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chore: update lc problems (#1588)
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index.html

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const isEn = () => location.hash.includes('README_EN');
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const isRoot = () => ['', '#/', '#/README', '#/README_EN'].includes(location.hash);
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const sidebar = () => isRoot() ? false : (isEn() ? 'summary_en.md' : 'summary.md');
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const cleanedHtml = (html) => {
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return html.replace(/<pre>([\s\S]*?)<\/pre>/g, function (match, group) {
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return "<pre>" + group.replace(/<code>([\s\S]*?)<\/code>/g, "$1") + "</pre>";
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});
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};
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window.addEventListener('hashchange', () => {
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window.$docsify.loadSidebar = sidebar();
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const github = `[GitHub](${url})`
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const gitee = `[Gitee](${url.replace("github", "gitee")})`
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html = cleanedHtml(html);
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const editHtml = isEn() ? `:memo: Edit on ${github} / ${gitee}\n` : `:memo: 在 ${github} / ${gitee} 编辑\n`
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return editHtml + html
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})

solution/0000-0099/0003.Longest Substring Without Repeating Characters/README.md

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<pre>
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<strong>输入: </strong>s = "abcabcbb"
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<strong>输出: </strong>3
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<strong>解释:</strong> 因为无重复字符的最长子串是 "abc",所以其长度为 3。
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<strong>解释:</strong> 因为无重复字符的最长子串是 <code>"abc"</code>,所以其长度为 3。
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre>
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<strong>输入: </strong>s = "bbbbb"
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<strong>输出: </strong>1
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<strong>解释: </strong>因为无重复字符的最长子串是 "b",所以其长度为 1。
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<strong>解释: </strong>因为无重复字符的最长子串是 <code>"b"</code>,所以其长度为 1。
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</pre>
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<p><strong>示例 3:</strong></p>
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<pre>
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<strong>输入: </strong>s = "pwwkew"
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<strong>输出: </strong>3
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<strong>解释: </strong>因为无重复字符的最长子串是 "wke",所以其长度为 3。
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请注意,你的答案必须是 <strong>子串 </strong>的长度,"pwke" 是一个<em>子序列,</em>不是子串。
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<strong>解释: </strong>因为无重复字符的最长子串是&nbsp;<code>"wke"</code>,所以其长度为 3。
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&nbsp; 请注意,你的答案必须是 <strong>子串 </strong>的长度,<code>"pwke"</code>&nbsp;是一个<em>子序列,</em>不是子串。
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</pre>
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<p>&nbsp;</p>

solution/0000-0099/0030.Substring with Concatenation of All Words/README.md

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<li>例如,如果&nbsp;<code>words = ["ab","cd","ef"]</code>, 那么&nbsp;<code>"abcdef"</code>,&nbsp;<code>"abefcd"</code>,<code>"cdabef"</code>,&nbsp;<code>"cdefab"</code>,<code>"efabcd"</code>, 和&nbsp;<code>"efcdab"</code> 都是串联子串。&nbsp;<code>"acdbef"</code> 不是串联子串,因为他不是任何&nbsp;<code>words</code>&nbsp;排列的连接。</li>
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</ul>
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<p>返回所有串联字串在&nbsp;<code>s</code><strong>&nbsp;</strong>中的开始索引。你可以以 <strong>任意顺序</strong> 返回答案。</p>
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<p>返回所有串联子串在&nbsp;<code>s</code><strong>&nbsp;</strong>中的开始索引。你可以以 <strong>任意顺序</strong> 返回答案。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<pre>
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<strong>输入:</strong>s = "barfoothefoobarman", words = ["foo","bar"]
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<strong>输出:</strong>[0,9]
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<strong>输出:</strong><code>[0,9]</code>
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<strong>解释:</strong>因为 words.length == 2 同时 words[i].length == 3,连接的子字符串的长度必须为 6。
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子串 "barfoo" 开始位置是 0。它是 words 中以 ["bar","foo"] 顺序排列的连接。
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子串 "foobar" 开始位置是 9。它是 words 中以 ["foo","bar"] 顺序排列的连接。
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<pre>
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<strong>输入:</strong>s = "wordgoodgoodgoodbestword", words = ["word","good","best","word"]
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<strong>输出:</strong>[]
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<code><strong>输出:</strong>[]</code>
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<strong>解释:</strong>因为<strong> </strong>words.length == 4 并且 words[i].length == 4,所以串联子串的长度必须为 16。
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s 中没有子串长度为 16 并且等于 words 的任何顺序排列的连接。
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所以我们返回一个空数组。

