leetcode-problemset/leetcode-cn/originData/reverse-prefix-of-word.json

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        "question": {
            "questionId": "2128",
            "questionFrontendId": "2000",
            "categoryTitle": "Algorithms",
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            "title": "Reverse Prefix of Word",
            "titleSlug": "reverse-prefix-of-word",
            "content": "<p>Given a <strong>0-indexed</strong> string <code>word</code> and a character <code>ch</code>, <strong>reverse</strong> the segment of <code>word</code> that starts at index <code>0</code> and ends at the index of the <strong>first occurrence</strong> of <code>ch</code> (<strong>inclusive</strong>). If the character <code>ch</code> does not exist in <code>word</code>, do nothing.</p>\n\n<ul>\n\t<li>For example, if <code>word = &quot;abcdefd&quot;</code> and <code>ch = &quot;d&quot;</code>, then you should <strong>reverse</strong> the segment that starts at <code>0</code> and ends at <code>3</code> (<strong>inclusive</strong>). The resulting string will be <code>&quot;<u>dcba</u>efd&quot;</code>.</li>\n</ul>\n\n<p>Return <em>the resulting string</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> word = &quot;<u>abcd</u>efd&quot;, ch = &quot;d&quot;\n<strong>Output:</strong> &quot;<u>dcba</u>efd&quot;\n<strong>Explanation:</strong>&nbsp;The first occurrence of &quot;d&quot; is at index 3. \nReverse the part of word from 0 to 3 (inclusive), the resulting string is &quot;dcbaefd&quot;.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> word = &quot;<u>xyxz</u>xe&quot;, ch = &quot;z&quot;\n<strong>Output:</strong> &quot;<u>zxyx</u>xe&quot;\n<strong>Explanation:</strong>&nbsp;The first and only occurrence of &quot;z&quot; is at index 3.\nReverse the part of word from 0 to 3 (inclusive), the resulting string is &quot;zxyxxe&quot;.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> word = &quot;abcd&quot;, ch = &quot;z&quot;\n<strong>Output:</strong> &quot;abcd&quot;\n<strong>Explanation:</strong>&nbsp;&quot;z&quot; does not exist in word.\nYou should not do any reverse operation, the resulting string is &quot;abcd&quot;.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= word.length &lt;= 250</code></li>\n\t<li><code>word</code> consists of lowercase English letters.</li>\n\t<li><code>ch</code> is a lowercase English letter.</li>\n</ul>\n",
            "translatedTitle": "反转单词前缀",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong> 开始的字符串 <code>word</code> 和一个字符 <code>ch</code> 。找出 <code>ch</code> 第一次出现的下标 <code>i</code> ，<strong>反转 </strong><code>word</code> 中从下标 <code>0</code> 开始、直到下标 <code>i</code> 结束（含下标 <code>i</code> ）的那段字符。如果 <code>word</code> 中不存在字符 <code>ch</code> ，则无需进行任何操作。</p>\n\n<ul>\n\t<li>例如，如果 <code>word = \"abcdefd\"</code> 且 <code>ch = \"d\"</code> ，那么你应该 <strong>反转</strong> 从下标 0 开始、直到下标 <code>3</code> 结束（含下标 <code>3</code> ）。结果字符串将会是 <code>\"<em><strong>dcba</strong></em>efd\"</code> 。</li>\n</ul>\n\n<p>返回 <strong>结果字符串</strong> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><strong>输入：</strong>word = \"<em><strong>abcd</strong></em>efd\", ch = \"d\"\n<strong>输出：</strong>\"<em><strong>dcba</strong></em>efd\"\n<strong>解释：</strong>\"d\" 第一次出现在下标 3 。 \n反转从下标 0 到下标 3（含下标 3）的这段字符，结果字符串是 \"dcbaefd\" 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><strong>输入：</strong>word = \"<em><strong>xyxz</strong></em>xe\", ch = \"z\"\n<strong>输出：</strong>\"<em><strong>zxyx</strong></em>xe\"\n<strong>解释：</strong>\"z\" 第一次也是唯一一次出现是在下标 3 。\n反转从下标 0 到下标 3（含下标 3）的这段字符，结果字符串是 \"zxyxxe\" 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre><strong>输入：</strong>word = \"abcd\", ch = \"z\"\n<strong>输出：</strong>\"abcd\"\n<strong>解释：</strong>\"z\" 不存在于 word 中。\n无需执行反转操作，结果字符串是 \"abcd\" 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= word.length &lt;= 250</code></li>\n\t<li><code>word</code> 由小写英文字母组成</li>\n\t<li><code>ch</code> 是一个小写英文字母</li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    string reversePrefix(string word, char ch) {\n\n    }\n};",
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                    "code": "class Solution(object):\n    def reversePrefix(self, word, ch):\n        \"\"\"\n        :type word: str\n        :type ch: str\n        :rtype: str\n        \"\"\"",
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                    "code": "defmodule Solution do\n  @spec reverse_prefix(word :: String.t, ch :: char) :: String.t\n  def reverse_prefix(word, ch) do\n\n  end\nend",
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            "hints": [
                "Find the first index where ch appears.",
                "Find a way to reverse a substring of word."
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