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feat: update lc problems
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‎solution/0000-0099/0015.3Sum/README.md‎

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<!-- 这里写题目描述 -->
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<p>给你一个包含 <code>n</code> 个整数的数组<code>nums</code>,判断<code>nums</code>中是否存在三个元素 <em>a,b,c ,</em>使得<em>a + b + c = </em>0 ?请你找出所有和为 <code>0</code> 且不重复的三元组。</p>
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<p>给你一个包含 <code>n</code> 个整数的数组&nbsp;<code>nums</code>,判断&nbsp;<code>nums</code>&nbsp;中是否存在三个元素 <em>a,b,c ,</em>使得&nbsp;<em>a + b + c = </em>0 ?请你找出所有和为 <code>0</code> 且不重复的三元组。</p>
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<p><strong>注意:</strong>答案中不可以包含重复的三元组。</p>
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<p></p>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<p><strong>示例 2:</strong></p>
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<pre>
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<strong>输入:</strong>nums = []
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<strong>输入:</strong>nums = [0,1,1]
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<strong>输出:</strong>[]
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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>nums = [0]
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<strong>输出:</strong>[]
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<strong>输入:</strong>nums = [0,0,0]
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<strong>输出:</strong>[[0,0,0]]
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</pre>
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<p></p>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>0 <= nums.length <= 3000</code></li>
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<li><code>-10<sup>5</sup> <= nums[i] <= 10<sup>5</sup></code></li>
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<li><code>3 &lt;= nums.length &lt;= 3000</code></li>
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<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
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</ul>
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## 解法

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

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## Description
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<p>You are given a string <code>s</code> and an array of strings <code>words</code> of <strong>the same length</strong>. Return&nbsp;all starting indices of substring(s) in <code>s</code>&nbsp;that is a concatenation of each word in <code>words</code> <strong>exactly once</strong>, <strong>in any order</strong>,&nbsp;and <strong>without any intervening characters</strong>.</p>
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<p>You are given a string <code>s</code> and an array of strings <code>words</code>. All the strings of <code>words</code> are of <strong>the same length</strong>.</p>
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<p>You can return the answer in <strong>any order</strong>.</p>
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<p>A <strong>concatenated substring</strong> in <code>s</code> is a substring that contains all the strings of any permutation of <code>words</code> concatenated.</p>
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<ul>
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<li>For example, if <code>words = [&quot;ab&quot;,&quot;cd&quot;,&quot;ef&quot;]</code>, then <code>&quot;abcdef&quot;</code>, <code>&quot;abefcd&quot;</code>, <code>&quot;cdabef&quot;</code>, <code>&quot;cdefab&quot;</code>, <code>&quot;efabcd&quot;</code>, and <code>&quot;efcdab&quot;</code> are all concatenated strings. <code>&quot;acdbef&quot;</code> is not a concatenated substring because it is not the concatenation of any permutation of <code>words</code>.</li>
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</ul>
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<p>Return <em>the starting indices of all the concatenated substrings in </em><code>s</code>. You can return the answer in <strong>any order</strong>.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> s = &quot;barfoothefoobarman&quot;, words = [&quot;foo&quot;,&quot;bar&quot;]
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<strong>Output:</strong> [0,9]
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<strong>Explanation:</strong> Substrings starting at index 0 and 9 are &quot;barfoo&quot; and &quot;foobar&quot; respectively.
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The output order does not matter, returning [9,0] is fine too.
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<strong>Explanation:</strong> Since words.length == 2 and words[i].length == 3, the concatenated substring has to be of length 6.
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The substring starting at 0 is &quot;barfoo&quot;. It is the concatenation of [&quot;bar&quot;,&quot;foo&quot;] which is a permutation of words.
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The substring starting at 9 is &quot;foobar&quot;. It is the concatenation of [&quot;foo&quot;,&quot;bar&quot;] which is a permutation of words.
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The output order does not matter. Returning [9,0] is fine too.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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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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We return an empty array.
2637
</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> s = &quot;barfoofoobarthefoobarman&quot;, words = [&quot;bar&quot;,&quot;foo&quot;,&quot;the&quot;]
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<strong>Output:</strong> [6,9,12]
44+
<strong>Explanation:</strong> Since words.length == 3 and words[i].length == 3, the concatenated substring has to be of length 9.
45+
The substring starting at 6 is &quot;foobarthe&quot;. It is the concatenation of [&quot;foo&quot;,&quot;bar&quot;,&quot;the&quot;] which is a permutation of words.
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The substring starting at 9 is &quot;barthefoo&quot;. It is the concatenation of [&quot;bar&quot;,&quot;the&quot;,&quot;foo&quot;] which is a permutation of words.
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The substring starting at 12 is &quot;thefoobar&quot;. It is the concatenation of [&quot;the&quot;,&quot;foo&quot;,&quot;bar&quot;] which is a permutation of words.
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</pre>
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<p>&nbsp;</p>