solution/0000-0099/0030.Substring with Concatenation of All Words/README_EN.md

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<strong>Input:</strong> s = &quot;wordgoodgoodgoodbestword&quot;, words = [&quot;word&quot;,&quot;good&quot;,&quot;best&quot;,&quot;word&quot;]
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<strong>Output:</strong> []
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<strong>Explanation:</strong> Since words.length == 4 and words[i].length == 4, the concatenated substring has to be of length 16.
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There is no substring of length 16 is s that is equal to the concatenation of any permutation of words.
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There is no substring of length 16 in s that is equal to the concatenation of any permutation of words.
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We return an empty array.
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</pre>
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solution/0000-0099/0033.Search in Rotated Sorted Array/README.md

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<p><strong>示例 1:</strong></p>
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<pre>
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<strong>输入:</strong>nums = [4,5,6,7,0,1,2], target = 0
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<strong>输入:</strong>nums = [<code>4,5,6,7,0,1,2]</code>, target = 0
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<strong>输出:</strong>4
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</pre>
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<p><strong>示例&nbsp;2:</strong></p>
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<pre>
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<strong>输入:</strong>nums = [4,5,6,7,0,1,2], target = 3
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<strong>输入:</strong>nums = [<code>4,5,6,7,0,1,2]</code>, target = 3
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<strong>输出:</strong>-1</pre>
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<p><strong>示例 3:</strong></p>

solution/0000-0099/0034.Find First and Last Position of Element in Sorted Array/README.md

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<p><strong>示例 1:</strong></p>
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<pre>
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<strong>输入:</strong>nums = [5,7,7,8,8,10], target = 8
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<strong>输入:</strong>nums = [<code>5,7,7,8,8,10]</code>, target = 8
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<strong>输出:</strong>[3,4]</pre>
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<p><strong>示例&nbsp;2:</strong></p>
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<pre>
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<strong>输入:</strong>nums = [5,7,7,8,8,10], target = 6
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<strong>输入:</strong>nums = [<code>5,7,7,8,8,10]</code>, target = 6
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<strong>输出:</strong>[-1,-1]</pre>
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<p><strong>示例 3:</strong></p>

solution/0000-0099/0038.Count and Say/README.md

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4. 1211
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5. 111221
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第一项是数字 1
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描述前一项,这个数是 1 即 “ 一 个 1 ”,记作 "11"
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描述前一项,这个数是 11 即 “ 二 个 1 ” ,记作 "21"
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描述前一项,这个数是 21 即 “ 一 个 2 + 一 个 1 ” ,记作 "1211"
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描述前一项,这个数是 1211 即 “ 一 个 1 + 一 个 2 + 二 个 1 ” ,记作 "111221"
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描述前一项,这个数是 <code>1</code> 即 “ 一 个 1 ”,记作 <code>"11"
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</code>描述前一项,这个数是 <code>11</code> 即 “ 二 个 1 ” ,记作 <code>"21"
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</code>描述前一项,这个数是 <code>21</code> 即 “ 一 个 2 + 一 个 1 ” ,记作 "<code>1211"
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</code>描述前一项,这个数是 <code>1211</code> 即 “ 一 个 1 + 一 个 2 + 二 个 1 ” ,记作 "<code>111221"</code>
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</pre>
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<p>要 <strong>描述</strong> 一个数字字符串,首先要将字符串分割为 <strong>最小</strong> 数量的组,每个组都由连续的最多 <strong>相同字符</strong> 组成。然后对于每个组,先描述字符的数量,然后描述字符,形成一个描述组。要将描述转换为数字字符串,先将每组中的字符数量用数字替换,再将所有描述组连接起来。</p>

