leetcode-problemset/leetcode-cn/originData/evaluate-the-bracket-pairs-of-a-string.json

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        "question": {
            "questionId": "1934",
            "questionFrontendId": "1807",
            "categoryTitle": "Algorithms",
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            "title": "Evaluate the Bracket Pairs of a String",
            "titleSlug": "evaluate-the-bracket-pairs-of-a-string",
            "content": "<p>You are given a string <code>s</code> that contains some bracket pairs, with each pair containing a <strong>non-empty</strong> key.</p>\n\n<ul>\n\t<li>For example, in the string <code>&quot;(name)is(age)yearsold&quot;</code>, there are <strong>two</strong> bracket pairs that contain the keys <code>&quot;name&quot;</code> and <code>&quot;age&quot;</code>.</li>\n</ul>\n\n<p>You know the values of a wide range of keys. This is represented by a 2D string array <code>knowledge</code> where each <code>knowledge[i] = [key<sub>i</sub>, value<sub>i</sub>]</code> indicates that key <code>key<sub>i</sub></code> has a value of <code>value<sub>i</sub></code>.</p>\n\n<p>You are tasked to evaluate <strong>all</strong> of the bracket pairs. When you evaluate a bracket pair that contains some key <code>key<sub>i</sub></code>, you will:</p>\n\n<ul>\n\t<li>Replace <code>key<sub>i</sub></code> and the bracket pair with the key&#39;s corresponding <code>value<sub>i</sub></code>.</li>\n\t<li>If you do not know the value of the key, you will replace <code>key<sub>i</sub></code> and the bracket pair with a question mark <code>&quot;?&quot;</code> (without the quotation marks).</li>\n</ul>\n\n<p>Each key will appear at most once in your <code>knowledge</code>. There will not be any nested brackets in <code>s</code>.</p>\n\n<p>Return <em>the resulting string after evaluating <strong>all</strong> of the bracket pairs.</em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;(name)is(age)yearsold&quot;, knowledge = [[&quot;name&quot;,&quot;bob&quot;],[&quot;age&quot;,&quot;two&quot;]]\n<strong>Output:</strong> &quot;bobistwoyearsold&quot;\n<strong>Explanation:</strong>\nThe key &quot;name&quot; has a value of &quot;bob&quot;, so replace &quot;(name)&quot; with &quot;bob&quot;.\nThe key &quot;age&quot; has a value of &quot;two&quot;, so replace &quot;(age)&quot; with &quot;two&quot;.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;hi(name)&quot;, knowledge = [[&quot;a&quot;,&quot;b&quot;]]\n<strong>Output:</strong> &quot;hi?&quot;\n<strong>Explanation:</strong> As you do not know the value of the key &quot;name&quot;, replace &quot;(name)&quot; with &quot;?&quot;.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;(a)(a)(a)aaa&quot;, knowledge = [[&quot;a&quot;,&quot;yes&quot;]]\n<strong>Output:</strong> &quot;yesyesyesaaa&quot;\n<strong>Explanation:</strong> The same key can appear multiple times.\nThe key &quot;a&quot; has a value of &quot;yes&quot;, so replace all occurrences of &quot;(a)&quot; with &quot;yes&quot;.\nNotice that the &quot;a&quot;s not in a bracket pair are not evaluated.\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>0 &lt;= knowledge.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>knowledge[i].length == 2</code></li>\n\t<li><code>1 &lt;= key<sub>i</sub>.length, value<sub>i</sub>.length &lt;= 10</code></li>\n\t<li><code>s</code> consists of lowercase English letters and round brackets <code>&#39;(&#39;</code> and <code>&#39;)&#39;</code>.</li>\n\t<li>Every open bracket <code>&#39;(&#39;</code> in <code>s</code> will have a corresponding close bracket <code>&#39;)&#39;</code>.</li>\n\t<li>The key in each bracket pair of <code>s</code> will be non-empty.</li>\n\t<li>There will not be any nested bracket pairs in <code>s</code>.</li>\n\t<li><code>key<sub>i</sub></code> and <code>value<sub>i</sub></code> consist of lowercase English letters.</li>\n\t<li>Each <code>key<sub>i</sub></code> in <code>knowledge</code> is unique.</li>\n</ul>\n",