‎solution/0000-0099/0041.First Missing Positive/README_EN.md‎

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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre><strong>Input:</strong> nums = [1,2,0]
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<pre>
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<strong>Input:</strong> nums = [1,2,0]
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<strong>Output:</strong> 3
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</pre><p><strong>Example 2:</strong></p>
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<pre><strong>Input:</strong> nums = [3,4,-1,1]
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<strong>Explanation:</strong> The numbers in the range [1,2] are all in the array.
18+
</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [3,4,-1,1]
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<strong>Output:</strong> 2
18-
</pre><p><strong>Example 3:</strong></p>
19-
<pre><strong>Input:</strong> nums = [7,8,9,11,12]
25+
<strong>Explanation:</strong> 1 is in the array but 2 is missing.
26+
</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [7,8,9,11,12]
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<strong>Output:</strong> 1
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<strong>Explanation:</strong> The smallest positive integer 1 is missing.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>5</sup></code></li>
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<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
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<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
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</ul>
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‎solution/0100-0199/0125.Valid Palindrome/README.md‎

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<!-- 这里写题目描述 -->
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<p>给定一个字符串,验证它是否是回文串,只考虑字母和数字字符,可以忽略字母的大小写。</p>
9+
<p>如果在将所有大写字符转换为小写字符、并移除所有非字母数字字符之后,短语正着读和反着读都一样。则可以认为该短语是一个回文串。</p>
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<p><strong>说明:</strong>本题中,我们将空字符串定义为有效的回文串。</p>
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<p>字母和数字都属于字母数字字符。</p>
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<p></p>
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<p>给你一个字符串 <code>s</code>,如果它是回文串,返回 <code>true</code><em> </em>;否则,返回<em> </em><code>false</code><em> </em>。</p>
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<p><strong>示例 1:</strong></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> "A man, a plan, a canal: Panama"
19-
<strong>输出:</strong> true
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<strong>解释:</strong>"amanaplanacanalpanama" 是回文串
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<strong>输出:</strong>true
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<strong>解释:</strong>"amanaplanacanalpanama" 是回文串。
23+
</pre>
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<p><strong>示例 2:</strong></p>
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<pre>
28+
<strong>输入:</strong>"race a car"
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<strong>输出:</strong>false
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<strong>解释:</strong>"raceacar" 不是回文串。
2131
</pre>
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<p><strong>示例 2:</strong></p>
33+
<p><strong>示例 3:</strong></p>
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<pre>
26-
<strong>输入:</strong> "race a car"
27-
<strong>输出:</strong> false
28-
<strong>解释:</strong>"raceacar" 不是回文串
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<strong>输入:</strong>s = " "
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<strong>输出:</strong>true
38+
<strong>解释:</strong>在移除非字母数字字符之后,s 是一个空字符串 "" 。
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由于空字符串正着反着读都一样,所以是回文串。
2940
</pre>
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<p></p>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>1 <= s.length <= 2 * 10<sup>5</sup></code></li>
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<li>字符串 <code>s</code> 由 ASCII 字符组成</li>
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<li><code>1 &lt;= s.length &lt;= 2 * 10<sup>5</sup></code></li>
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<li><code>s</code> 仅由可打印的 ASCII 字符组成</li>
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</ul>
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## 解法

