LeetCode 10 Regular Expression Matching (C,C++,Java,Python)

来源：互联网发布：网络用语开车什么意思编辑：程序博客网时间：2024/06/17 03:32

Problem:

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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true

Solution:

多种方法解决，可以用DFS，也可以用DP，参考这里：http://blog.csdn.net/hcbbt/article/details/44016237

题目大意：

给定一个字符串和一个正则表达式，给出正则是否匹配字符串

解题思路：

关键问题在*上面，*可以占用零个或者多个位置，因此用DFS的办法就是尝试遍所有的可能匹配的情况，比如AAAAAB和A*B，就要看AAAAAB和B还有AAAB和B和AB和B

Java源代码（用时302ms）：

public class Solution {    public boolean isMatch(String s, String p) {        char[] chs = s.toCharArray();        char[] chp = p.toCharArray();        return Match(chs,0,chp,0);    }    public boolean Match(char[] chs,int index1,char[] chp,int index2){        if(index2>=chp.length)return index1>=chs.length;        if(index2+1<chp.length && chp[index2+1]=='*'){            while(index1<chs.length && (chp[index2]=='.' || chp[index2]==chs[index1])){                if(Match(chs,index1,chp,index2+2))return true;                index1++;            }            return Match(chs,index1,chp,index2+2);        }else if(index1<chs.length && (chp[index2]=='.' || chs[index1]==chp[index2])){            return Match(chs,index1+1,chp,index2+1);        }        return false;    }}

C语言源代码（用时21ms）：

bool isMatch(char* s, char* p) {    if(s==NULL || p==NULL)return false;    if(!*p) return !*s;    if(*(p+1)=='*'){        while((*p==*s)||(*s && *p=='.')){            if(isMatch(s,p+2))return true;            s++;        }        return isMatch(s,p+2);    }else if((*p==*s)||(*s && *p=='.')){        return isMatch(s+1,p+1);    }    return false;}

C++源代码（用时407ms）：

class Solution {public:    bool isMatch(string s, string p) {        return Match(s,0,p,0);    }    bool Match(string s,int index1,string p,int index2){        if(index2>=p.size())return index1>=s.size();        if(index2+1<p.size() && p[index2+1]=='*'){            while(index1<s.size() && (p[index2]=='.' || p[index2]==s[index1])){                if(Match(s,index1,p,index2+2))return true;                index1++;            }            return Match(s,index1,p,index2+2);        }else if(index1<s.size() && (p[index2]=='.' || p[index2]==s[index1])){            return Match(s,index1+1,p,index2+1);        }        return false;    }};

Python源代码（调用自带函数用时140ms）：

class Solution:    # @param {string} s    # @param {string} p    # @return {boolean}    def isMatch(self, s, p):        return re.match('^' + p + '$', s) != None

Python源代码（DFS超时）：

class Solution:    # @param {string} s    # @param {string} p    # @return {boolean}    def isMatch(self, s, p):        return self.Match(s,0,p,0)    def Match(self,s,index1,p,index2):        if index2>=len(p):return index1>=len(s)        if index2+1<len(p) and p[index2+1]=='*':            while index1<len(s) and (p[index2]=='.' or p[index2]==s[index1]):                if self.Match(s,index1,p,index2+2):return True                index1+=1            return self.Match(s,index1,p,index2+2)        elif index1<len(s) and (p[index2]=='.' or p[index2]==s[index1]):            return self.Match(s,index1+1,p,index2+1)        return False

0 0