Wildcard Matching
<p>Given an input string (<code>s</code>) and a pattern (<code>p</code>), implement wildcard pattern 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 any sequence of characters (including the empty sequence).</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;*&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong>&nbsp;&#39;*&#39; matches any sequence.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;cb&quot;, p = &quot;?a&quot;
<strong>Output:</strong> false
<strong>Explanation:</strong>&nbsp;&#39;?&#39; matches &#39;c&#39;, but the second letter is &#39;a&#39;, which does not match &#39;b&#39;.
</pre>

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

<ul>
	<li><code>0 &lt;= s.length, p.length &lt;= 2000</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> or <code>&#39;*&#39;</code>.</li>
</ul>
Greedy; Recursion; String; Dynamic Programming
class Solution {
    private Boolean[][] f;
    private char[] s;
    private char[] p;
    private int m;
    private int n;

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

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