update

Author: coder-xiaomo

README.md

@@ -1,6 +1,6 @@
 # 力扣题库（完整版）
 
-> 最后更新日期： **2022.10.26**
+> 最后更新日期： **2022.11.01**
 >
 > 使用脚本前请务必仔细完整阅读本 `README.md` 文件
 

leetcode-cn/origin-data.json

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+<p>给你一个下标从 <strong>0</strong>&nbsp;开始的非负整数数组&nbsp;<code>nums</code>&nbsp;。对于&nbsp;<code>nums</code>&nbsp;中每一个整数，你必须找到对应元素的&nbsp;<strong>第二大</strong>&nbsp;整数。</p>
+
+<p>如果&nbsp;<code>nums[j]</code>&nbsp;满足以下条件，那么我们称它为&nbsp;<code>nums[i]</code>&nbsp;的&nbsp;<strong>第二大</strong>&nbsp;整数：</p>
+
+<ul>
+	<li><code>j &gt; i</code></li>
+	<li><code>nums[j] &gt; nums[i]</code></li>
+	<li>恰好存在 <strong>一个</strong>&nbsp;<code>k</code>&nbsp;满足 <code>i &lt; k &lt; j</code>&nbsp;且&nbsp;<code>nums[k] &gt; nums[i]</code>&nbsp;。</li>
+</ul>
+
+<p>如果不存在&nbsp;<code>nums[j]</code>&nbsp;，那么第二大整数为&nbsp;<code>-1</code>&nbsp;。</p>
+
+<ul>
+	<li>比方说，数组&nbsp;<code>[1, 2, 4, 3]</code>&nbsp;中，<code>1</code>&nbsp;的第二大整数是&nbsp;<code>4</code>&nbsp;，<code>2</code>&nbsp;的第二大整数是&nbsp;<code>3</code>&nbsp;，<code>3</code> 和&nbsp;<code>4</code>&nbsp;的第二大整数是&nbsp;<code>-1</code>&nbsp;。</li>
+</ul>
+
+<p>请你返回一个整数数组<em>&nbsp;</em><code>answer</code>&nbsp;，其中<em>&nbsp;</em><code>answer[i]</code>是<em>&nbsp;</em><code>nums[i]</code>&nbsp;的第二大整数。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<b>输入：</b>nums = [2,4,0,9,6]
+<b>输出：</b>[9,6,6,-1,-1]
+<strong>解释：</strong>
+下标为 0 处：2 的右边，4 是大于 2 的第一个整数，9 是第二个大于 2 的整数。
+下标为 1 处：4 的右边，9 是大于 4 的第一个整数，6 是第二个大于 4 的整数。
+下标为 2 处：0 的右边，9 是大于 0 的第一个整数，6 是第二个大于 0 的整数。
+下标为 3 处：右边不存在大于 9 的整数，所以第二大整数为 -1 。
+下标为 4 处：右边不存在大于 6 的整数，所以第二大整数为 -1 。
+所以我们返回 [9,6,6,-1,-1] 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [3,3]
+<b>输出：</b>[-1,-1]
+<strong>解释：</strong>
+由于每个数右边都没有更大的数，所以我们返回 [-1,-1] 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
+</ul>

leetcode-cn/originData/next-greater-element-iv.json

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+<p>给你两个字符串数组&nbsp;<code>queries</code> 和&nbsp;<code>dictionary</code>&nbsp;。数组中所有单词都只包含小写英文字母，且长度都相同。</p>
+
+<p>一次 <strong>编辑</strong>&nbsp;中，你可以从 <code>queries</code>&nbsp;中选择一个单词，将任意一个字母修改成任何其他字母。从&nbsp;<code>queries</code>&nbsp;中找到所有满足以下条件的字符串：<strong>不超过</strong>&nbsp;两次编辑内，字符串与&nbsp;<code>dictionary</code>&nbsp;中某个字符串相同。</p>
+
+<p>请你返回<em>&nbsp;</em><code>queries</code>&nbsp;中的单词列表，这些单词距离&nbsp;<code>dictionary</code>&nbsp;中的单词&nbsp;<strong>编辑次数</strong>&nbsp;不超过&nbsp;<strong>两次</strong>&nbsp;。单词返回的顺序需要与&nbsp;<code>queries</code>&nbsp;中原本顺序相同。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><b>输入：</b>queries = ["word","note","ants","wood"], dictionary = ["wood","joke","moat"]
+<b>输出：</b>["word","note","wood"]
+<strong>解释：</strong>
+- 将 "word" 中的 'r' 换成 'o' ，得到 dictionary 中的单词 "wood" 。
+- 将 "note" 中的 'n' 换成 'j' 且将 't' 换成 'k' ，得到 "joke" 。
+- "ants" 需要超过 2 次编辑才能得到 dictionary 中的单词。
+- "wood" 不需要修改（0 次编辑），就得到 dictionary 中相同的单词。
+所以我们返回 ["word","note","wood"] 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><b>输入：</b>queries = ["yes"], dictionary = ["not"]
+<b>输出：</b>[]
+<strong>解释：</strong>
+"yes" 需要超过 2 次编辑才能得到 "not" 。
+所以我们返回空数组。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= queries.length, dictionary.length &lt;= 100</code></li>
+	<li><code>n == queries[i].length == dictionary[j].length</code></li>
+	<li><code>1 &lt;= n &lt;= 100</code></li>
+	<li>所有&nbsp;<code>queries[i]</code> 和&nbsp;<code>dictionary[j]</code>&nbsp;都只包含小写英文字母。</li>
+</ul>

