10. Regular Expression Matching *HARD*

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

1.

bool isMatch(string s, string p) {
    int ls = s.length(), lp = p.length(), i, j;
    vector<vector<int>> dp(, vector<int>(lp + , ));
    bool k = ;
    dp[][] = ; dp[][] = ;
    for (i = ; i <= lp; i++)
        dp[][i] = (dp[][i - ] && (p[i - ] == '*'));
    for (i = ; i <= ls; i++)
    {
        dp[k][] = ;
        for (j = ; j <= lp; j++)
        {
            if('*' == p[j-] && j > )
            {
                if(p[j-] == s[i-] || '.' == p[j-])
                    dp[k][j] = dp[k][j-] | dp[!k][j];
                else
                    dp[k][j] = dp[k][j-];
            }
            else if(p[j-] == s[i-] || '.' == p[j-])
                dp[k][j] = dp[!k][j-];
            else
                dp[k][j] = ;
        }
        k = !k;
    }
    return dp[!k][lp];
}

2.

bool isMatch(string s, string p) {
    int ls = s.length(), lp = p.length(), i, j;
    if(p == "")
        return s == "";
    if( == lp || p[] != '*')
    {
        if(s == "" || (p[] != '.' && s[] != p[]))
            return false;
        return isMatch(s.substr(), p.substr());
    }
    //p[1] == '*'
    for(i = -; i < ls && (- == i || '.' == p[] || s[i] == p[]); i++ )
    {
        if(isMatch(s.substr(i+), p.substr()))
            return true;
    }
}