刷题10. Regular Expression Matching

一、题目说明

这个题目是10. Regular Expression Matching，乍一看不是很难。

但我实现提交后，总是报错。不得已查看了答案。

二、我的做法

我的实现，最大的问题在于对.*的处理有问题，始终无法成功。

#include<iostream>

using namespace std;

class Solution{
    public:
	bool isMatch(string s,string p){
		bool result = true;
		if(s.length()<=0 && p.length()<=0){
			return true;
		}
		if(p==".*"){
			return true;
		}
		
		int sCurr=0,pCurr=0;
		int lenS = s.length();
		int lenP = p.length();
				
		//count the num of .*
		int numOfWildCard = 0;
		while(pCurr<lenP){
			if(p[pCurr]=='.' && pCurr+1<lenP && p[pCurr+1]=='*'){
				numOfWildCard++;
			}
			
			pCurr++;
		}
		//cout<<numOfWildCard<<":";
		
		pCurr = 0;
		while(sCurr<lenS && pCurr<lenP){
			if((pCurr+1<lenP) && p[pCurr]=='.' && p[pCurr+1]=='*'){
				if(pCurr+2<lenP){
					pCurr = pCurr+2;
					while(sCurr<lenS && s[sCurr]!=p[pCurr]){
						sCurr++;
					}
				}
				
			}
			if((pCurr+1<lenP) && p[pCurr+1]=='*'){
				while(sCurr<lenS && s[sCurr]==p[pCurr]){
					sCurr++;
				}
				pCurr = pCurr+2;
			}
			if(sCurr<lenS && pCurr<lenP && p[pCurr+1]!='*'){
				if(s[sCurr]==p[pCurr] || p[pCurr]=='.'){
					sCurr++;
					pCurr++;
				}
			}
		}
		
		if(sCurr==lenS && pCurr==lenP){
			result = true;
		}else{
			result = false;
		}
		return result;
	}	
};


int main(){
	Solution s;
	cout<<(false==s.isMatch("aa","a"))<<endl;
	cout<<(true==s.isMatch("aa","a*"))<<endl;
	cout<<(true==s.isMatch("ab",".*"))<<endl;
	cout<<(true==s.isMatch("aab","c*a*b"))<<endl;
	cout<<(false==s.isMatch("mississippi","mis*is*p*."))<<endl;
	return 0;
}

三、正确的做法

1.递归方法

#include<iostream>
#include<vector>

using namespace std;

class Solution{
    public:
    bool isMatch(string s, string p) {//aa a
        if(p.empty()) return s.empty();
        if(s.empty()) return p.empty() || (p[1] == '*' ? isMatch(s, p.substr(2)) : false);
        if(p[0] != '.' && s[0] != p[0]) return p[1] == '*' ? isMatch(s, p.substr(2)) : false;
        if(p[1] == '*') return isMatch(s.substr(1), p) || isMatch(s, p.substr(2));
        return isMatch(s.substr(1), p.substr(1));
    }
};

int main(){
	Solution s;
	cout<<(false==s.isMatch("aa","a"))<<endl;
	cout<<(true==s.isMatch("aa","a*"))<<endl;
	cout<<(true==s.isMatch("ab",".*"))<<endl;
	cout<<(true==s.isMatch("aab","c*a*b"))<<endl;
	cout<<(false==s.isMatch("mississippi","mis*is*p*."))<<endl;
	cout<<(false==s.isMatch("ab",".*c"))<<endl;
	cout<<(true==s.isMatch("aaa","a*a"))<<endl;
	return 0;
}

2.DP方法

dp是什么？动态规划啊，

#include<iostream>
#include<vector>
#include <mem.h>

using namespace std;

class Solution {
public:
    bool isMatch(string s, string p) {
        int ssize = s.size(),psize = p.size();
		string pp="";
		vector<bool> star;
		
		for(int i=0;i<p.size();i++){
			if(p[i]=='*'){
				star.back()=true;
			}else{
				star.push_back(false);
				pp+= p[i];
			}
		}
		
		psize = pp.size();
		
		bool dp[psize+1][ssize+1];
		memset(dp,false,sizeof(dp));
		
		dp[0][0] = true;
		for(int i=1;i<=psize;i++){
			if(star[i-1]==true){
				dp[i][0] = true;
			}else{
				break;
			}
		}
		
		for(int i=1;i<=psize;i++)
		    for(int j=1;j<=ssize;j++){
		    	if(dp[i-1][j-1]== true){
		    		if(pp[i-1]==s[j-1] || pp[i-1]=='.'){
		    			dp[i][j] = true;
		    			continue;
					}
				}
				
				if(dp[i-1][j]== true){
		    		if(star[i-1]==true){
		    			dp[i][j] = true;
		    			continue;
					}
				}
				
				if(dp[i][j-1]== true){
		    		if(star[i-1]==true && (pp[i-1]==s[j-1] || pp[i-1]=='.')){
		    			dp[i][j] = true;
		    			continue;
					}
				}
			}
		return dp[psize][ssize];
    }
};

int main(){
	Solution s;
	cout<<(false==s.isMatch("aa","a"))<<endl;
	cout<<(true==s.isMatch("aa","a*"))<<endl;
	cout<<(true==s.isMatch("ab",".*"))<<endl;
	cout<<(true==s.isMatch("aab","c*a*b"))<<endl;
	cout<<(false==s.isMatch("mississippi","mis*is*p*."))<<endl;
	cout<<(false==s.isMatch("ab",".*c"))<<endl;
	cout<<(true==s.isMatch("aaa","a*a"))<<endl;
	cout<<(false==s.isMatch("a",""))<<endl;
	return 0;
}

四、总结

看来基础知识还需要恶补，加油！