Regular Expression Matching
<p>Given an input string <code>s</code>&nbsp;and a pattern <code>p</code>, implement regular expression matching with support for <code>&#39;.&#39;</code> and <code>&#39;*&#39;</code> where:</p>

<ul>
	<li><code>&#39;.&#39;</code> Matches any single character.​​​​</li>
	<li><code>&#39;*&#39;</code> Matches zero or more of the preceding element.</li>
</ul>

<p>The matching should cover the <strong>entire</strong> input string (not partial).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aa&quot;, p = &quot;a&quot;
<strong>Output:</strong> false
<strong>Explanation:</strong> &quot;a&quot; does not match the entire string &quot;aa&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aa&quot;, p = &quot;a*&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> &#39;*&#39; means zero or more of the preceding element, &#39;a&#39;. Therefore, by repeating &#39;a&#39; once, it becomes &quot;aa&quot;.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;ab&quot;, p = &quot;.*&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> &quot;.*&quot; means &quot;zero or more (*) of any character (.)&quot;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length&nbsp;&lt;= 20</code></li>
	<li><code>1 &lt;= p.length&nbsp;&lt;= 20</code></li>
	<li><code>s</code> contains only lowercase English letters.</li>
	<li><code>p</code> contains only lowercase English letters, <code>&#39;.&#39;</code>, and&nbsp;<code>&#39;*&#39;</code>.</li>
	<li>It is guaranteed for each appearance of the character <code>&#39;*&#39;</code>, there will be a previous valid character to match.</li>
</ul>
Recursion; String; Dynamic Programming
#define MAX_LEN 1000

char *ss, *pp;
int m, n;
int f[MAX_LEN + 1][MAX_LEN + 1];

bool dfs(int i, int j) {
    if (j >= n) {
        return i == m;
    }
    if (f[i][j] != 0) {
        return f[i][j] == 1;
    }
    int res = -1;
    if (j + 1 < n && pp[j + 1] == '*') {
        if (dfs(i, j + 2) || (i < m && (ss[i] == pp[j] || pp[j] == '.') && dfs(i + 1, j))) {
            res = 1;
        }
    } else if (i < m && (ss[i] == pp[j] || pp[j] == '.') && dfs(i + 1, j + 1)) {
        res = 1;
    }
    f[i][j] = res;
    return res == 1;
}

bool isMatch(char* s, char* p) {
    ss = s;
    pp = p;
    m = strlen(s);
    n = strlen(p);
    memset(f, 0, sizeof(f));
    return dfs(0, 0);
}
