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
class Solution {
    private Boolean[][] f;
    private String s;
    private String p;
    private int m;
    private int n;

    public boolean isMatch(String s, String p) {
        m = s.length();
        n = p.length();
        f = new Boolean[m + 1][n + 1];
        this.s = s;
        this.p = p;
        return dfs(0, 0);
    }

    private boolean dfs(int i, int j) {
        if (j >= n) {
            return i == m;
        }
        if (f[i][j] != null) {
            return f[i][j];
        }
        boolean res = false;
        if (j + 1 < n && p.charAt(j + 1) == '*') {
            res = dfs(i, j + 2)
                || (i < m && (s.charAt(i) == p.charAt(j) || p.charAt(j) == '.') && dfs(i + 1, j));
        } else {
            res = i < m && (s.charAt(i) == p.charAt(j) || p.charAt(j) == '.') && dfs(i + 1, j + 1);
        }
        return f[i][j] = res;
    }
}