[Leetcode 10, Hard] Regular Expression Matching

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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

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:

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