;
		while(s[i] != '\0')
		{
			temp = temp + s[i];
			i++;
		}
		return temp;
	}
};
原文链接:https://www.f2er.com/regex/360871.html

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