solution/0000-0099/0039.Combination Sum/README.md

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<p><strong>示例&nbsp;1:</strong></p>
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<pre>
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<strong>输入:</strong>candidates = [2,3,6,7], target = 7
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<strong>输入:</strong>candidates = <code>[2,3,6,7], </code>target = <code>7</code>
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<strong>输出:</strong>[[2,2,3],[7]]
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<strong>解释:</strong>
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2 和 3 可以形成一组候选,2 + 2 + 3 = 7 。注意 2 可以使用多次。
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<p><strong>示例&nbsp;2:</strong></p>
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<pre>
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<strong>输入: </strong>candidates = [2,3,5], target = 8
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<strong>输入: </strong>candidates = [2,3,5]<code>, </code>target = 8
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<strong>输出: </strong>[[2,2,2,2],[2,3,3],[3,5]]</pre>
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<p><strong>示例 3:</strong></p>
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<pre>
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<strong>输入: </strong>candidates = [2], target = 1
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<strong>输入: </strong>candidates = <code>[2], </code>target = 1
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<strong>输出: </strong>[]
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</pre>
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solution/0000-0099/0040.Combination Sum II/README.md

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<p><strong>示例&nbsp;1:</strong></p>
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<pre>
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<strong>输入:</strong> candidates =&nbsp;[10,1,2,7,6,1,5], target =&nbsp;8,
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<strong>输入:</strong> candidates =&nbsp;<code>[10,1,2,7,6,1,5]</code>, target =&nbsp;<code>8</code>,
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<strong>输出:</strong>
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[
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[1,1,6],

solution/0000-0099/0057.Insert Interval/README.md

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<pre>
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<strong>输入:</strong>intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
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<strong>输出:</strong>[[1,2],[3,10],[12,16]]
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<strong>解释:</strong>这是因为新的区间 [4,8] 与 [3,5],[6,7],[8,10] 重叠。</pre>
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<strong>解释:</strong>这是因为新的区间 <code>[4,8]</code><code>[3,5],[6,7],[8,10]</code> 重叠。</pre>
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<p><strong>示例 3:</strong></p>
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solution/0100-0199/0115.Distinct Subsequences/README.md

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<p><strong>示例&nbsp;1:</strong></p>
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<pre>
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<strong>输入:</strong>s = "rabbbit", t = "rabbit"
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<strong>输出</strong><strong>:</strong>3
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<strong>解释:</strong>
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如下所示, 有 3 种可以从 s 中得到 "rabbit" 的方案。
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<strong><u>rabb</u></strong>b<strong><u>it</u></strong>
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<strong><u>ra</u></strong>b<strong><u>bbit</u></strong>
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<strong><u>rab</u></strong>b<strong><u>bit</u></strong></pre>
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<strong>输入:</strong>s = "rabbbit", t = "rabbit"<code>
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<strong>输出</strong></code><strong>:</strong><code>3
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</code><strong>解释:</strong>
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如下所示, 有 3 种可以从 s 中得到 <code>"rabbit" 的方案</code>
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<code><strong><u>rabb</u></strong>b<strong><u>it</u></strong></code>
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<code><strong><u>ra</u></strong>b<strong><u>bbit</u></strong></code>
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<code><strong><u>rab</u></strong>b<strong><u>bit</u></strong></code></pre>
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<p><strong>示例&nbsp;2:</strong></p>
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<pre>
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<strong>输入:</strong>s = "babgbag", t = "bag"
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<strong>输出</strong><strong>:</strong>5
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<strong>解释:</strong>
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如下所示, 有 5 种可以从 s 中得到 "bag" 的方案。
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<strong><u>ba</u></strong>b<u><strong>g</strong></u>bag
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<strong><u>ba</u></strong>bgba<strong><u>g</u></strong>
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<u><strong>b</strong></u>abgb<strong><u>ag</u></strong>
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ba<u><strong>b</strong></u>gb<u><strong>ag</strong></u>
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babg<strong><u>bag</u></strong>
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<code><strong>输出</strong></code><strong>:</strong><code>5
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</code><strong>解释:</strong>
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如下所示, 有 5 种可以从 s 中得到 <code>"bag" 的方案</code>
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<code><strong><u>ba</u></strong>b<u><strong>g</strong></u>bag</code>
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<code><strong><u>ba</u></strong>bgba<strong><u>g</u></strong></code>
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<code><u><strong>b</strong></u>abgb<strong><u>ag</u></strong></code>
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<code>ba<u><strong>b</strong></u>gb<u><strong>ag</strong></u></code>
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<code>babg<strong><u>bag</u></strong></code>
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</pre>
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<p>&nbsp;</p>

solution/0100-0199/0128.Longest Consecutive Sequence/README.md

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<pre>
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<strong>输入:</strong>nums = [100,4,200,1,3,2]
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<strong>输出:</strong>4
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<strong>解释:</strong>最长数字连续序列是 [1, 2, 3, 4]。它的长度为 4。</pre>
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<strong>解释:</strong>最长数字连续序列是 <code>[1, 2, 3, 4]。它的长度为 4。</code></pre>
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<p><strong>示例 2:</strong></p>
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solution/0200-0299/0235.Lowest Common Ancestor of a Binary Search Tree/README.md