            "translatedTitle": "替换字符串中的括号内容",
            "translatedContent": "<p>给你一个字符串&nbsp;<code>s</code>&nbsp;，它包含一些括号对，每个括号中包含一个 <strong>非空</strong>&nbsp;的键。</p>\n\n<ul>\n\t<li>比方说，字符串&nbsp;<code>\"(name)is(age)yearsold\"</code>&nbsp;中，有&nbsp;<strong>两个</strong>&nbsp;括号对，分别包含键&nbsp;<code>\"name\"</code> 和&nbsp;<code>\"age\"</code>&nbsp;。</li>\n</ul>\n\n<p>你知道许多键对应的值，这些关系由二维字符串数组&nbsp;<code>knowledge</code>&nbsp;表示，其中&nbsp;<code>knowledge[i] = [key<sub>i</sub>, value<sub>i</sub>]</code>&nbsp;，表示键&nbsp;<code>key<sub>i</sub></code>&nbsp;对应的值为&nbsp;<code>value<sub>i</sub></code><sub>&nbsp;</sub>。</p>\n\n<p>你需要替换 <strong>所有</strong>&nbsp;的括号对。当你替换一个括号对，且它包含的键为&nbsp;<code>key<sub>i</sub></code>&nbsp;时，你需要：</p>\n\n<ul>\n\t<li>将&nbsp;<code>key<sub>i</sub></code>&nbsp;和括号用对应的值&nbsp;<code>value<sub>i</sub></code>&nbsp;替换。</li>\n\t<li>如果从 <code>knowledge</code>&nbsp;中无法得知某个键对应的值，你需要将&nbsp;<code>key<sub>i</sub></code>&nbsp;和括号用问号&nbsp;<code>\"?\"</code>&nbsp;替换（不需要引号）。</li>\n</ul>\n\n<p><code>knowledge</code>&nbsp;中每个键最多只会出现一次。<code>s</code>&nbsp;中不会有嵌套的括号。</p>\n\n<p>请你返回替换 <strong>所有</strong>&nbsp;括号对后的结果字符串。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>s = \"(name)is(age)yearsold\", knowledge = [[\"name\",\"bob\"],[\"age\",\"two\"]]\n<b>输出：</b>\"bobistwoyearsold\"\n<strong>解释：</strong>\n键 \"name\" 对应的值为 \"bob\" ，所以将 \"(name)\" 替换为 \"bob\" 。\n键 \"age\" 对应的值为 \"two\" ，所以将 \"(age)\" 替换为 \"two\" 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>s = \"hi(name)\", knowledge = [[\"a\",\"b\"]]\n<b>输出：</b>\"hi?\"\n<b>解释：</b>由于不知道键 \"name\" 对应的值，所以用 \"?\" 替换 \"(name)\" 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<b>输入：</b>s = \"(a)(a)(a)aaa\", knowledge = [[\"a\",\"yes\"]]\n<b>输出：</b>\"yesyesyesaaa\"\n<b>解释：</b>相同的键在 s 中可能会出现多次。\n键 \"a\" 对应的值为 \"yes\" ，所以将所有的 \"(a)\" 替换为 \"yes\" 。\n注意，不在括号里的 \"a\" 不需要被替换。\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 &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= knowledge.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>knowledge[i].length == 2</code></li>\n\t<li><code>1 &lt;= key<sub>i</sub>.length, value<sub>i</sub>.length &lt;= 10</code></li>\n\t<li><code>s</code>&nbsp;只包含小写英文字母和圆括号&nbsp;<code>'('</code>&nbsp;和&nbsp;<code>')'</code>&nbsp;。</li>\n\t<li><code>s</code>&nbsp;中每一个左圆括号&nbsp;<code>'('</code>&nbsp;都有对应的右圆括号&nbsp;<code>')'</code>&nbsp;。</li>\n\t<li><code>s</code>&nbsp;中每对括号内的键都不会为空。</li>\n\t<li><code>s</code>&nbsp;中不会有嵌套括号对。</li>\n\t<li><code>key<sub>i</sub></code>&nbsp;和&nbsp;<code>value<sub>i</sub></code>&nbsp;只包含小写英文字母。</li>\n\t<li><code>knowledge</code>&nbsp;中的&nbsp;<code>key<sub>i</sub></code>&nbsp;不会重复。</li>\n</ul>\n",
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href=\\\"https:\\/\\/github.com\\/emirpasic\\/gods\\/tree\\/v1.18.1\\\" target=\\\"_blank\\\">https:\\/\\/godoc.org\\/github.com\\/emirpasic\\/gods@v1.18.1<\\/a> \\u7b2c\\u4e09\\u65b9\\u5e93\\u3002<\\/p>\"],\"python3\":[\"Python3\",\"<p>\\u7248\\u672c\\uff1a<code>Python 3.10<\\/code><\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982<a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/array.html\\\" target=\\\"_blank\\\">array<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/bisect.html\\\" target=\\\"_blank\\\">bisect<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/collections.html\\\" target=\\\"_blank\\\">collections<\\/a>\\u3002 \\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5982\\u9700\\u4f7f\\u7528 Map\\/TreeMap 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