leetcode-problemset/leetcode-cn/originData/remove-outermost-parentheses.json

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        "question": {
            "questionId": "1078",
            "questionFrontendId": "1021",
            "categoryTitle": "Algorithms",
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            "title": "Remove Outermost Parentheses",
            "titleSlug": "remove-outermost-parentheses",
            "content": "<p>A valid parentheses string is either empty <code>&quot;&quot;</code>, <code>&quot;(&quot; + A + &quot;)&quot;</code>, or <code>A + B</code>, where <code>A</code> and <code>B</code> are valid parentheses strings, and <code>+</code> represents string concatenation.</p>\n\n<ul>\n\t<li>For example, <code>&quot;&quot;</code>, <code>&quot;()&quot;</code>, <code>&quot;(())()&quot;</code>, and <code>&quot;(()(()))&quot;</code> are all valid parentheses strings.</li>\n</ul>\n\n<p>A valid parentheses string <code>s</code> is primitive if it is nonempty, and there does not exist a way to split it into <code>s = A + B</code>, with <code>A</code> and <code>B</code> nonempty valid parentheses strings.</p>\n\n<p>Given a valid parentheses string <code>s</code>, consider its primitive decomposition: <code>s = P<sub>1</sub> + P<sub>2</sub> + ... + P<sub>k</sub></code>, where <code>P<sub>i</sub></code> are primitive valid parentheses strings.</p>\n\n<p>Return <code>s</code> <em>after removing the outermost parentheses of every primitive string in the primitive decomposition of </em><code>s</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;(()())(())&quot;\n<strong>Output:</strong> &quot;()()()&quot;\n<strong>Explanation:</strong> \nThe input string is &quot;(()())(())&quot;, with primitive decomposition &quot;(()())&quot; + &quot;(())&quot;.\nAfter removing outer parentheses of each part, this is &quot;()()&quot; + &quot;()&quot; = &quot;()()()&quot;.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;(()())(())(()(()))&quot;\n<strong>Output:</strong> &quot;()()()()(())&quot;\n<strong>Explanation:</strong> \nThe input string is &quot;(()())(())(()(()))&quot;, with primitive decomposition &quot;(()())&quot; + &quot;(())&quot; + &quot;(()(()))&quot;.\nAfter removing outer parentheses of each part, this is &quot;()()&quot; + &quot;()&quot; + &quot;()(())&quot; = &quot;()()()()(())&quot;.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;()()&quot;\n<strong>Output:</strong> &quot;&quot;\n<strong>Explanation:</strong> \nThe input string is &quot;()()&quot;, with primitive decomposition &quot;()&quot; + &quot;()&quot;.\nAfter removing outer parentheses of each part, this is &quot;&quot; + &quot;&quot; = &quot;&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 &lt;= 10<sup>5</sup></code></li>\n\t<li><code>s[i]</code> is either <code>&#39;(&#39;</code> or <code>&#39;)&#39;</code>.</li>\n\t<li><code>s</code> is a valid parentheses string.</li>\n</ul>\n",
            "translatedTitle": "删除最外层的括号",
            "translatedContent": "<p>有效括号字符串为空 <code>\"\"</code>、<code>\"(\" + A + \")\"</code> 或 <code>A + B</code> ，其中 <code>A</code> 和 <code>B</code> 都是有效的括号字符串，<code>+</code> 代表字符串的连接。</p>\n\n<ul>\n\t<li>例如，<code>\"\"</code>，<code>\"()\"</code>，<code>\"(())()\"</code> 和 <code>\"(()(()))\"</code> 都是有效的括号字符串。</li>\n</ul>\n\n<p>如果有效字符串 <code>s</code> 非空，且不存在将其拆分为 <code>s = A + B</code> 的方法，我们称其为<strong>原语（primitive）</strong>，其中 <code>A</code> 和 <code>B</code> 都是非空有效括号字符串。</p>\n\n<p>给出一个非空有效字符串 <code>s</code>，考虑将其进行原语化分解，使得：<code>s = P_1 + P_2 + ... + P_k</code>，其中 <code>P_i</code> 是有效括号字符串原语。</p>\n\n<p>对 <code>s</code> 进行原语化分解，删除分解中每个原语字符串的最外层括号，返回 <code>s</code> 。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"(()())(())\"\n<strong>输出：</strong>\"()()()\"\n<strong>解释：\n</strong>输入字符串为 \"(()())(())\"，原语化分解得到 \"(()())\" + \"(())\"，\n删除每个部分中的最外层括号后得到 \"()()\" + \"()\" = \"()()()\"。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"(()())(())(()(()))\"\n<strong>输出：</strong>\"()()()()(())\"\n<strong>解释：</strong>\n输入字符串为 \"(()())(())(()(()))\"，原语化分解得到 \"(()())\" + \"(())\" + \"(()(()))\"，\n删除每个部分中的最外层括号后得到 \"()()\" + \"()\" + \"()(())\" = \"()()()()(())\"。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"()()\"\n<strong>输出：</strong>\"\"\n<strong>解释：</strong>\n输入字符串为 \"()()\"，原语化分解得到 \"()\" + \"()\"，\n删除每个部分中的最外层括号后得到 \"\" + \"\" = \"\"。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= s.length <= 10<sup>5</sup></code></li>\n\t<li><code>s[i]</code> 为 <code>'('</code> 或 <code>')'</code></li>\n\t<li><code>s</code> 是一个有效括号字符串</li>\n</ul>\n",
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