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<pre><strong>输入:</strong> root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
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<strong>输出:</strong> 6
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<strong>解释: </strong>节点 2 和节点 8 的最近公共祖先是 6。
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<strong>解释: </strong>节点 <code>2 </code>和节点 <code>8 </code>的最近公共祖先是 <code>6。</code>
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre><strong>输入:</strong> root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
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<strong>输出:</strong> 2
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<strong>解释: </strong>节点 2 和节点 4 的最近公共祖先是 2, 因为根据定义最近公共祖先节点可以为节点本身。</pre>
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<strong>解释: </strong>节点 <code>2</code> 和节点 <code>4</code> 的最近公共祖先是 <code>2</code>, 因为根据定义最近公共祖先节点可以为节点本身。</pre>
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<p>&nbsp;</p>
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solution/0200-0299/0236.Lowest Common Ancestor of a Binary Tree/README.md

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<pre>
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<strong>输入:</strong>root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
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<strong>输出:</strong>3
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<strong>解释:</strong>节点 5 和节点 1 的最近公共祖先是节点 3 。
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<strong>解释:</strong>节点 <code>5 </code>和节点 <code>1 </code>的最近公共祖先是节点 <code>3 。</code>
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</pre>
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<p><strong>示例 2:</strong></p>
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0236.Lowest%20Common%20Ancestor%20of%20a%20Binary%20Tree/images/binarytree.png" style="width: 200px; height: 190px;" />
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<pre>
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<strong>输入:</strong>root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
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<strong>输出:</strong>5
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<strong>解释:</strong>节点 5 和节点 4 的最近公共祖先是节点 5 。因为根据定义最近公共祖先节点可以为节点本身。
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<strong>解释:</strong>节点 <code>5 </code>和节点 <code>4 </code>的最近公共祖先是节点 <code>5 。</code>因为根据定义最近公共祖先节点可以为节点本身。
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</pre>
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<p><strong>示例 3:</strong></p>

solution/0200-0299/0238.Product of Array Except Self/README.md

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<p><strong>示例 1:</strong></p>
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<pre>
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<strong>输入:</strong> nums = [1,2,3,4]
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<strong>输出:</strong> [24,12,8,6]
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<strong>输入:</strong> nums = <code>[1,2,3,4]</code>
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<strong>输出:</strong> <code>[24,12,8,6]</code>
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</pre>
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<p><strong>示例 2:</strong></p>

solution/0200-0299/0258.Add Digits/README.md

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<p><strong>示例 1:</strong></p>
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<pre>
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<strong>输入:</strong> num =<strong> </strong>38
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<strong>输入:</strong> num =<strong> </strong><code>38</code>
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<strong>输出:</strong> 2
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<strong>解释: </strong>各位相加的过程为<strong>:
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</strong>38 --&gt; 3 + 8 --&gt; 11
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11 --&gt; 1 + 1 --&gt; 2
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由于&nbsp;2 是一位数,所以返回 2。
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由于&nbsp;<code>2</code> 是一位数,所以返回 2。
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</pre>
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<p><strong>示例 2:</strong></p>

solution/0200-0299/0259.3Sum Smaller/README.md

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<p><strong>示例 1:</strong></p>
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<pre>
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<strong>输入: </strong><em>nums</em> = [-2,0,1,3], <em>target</em> = 2
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<strong>输入: </strong><em>nums</em> = <code>[-2,0,1,3]</code>, <em>target</em> = 2
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<strong>输出: </strong>2
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<strong>解释: </strong>因为一共有两个三元组满足累加和小于 2:
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&nbsp; [-2,0,1]
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<p><strong>示例 2:</strong></p>
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<pre>
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<strong>输入: </strong><em>nums</em> = [], <em>target</em> = 0
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<strong>输入: </strong><em>nums</em> = <code>[]</code>, <em>target</em> = 0
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<strong>输出: </strong>0 </pre>
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<p><strong>示例 3:</strong></p>
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<pre>
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<strong>输入: </strong><em>nums</em> = [0], <em>target</em> = 0
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<strong>输入: </strong><em>nums</em> = <code>[0]</code>, <em>target</em> = 0
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<strong>输出: </strong>0 </pre>
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<p>&nbsp;</p>

