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"question": {
"questionId": "841",
"questionFrontendId": "821",
"categoryTitle": "Algorithms",
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"title": "Shortest Distance to a Character",
"titleSlug": "shortest-distance-to-a-character",
"content": "<p>Given a string <code>s</code> and a character <code>c</code> that occurs in <code>s</code>, return <em>an array of integers </em><code>answer</code><em> where </em><code>answer.length == s.length</code><em> and </em><code>answer[i]</code><em> is the <strong>distance</strong> from index </em><code>i</code><em> to the <strong>closest</strong> occurrence of character </em><code>c</code><em> in </em><code>s</code>.</p>\n\n<p>The <strong>distance</strong> between two indices <code>i</code> and <code>j</code> is <code>abs(i - j)</code>, where <code>abs</code> is the absolute value function.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;loveleetcode&quot;, c = &quot;e&quot;\n<strong>Output:</strong> [3,2,1,0,1,0,0,1,2,2,1,0]\n<strong>Explanation:</strong> The character &#39;e&#39; appears at indices 3, 5, 6, and 11 (0-indexed).\nThe closest occurrence of &#39;e&#39; for index 0 is at index 3, so the distance is abs(0 - 3) = 3.\nThe closest occurrence of &#39;e&#39; for index 1 is at index 3, so the distance is abs(1 - 3) = 2.\nFor index 4, there is a tie between the &#39;e&#39; at index 3 and the &#39;e&#39; at index 5, but the distance is still the same: abs(4 - 3) == abs(4 - 5) = 1.\nThe closest occurrence of &#39;e&#39; for index 8 is at index 6, so the distance is abs(8 - 6) = 2.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;aaab&quot;, c = &quot;b&quot;\n<strong>Output:</strong> [3,2,1,0]\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>4</sup></code></li>\n\t<li><code>s[i]</code> and <code>c</code> are lowercase English letters.</li>\n\t<li>It is guaranteed that <code>c</code> occurs at least once in <code>s</code>.</li>\n</ul>\n",
"translatedTitle": "字符的最短距离",
"translatedContent": "<p>给你一个字符串 <code>s</code> 和一个字符 <code>c</code> ,且 <code>c</code> 是 <code>s</code> 中出现过的字符。</p>\n\n<p>返回一个整数数组 <code>answer</code> ,其中 <code>answer.length == s.length</code> 且 <code>answer[i]</code> 是 <code>s</code> 中从下标 <code>i</code> 到离它 <strong>最近</strong> 的字符 <code>c</code> 的 <strong>距离</strong> 。</p>\n\n<p>两个下标&nbsp;<code>i</code> 和 <code>j</code> 之间的 <strong>距离</strong> 为 <code>abs(i - j)</code> ,其中 <code>abs</code> 是绝对值函数。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<pre>\n<strong>输入:</strong>s = \"loveleetcode\", c = \"e\"\n<strong>输出:</strong>[3,2,1,0,1,0,0,1,2,2,1,0]\n<strong>解释:</strong>字符 'e' 出现在下标 3、5、6 和 11 处(下标从 0 开始计数)。\n距下标 0 最近的 'e' 出现在下标 3 ,所以距离为 abs(0 - 3) = 3 。\n距下标 1 最近的 'e' 出现在下标 3 ,所以距离为 abs(1 - 3) = 2 。\n对于下标 4 ,出现在下标 3 和下标 5 处的 'e' 都离它最近,但距离是一样的 abs(4 - 3) == abs(4 - 5) = 1 。\n距下标 8 最近的 'e' 出现在下标 6 ,所以距离为 abs(8 - 6) = 2 。\n</pre>\n\n<p><strong>示例 2</strong></p>\n\n<pre>\n<strong>输入:</strong>s = \"aaab\", c = \"b\"\n<strong>输出:</strong>[3,2,1,0]\n</pre>\n\n<p>&nbsp;</p>\n<strong>提示:</strong>\n\n<ul>\n\t<li><code>1 &lt;= s.length &lt;= 10<sup>4</sup></code></li>\n\t<li><code>s[i]</code> 和 <code>c</code> 均为小写英文字母</li>\n\t<li>题目数据保证 <code>c</code> 在 <code>s</code> 中至少出现一次</li>\n</ul>\n",
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