‎solution/0100-0199/0151.Reverse Words in a String/README.md‎

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# [151. 颠倒字符串中的单词](https://leetcode.cn/problems/reverse-words-in-a-string)
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# [151. 反转字符串中的单词](https://leetcode.cn/problems/reverse-words-in-a-string)
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[English Version](/solution/0100-0199/0151.Reverse%20Words%20in%20a%20String/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>给你一个字符串 <code>s</code> ,颠倒字符串中 <strong>单词</strong> 的顺序。</p>
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<p>给你一个字符串 <code>s</code> ,请你反转字符串中 <strong>单词</strong> 的顺序。</p>
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<p><strong>单词</strong> 是由非空格字符组成的字符串。<code>s</code> 中使用至少一个空格将字符串中的 <strong>单词</strong> 分隔开。</p>
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<pre>
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<strong>输入:</strong>s = " &nbsp;hello world &nbsp;"
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<strong>输出:</strong>"world hello"
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<strong>解释:</strong>颠倒后的字符串中不能存在前导空格和尾随空格。
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<strong>解释:</strong>反转后的字符串中不能存在前导空格和尾随空格。
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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 = "a good &nbsp; example"
3838
<strong>输出:</strong>"example good a"
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<strong>解释:</strong>如果两个单词间有多余的空格,颠倒后的字符串需要将单词间的空格减少到仅有一个。
39+
<strong>解释:</strong>如果两个单词间有多余的空格,反转后的字符串需要将单词间的空格减少到仅有一个。
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</pre>
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<p>&nbsp;</p>

‎solution/0100-0199/0160.Intersection of Two Linked Lists/README_EN.md‎

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<li><code>skipB</code> - The number of nodes to skip ahead in <code>listB</code> (starting from the head) to get to the intersected node.</li>
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</ul>
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<p>The judge will then create the linked structure based on these inputs and pass the two heads, <code>headA</code> and <code>headB</code>&nbsp;to your program. If you correctly return the intersected node, then your solution will be <strong>accepted</strong>.</p>
27+
<p>The judge will then create the linked structure based on these inputs and pass the two heads, <code>headA</code> and <code>headB</code>to your program. If you correctly return the intersected node, then your solution will be <strong>accepted</strong>.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<strong>Output:</strong> Intersected at &#39;8&#39;
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<strong>Explanation:</strong> The intersected node&#39;s value is 8 (note that this must not be 0 if the two lists intersect).
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From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,6,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.
37+
- Note that the intersected node&#39;s value is not 1 because the nodes with value 1 in A and B (2<sup>nd</sup> node in A and 3<sup>rd</sup> node in B) are different node references. In other words, they point to two different locations in memory, while the nodes with value 8 in A and B (3<sup>rd</sup> node in A and 4<sup>th</sup> node in B) point to the same location in memory.
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</pre>
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<p><strong>Example 2:</strong></p>

