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
func isMatch(s string, p string) bool {
	m, n := len(s), len(p)
	f := make([][]int, m+1)
	for i := range f {
		f[i] = make([]int, n+1)
	}
	var dfs func(i, j int) bool
	dfs = func(i, j int) bool {
		if i >= m {
			return j >= n || p[j] == '*' && dfs(i, j+1)
		}
		if j >= n {
			return false
		}
		if f[i][j] != 0 {
			return f[i][j] == 1
		}
		f[i][j] = 2
		ok := false
		if p[j] == '*' {
			ok = dfs(i+1, j) || dfs(i+1, j+1) || dfs(i, j+1)
		} else {
			ok = (p[j] == '?' || s[i] == p[j]) && dfs(i+1, j+1)
		}
		if ok {
			f[i][j] = 1
		}
		return ok
	}
	return dfs(0, 0)
}