solution/0200-0299/0261.Graph Valid Tree/README.md

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<p><img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0261.Graph%20Valid%20Tree/images/tree1-graph.jpg" /></p>
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<pre>
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<strong>输入:</strong> n = 5, edges = [[0,1],[0,2],[0,3],[1,4]]
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<strong>输入:</strong> <code>n = 5</code>, edges<code> = [[0,1],[0,2],[0,3],[1,4]]</code>
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<strong>输出:</strong> true</pre>
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<p><strong>示例 2:</strong></p>
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<p><img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0261.Graph%20Valid%20Tree/images/tree2-graph.jpg" /></p>
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<pre>
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<strong>输入:</strong> n = 5, edges = [[0,1],[1,2],[2,3],[1,3],[1,4]]
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<strong>输入:</strong> <code>n = 5, </code>edges<code> = [[0,1],[1,2],[2,3],[1,3],[1,4]]</code>
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<strong>输出:</strong> false</pre>
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<p>&nbsp;</p>

solution/0200-0299/0263.Ugly Number/README.md

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@@ -31,7 +31,7 @@
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<pre>
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<strong>输入:</strong>n = 14
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<strong>输出:</strong>false
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<strong>解释:</strong>14 不是丑数,因为它包含了另外一个质因数&nbsp;7
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<strong>解释:</strong>14 不是丑数,因为它包含了另外一个质因数&nbsp;<code>7 </code>
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</pre>
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<p>&nbsp;</p>

solution/0200-0299/0266.Palindrome Permutation/README.md

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@@ -10,17 +10,17 @@
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<p><strong>示例 1:</strong></p>
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<pre><strong>输入:</strong> &quot;code&quot;
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<pre><strong>输入:</strong> <code>&quot;code&quot;</code>
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<strong>输出:</strong> false</pre>
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<p><strong>示例 2:</strong></p>
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<pre><strong>输入:</strong> &quot;aab&quot;
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<pre><strong>输入:</strong> <code>&quot;aab&quot;</code>
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<strong>输出:</strong> true</pre>
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<p><strong>示例 3:</strong></p>
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<pre><strong>输入:</strong> &quot;carerac&quot;
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<pre><strong>输入:</strong> <code>&quot;carerac&quot;</code>
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<strong>输出:</strong> true</pre>
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## 解法

solution/0200-0299/0267.Palindrome Permutation II/README.md

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@@ -15,14 +15,14 @@
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<p><strong>示例 1:</strong></p>
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<pre>
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<strong>输入: </strong>s = "aabb"
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<strong>输出: </strong>["abba", "baab"]></pre>
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<strong>输入: </strong>s = <code>"aabb"</code>
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<strong>输出: </strong><code>["abba", "baab"]</code></pre>
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<p><strong>示例 2:</strong></p>
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<pre>
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<strong>输入: </strong>s = "abc"
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<strong>输出: </strong>[]
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<strong>输入: </strong>s = <code>"abc"</code>
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<strong>输出: </strong><code>[]</code>
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</pre>
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<p>&nbsp;</p>

solution/0200-0299/0269.Alien Dictionary/README.md

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@@ -35,7 +35,7 @@
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<pre>
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<strong>输入:</strong>words = ["z","x","z"]
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<strong>输出:</strong>""
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<strong>解释:</strong>不存在合法字母顺序,因此返回 "" 。
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<strong>解释:</strong>不存在合法字母顺序,因此返回 <code>"" 。</code>
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</pre>
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<p>&nbsp;</p>

solution/0200-0299/0269.Alien Dictionary/README_EN.md

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<pre>
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<strong>Input:</strong> words = [&quot;z&quot;,&quot;x&quot;,&quot;z&quot;]
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<strong>Output:</strong> &quot;&quot;
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<strong>Explanation:</strong> The order is invalid, so return &quot;&quot;.
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<strong>Explanation:</strong> The order is invalid, so return <code>&quot;&quot;</code>.
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</pre>
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<p>&nbsp;</p>

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