<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>

class Solution {
public:
    bool isMatch(string s, string p) {
        int n, m, f[2005][2005], g[2005];
        //f[i][j]:the first i from s and first j from p is it matchable?
        //g[j]: is there any i for which f[i][j] is true
        memset(f, 0, sizeof f);
        memset(g, 0, sizeof g);
        n = s.size();
        m = p.size();
        //f[1][1] = 1;
        //g[1] = 1;
        f[0][0] = 1;
        g[0] = 1;
        if(p[0] != s[0]){
            if(p[0] != '*' && p[0] != '?'){
                //cout<<"false"<<endl;
                return false;
            }
        }
        for(int i = 0; i <= n; i++){
            for(int j = 1; j <= m; j++){
                if(i == 0){
                    if(p[j - 1] == '*')f[i][j] = f[i][j - 1];
                    if(f[i][j] == 1)g[j] = 1;
                    continue;
                }
                if(s[i - 1] == p[j - 1]){
                    f[i][j] = f[i - 1][j - 1];
                }else if(p[j - 1] == '?'){
                    f[i][j] = f[i - 1][j - 1];
                }else if(p[j - 1] == '*'){
                    f[i][j] = g[j - 1];//f[i - k][j - 1]
                    //cout<<i<<" "<<j<<" "<<g[j - 1]<<endl;
                }else{
                    f[i][j] = 0;
                }
                if(f[i][j] == 1)g[j] = 1;
            }
        }
        return bool(f[n][m]);
    }
};
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