<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 {
public:
    bool matches(char char_s, char char_p) {
        return char_p == '.' || char_s == char_p;
    }
    bool isMatch(string s, string p) {
        int m = s.size(), n = p.size();
        vector<bool> prev(n + 1, false); // 前一行
        vector<bool> curr(n + 1, false); // 當前行
        
        prev[0] = true; // 空字串與空模式匹配

        // 初始化 prev 陣列
        for (int j = 2; j <= n; j++) {
            if (p[j - 1] == '*') {
                prev[j] = prev[j - 2];
            }
        }

        // 填充 DP 陣列
        for (int i = 1; i <= m; i++) {
            curr[0] = false;
            for (int j = 1; j <= n; j++) {
                if (matches(s[i - 1], p[j - 1])) {
                    curr[j] = prev[j - 1];
                } else if (p[j - 1] == '*') {
                    curr[j] = curr[j - 2] || (matches(s[i - 1], p[j - 2]) && prev[j]);
                } else {
                    curr[j] = false;
                }
            }
            prev.swap(curr); // 更新 prev 為當前計算行
        }

        return prev[n];
    }
};
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