leetcode-problemset/leetcode-cn/originData/regular-expression-matching.json

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            "title": "Regular Expression Matching",
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            "content": "<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>\n\n<ul>\n\t<li><code>&#39;.&#39;</code> Matches any single character.​​​​</li>\n\t<li><code>&#39;*&#39;</code> Matches zero or more of the preceding element.</li>\n</ul>\n\n<p>The matching should cover the <strong>entire</strong> input string (not partial).</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;aa&quot;, p = &quot;a&quot;\n<strong>Output:</strong> false\n<strong>Explanation:</strong> &quot;a&quot; does not match the entire string &quot;aa&quot;.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;aa&quot;, p = &quot;a*&quot;\n<strong>Output:</strong> true\n<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;.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;ab&quot;, p = &quot;.*&quot;\n<strong>Output:</strong> true\n<strong>Explanation:</strong> &quot;.*&quot; means &quot;zero or more (*) of any character (.)&quot;.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length&nbsp;&lt;= 20</code></li>\n\t<li><code>1 &lt;= p.length&nbsp;&lt;= 30</code></li>\n\t<li><code>s</code> contains only lowercase English letters.</li>\n\t<li><code>p</code> contains only lowercase English letters, <code>&#39;.&#39;</code>, and&nbsp;<code>&#39;*&#39;</code>.</li>\n\t<li>It is guaranteed for each appearance of the character <code>&#39;*&#39;</code>, there will be a previous valid character to match.</li>\n</ul>\n",
            "translatedTitle": "正则表达式匹配",
            "translatedContent": "<p>给你一个字符串&nbsp;<code>s</code>&nbsp;和一个字符规律&nbsp;<code>p</code>，请你来实现一个支持 <code>'.'</code>&nbsp;和&nbsp;<code>'*'</code>&nbsp;的正则表达式匹配。</p>\n\n<ul>\n\t<li><code>'.'</code> 匹配任意单个字符</li>\n\t<li><code>'*'</code> 匹配零个或多个前面的那一个元素</li>\n</ul>\n\n<p>所谓匹配，是要涵盖&nbsp;<strong>整个&nbsp;</strong>字符串&nbsp;<code>s</code>的，而不是部分字符串。</p>\n&nbsp;\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"aa\", p = \"a\"\n<strong>输出：</strong>false\n<strong>解释：</strong>\"a\" 无法匹配 \"aa\" 整个字符串。\n</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"aa\", p = \"a*\"\n<strong>输出：</strong>true\n<strong>解释：</strong>因为 '*' 代表可以匹配零个或多个前面的那一个元素, 在这里前面的元素就是 'a'。因此，字符串 \"aa\" 可被视为 'a' 重复了一次。\n</pre>\n\n<p><strong>示例&nbsp;3：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"ab\", p = \".*\"\n<strong>输出：</strong>true\n<strong>解释：</strong>\".*\" 表示可匹配零个或多个（'*'）任意字符（'.'）。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length&nbsp;&lt;= 20</code></li>\n\t<li><code>1 &lt;= p.length&nbsp;&lt;= 30</code></li>\n\t<li><code>s</code>&nbsp;只包含从&nbsp;<code>a-z</code>&nbsp;的小写字母。</li>\n\t<li><code>p</code>&nbsp;只包含从&nbsp;<code>a-z</code>&nbsp;的小写字母，以及字符&nbsp;<code>.</code>&nbsp;和&nbsp;<code>*</code>。</li>\n\t<li>保证每次出现字符&nbsp;<code>*</code> 时，前面都匹配到有效的字符</li>\n</ul>\n",
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