‎solution/0100-0199/0186.Reverse Words in a String II/README.md‎

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# [186. 翻转字符串里的单词 II](https://leetcode.cn/problems/reverse-words-in-a-string-ii)
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# [186. 反转字符串中的单词 II](https://leetcode.cn/problems/reverse-words-in-a-string-ii)
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[English Version](/solution/0100-0199/0186.Reverse%20Words%20in%20a%20String%20II/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>给定一个字符串,逐个翻转字符串中的每个单词。</p>
9+
<p>给你一个字符数组 <code>s</code> ,反转其中 <strong>单词</strong> 的顺序。</p>
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<p><strong>示例:</strong></p>
11+
<p><strong>单词</strong> 的定义为:单词是一个由非空格字符组成的序列。<code>s</code> 中的单词将会由单个空格分隔。</p>
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<pre><strong>输入: </strong>[&quot;t&quot;,&quot;h&quot;,&quot;e&quot;,&quot; &quot;,&quot;s&quot;,&quot;k&quot;,&quot;y&quot;,&quot; &quot;,&quot;i&quot;,&quot;s&quot;,&quot; &quot;,&quot;b&quot;,&quot;l&quot;,&quot;u&quot;,&quot;e&quot;]
14-
<strong>输出: </strong>[&quot;b&quot;,&quot;l&quot;,&quot;u&quot;,&quot;e&quot;,&quot; &quot;,&quot;i&quot;,&quot;s&quot;,&quot; &quot;,&quot;s&quot;,&quot;k&quot;,&quot;y&quot;,&quot; &quot;,&quot;t&quot;,&quot;h&quot;,&quot;e&quot;]</pre>
13+
<div class="original__bRMd">
14+
<div>
15+
<p>必须设计并实现 <strong>原地</strong> 解法来解决此问题,即不分配额外的空间。</p>
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<p><strong>注意:</strong></p>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<pre>
22+
<strong>输入:</strong>s = ["t","h","e"," ","s","k","y"," ","i","s"," ","b","l","u","e"]
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<strong>输出:</strong>["b","l","u","e"," ","i","s"," ","s","k","y"," ","t","h","e"]
24+
</pre>
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<p><strong>示例 2:</strong></p>
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<pre>
29+
<strong>输入:</strong>s = ["a"]
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<strong>输出:</strong>["a"]
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</pre>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<ul>
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<li>单词的定义是不包含空格的一系列字符</li>
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<li>输入字符串中不会包含前置或尾随的空格</li>
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<li>单词与单词之间永远是以单个空格隔开的</li>
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<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
39+
<li><code>s[i]</code> 可以是一个英文字母(大写或小写)、数字、或是空格 <code>' '</code> 。</li>
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<li><code>s</code> 中至少存在一个单词</li>
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<li><code>s</code> 不含前导或尾随空格</li>
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<li>题目数据保证:<code>s</code> 中的每个单词都由单个空格分隔</li>
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</ul>
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<p><strong>进阶:</strong>使用&nbsp;<em>O</em>(1) 额外空间复杂度的原地解法。</p>
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</div>
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</div>
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## 解法
2748

‎solution/0200-0299/0277.Find the Celebrity/README.md‎

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ans not knows 4
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ans not knows 5
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ans not knows 6
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ans = 6
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ans = 6
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```
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这里 $ans$ 认识 3ドル,ドル说明 $ans$ 不是名人(即 0ドル$ 不是名人),那么名人会是 1ドル$ 或者 2ドル$ 吗?不会!因为若 1ドル$ 或者 2ドル$ 是名人,那么 0ドル$ 应该认识 1ドル$ 或者 2ドル$ 才对,与前面的例子冲突。因此,我们可以直接将 $ans$ 更新为 $i$。

‎solution/0400-0499/0421.Maximum XOR of Two Numbers in an Array/README.md‎

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<p>给你一个整数数组 <code>nums</code> ,返回<em> </em><code>nums[i] XOR nums[j]</code> 的最大运算结果,其中 <code>0 ≤ i ≤ j &lt; n</code> 。</p>
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<p><strong>进阶:</strong>你可以在 <code>O(n)</code> 的时间解决这个问题吗?</p>
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<p>&nbsp;</p>
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<div class="original__bRMd">
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<p><strong>示例 2:</strong></p>
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<pre>
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<strong>输入:</strong>nums = [0]
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<strong>输出:</strong>0
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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>nums = [2,4]
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<strong>输出:</strong>6
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</pre>
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<p><strong>示例 4:</strong></p>
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<pre>
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<strong>输入:</strong>nums = [8,10,2]
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<strong>输出:</strong>10
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</pre>
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<p><strong>示例 5:</strong></p>
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<pre>
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<strong>输入:</strong>nums = [14,70,53,83,49,91,36,80,92,51,66,70]
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<strong>输出:</strong>127

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AltStyle によって変換されたページ (->オリジナル) /