<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>

class Solution {
    int dp[21][21];
    private:
    bool solve(string s,string p,int n,int m){
        if(n<0&&m<0) return true;
        if(m<0&&n>=0) return false;
        if(n<0){
            for(int i=m;i>=0;i-=2){
                if(p[i]!='*') return false;
            }
            return true;
        }
        if(dp[n][m]!=-1) return dp[n][m];
        if(s[n]==p[m]||p[m]=='.'){
            return dp[n][m]=solve(s,p,n-1,m-1);
        }else if(p[m]=='*'){
            bool flag1=solve(s,p,n,m-2);//not take
            bool flag2=false;
            if(s[n]==p[m-1]||p[m-1]=='.'){
                flag2=solve(s,p,n-1,m);//take
            }
            return dp[n][m]=flag1||flag2;
        }else{
            return dp[n][m]=false;
        }
    }
public:
    bool isMatch(string s, string p) {
        int n=s.length(),m=p.length();
        memset(dp,-1,sizeof(dp));
        return solve(s,p,n-1,m-1);
    }
};
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