[Leetcode 10, Hard] Regular Expression Matching

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") → 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

Analysis:

The solutions come from internet. 

Solutions:

C++:

(1) DFS

    bool isMatchAux(const char *s, const char *p) 
    {
        if (!p[0])
            return !s[0];

        int slen = strlen(s), plen = strlen(p);
        if (plen == 1 || p[1] != '*') {
            return slen &&
                    (p[0] == '.' || s[0] == p[0]) &&
                    isMatchAux(s + 1, p + 1);
        }

        while (s[0] && (p[0] == '.' || s[0] == p[0])) {
            if (isMatchAux(s++, p + 2))
                return true;
        }

        return isMatchAux(s, p + 2);
    }

    bool isMatch(string s, string p) {
        if (p.empty())
            return s.empty();

        if(p[0] == '*')
            return true;

        bool r = isMatchAux(s.c_str(), p.c_str());
        return r;
    }

(2) Dynamic programming:

    bool isMatch(string s, string p) 
    {
        /**
         * f[i][j]: if s[0..i-1] matches p[0..j-1]
         * if p[j - 1] != '*'
         *      f[i][j] = f[i - 1][j - 1] && s[i - 1] == p[j - 1]
         * if p[j - 1] == '*', denote p[j - 2] with x
         *      f[i][j] is true iff any of the following is true
         *      1) "x*" repeats 0 time and matches empty: f[i][j - 2]
         *      2) "x*" repeats >= 1 times and matches "x*x": s[i - 1] == x && f[i - 1][j]
         * '.' matches any single character
         */
        int m = s.size(), n = p.size();
        vector<vector<bool>> f(m + 1, vector<bool>(n + 1, false));

        f[0][0] = true;
        for (int i = 1; i <= m; i++)
            f[i][0] = false;
        // p[0.., j - 3, j - 2, j - 1] matches empty iff p[j - 1] is '*' and p[0..j - 3] matches empty
        for (int j = 1; j <= n; j++)
            f[0][j] = j > 1 && '*' == p[j - 1] && f[0][j - 2];

        for (int i = 1; i <= m; i++)
            for (int j = 1; j <= n; j++)
                if (p[j - 1] != '*')
                    f[i][j] = f[i - 1][j - 1] && (s[i - 1] == p[j - 1] || '.' == p[j - 1]);
                else
                    // p[0] cannot be '*' so no need to check "j > 1" here
                    f[i][j] = f[i][j - 2] || (s[i - 1] == p[j - 2] || '.' == p[j - 2]) && f[i - 1][j];

        return f[m][n];
    }

Java:

Python: