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

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;    }};

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];    }};