Regular Expression Matching

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

思路：对于字符串匹配，s[i]和p[j].此题的关键就在于p[j+1]是否为'*'.递归思路解题。

如果p[j+1]!='*',s[i]==p[j]，则匹配下一位(i+1,j+1)，如果s[i]!=p[j]，匹配失败

如果p[j+1]=='*',s[i]==p[j]，则匹配(i,j+2)或者(i+1,j+2)，如果s[i]!=p[j]，则匹配(i,j+2)

class Solution {

public:

    bool isMatch(const char *s, const char *p) {

        if(*p=='\0')

            return *s=='\0';

        if(*(p+1)!='*')

        {

            while(*s!='\0' && (*s==*p || *p=='.'))

            {

                return isMatch(s+1,p+1);

            }

        }

        else

        {

            while(*s!='\0' && (*s==*p||*p=='.'))

            {

                if(isMatch(s,p+2))

                    return true;

                s++;

            }

            return isMatch(s,p+2);

        }

        return false;

    }

};

 使用动态规划解题，思路和上述很相似，dp[j][i]表示s[0..j]和p[0...j]是否匹配；

class Solution {

public:

    bool isMatch(const char *s, const char *p) {

        int slen=strlen(s);

    int plen=strlen(p);

    vector<vector<bool> > dp(slen+1,vector<bool>(plen+1,false));

    dp[0][0]=true;

    for(int i=1;i<=plen;++i)

    {

        if(p[i]!='*')

        {

            for(int j=1;j<=slen;++j)

            {

                dp[j][i]=(dp[j-1][i-1] && (s[j-1]==p[i-1]||p[i-1]=='.'));

            }

        }

        else

        {

            dp[0][i]=dp[0][i+1]=dp[0][i-1];

            for(int j=1;j<=slen;++j)

            {

                dp[j][i]=dp[j][i+1]=(dp[j-1][i]&&(s[j-1]==p[i-1] || p[i-1]=='.')||dp[j][i-1]);

            }

            ++i;

        }

    }

    return dp[slen][plen];

    }

};