Regular Expression Matching

Recursive Solution

class Solution {
public:
    bool isMatch(string s, string p) {
        if (p.empty())    return s.empty();

        if ('*' == p[1])
            // x* matches empty string or at least one character: x* -> xx*
            // *s is to ensure s is non-empty
            return (isMatch(s, p.substr(2)) || !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p));
        else
            return !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p.substr(1));
    }
};

DP Solution

class Solution {
public:
    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];
    }
};

To be honest, this algorithm bothers me for a very long time cause I simply don’t get it. Here is the explanation:
Consider s and p are presented as follows:

$$s: s_0, s_1, s_2, ..., s_{i -2}, s_{i - 1}$$ $$p: p_0, p_1, p_2, ..., p_{j -2}, p_{j - 1}$$
1. If $p_{j - 1}$ is not *, like this:
$$s: s_0, s_1, s_2, ..., s_{i -2}, s_{i - 1}$$ $$p: p_0, p_1, p_2, ..., p_{j -2}, a$$
then the only way $f[i][j]$ can be true is that: $s_0, s_1, s_2, ..., s_{i -2}$ matches $p_0, p_1, p_2, ..., p_{j -2}$ as well as $s_{i - 1}$ matches $p_{j - 1}$.
2. If $p_{j - 1}$ is *, like this:
$$s: s_0, s_1, s_2, ..., s_{i -2}, s_{i - 1}$$ $$p: p_0, p_1, p_2, ..., a, *$$
then the only way $f[i][j]$ can be true is that:
– For any length of $s$, it is equal to $p_0, p_1, p_2, ..., p_{j-3}$, which means $a, *$ will not appear in s.
– For string $s_0, s_1, s_2, ..., s_{i -2}$, it is equal to $p_0, p_1, p_2, ..., p_{j-2}, p_{j - 1}$, and any character after $s_{i - 2}$, it is a match of the character before *