leetcode-cn/originData/words-within-two-edits-of-dictionary.json

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+<p>You are given a <strong>0-indexed</strong> array of non-negative integers <code>nums</code>. For each integer in <code>nums</code>, you must find its respective <strong>second greater</strong> integer.</p>
+
+<p>The <strong>second greater</strong> integer of <code>nums[i]</code> is <code>nums[j]</code> such that:</p>
+
+<ul>
+	<li><code>j &gt; i</code></li>
+	<li><code>nums[j] &gt; nums[i]</code></li>
+	<li>There exists <strong>exactly one</strong> index <code>k</code> such that <code>nums[k] &gt; nums[i]</code> and <code>i &lt; k &lt; j</code>.</li>
+</ul>
+
+<p>If there is no such <code>nums[j]</code>, the second greater integer is considered to be <code>-1</code>.</p>
+
+<ul>
+	<li>For example, in the array <code>[1, 2, 4, 3]</code>, the second greater integer of <code>1</code> is <code>4</code>, <code>2</code> is <code>3</code>,&nbsp;and that of <code>3</code> and <code>4</code> is <code>-1</code>.</li>
+</ul>
+
+<p>Return<em> an integer array </em><code>answer</code><em>, where </em><code>answer[i]</code><em> is the second greater integer of </em><code>nums[i]</code><em>.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [2,4,0,9,6]
+<strong>Output:</strong> [9,6,6,-1,-1]
+<strong>Explanation:</strong>
+0th index: 4 is the first integer greater than 2, and 9 is the second integer greater than 2, to the right of 2.
+1st index: 9 is the first, and 6 is the second integer greater than 4, to the right of 4.
+2nd index: 9 is the first, and 6 is the second integer greater than 0, to the right of 0.
+3rd index: There is no integer greater than 9 to its right, so the second greater integer is considered to be -1.
+4th index: There is no integer greater than 6 to its right, so the second greater integer is considered to be -1.
+Thus, we return [9,6,6,-1,-1].
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [3,3]
+<strong>Output:</strong> [-1,-1]
+<strong>Explanation:</strong>
+We return [-1,-1] since neither integer has any integer greater than it.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
+</ul>

leetcode-cn/problem (Chinese)/下一个更大元素 IV [next-greater-element-iv].html

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+<p>You are given two string arrays, <code>queries</code> and <code>dictionary</code>. All words in each array comprise of lowercase English letters and have the same length.</p>
+
+<p>In one <strong>edit</strong> you can take a word from <code>queries</code>, and change any letter in it to any other letter. Find all words from <code>queries</code> that, after a <strong>maximum</strong> of two edits, equal some word from <code>dictionary</code>.</p>
+
+<p>Return<em> a list of all words from </em><code>queries</code><em>, </em><em>that match with some word from </em><code>dictionary</code><em> after a maximum of <strong>two edits</strong></em>. Return the words in the <strong>same order</strong> they appear in <code>queries</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> queries = [&quot;word&quot;,&quot;note&quot;,&quot;ants&quot;,&quot;wood&quot;], dictionary = [&quot;wood&quot;,&quot;joke&quot;,&quot;moat&quot;]
+<strong>Output:</strong> [&quot;word&quot;,&quot;note&quot;,&quot;wood&quot;]
+<strong>Explanation:</strong>
+- Changing the &#39;r&#39; in &quot;word&quot; to &#39;o&#39; allows it to equal the dictionary word &quot;wood&quot;.
+- Changing the &#39;n&#39; to &#39;j&#39; and the &#39;t&#39; to &#39;k&#39; in &quot;note&quot; changes it to &quot;joke&quot;.
+- It would take more than 2 edits for &quot;ants&quot; to equal a dictionary word.
+- &quot;wood&quot; can remain unchanged (0 edits) and match the corresponding dictionary word.
+Thus, we return [&quot;word&quot;,&quot;note&quot;,&quot;wood&quot;].
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> queries = [&quot;yes&quot;], dictionary = [&quot;not&quot;]
+<strong>Output:</strong> []
+<strong>Explanation:</strong>
+Applying any two edits to &quot;yes&quot; cannot make it equal to &quot;not&quot;. Thus, we return an empty array.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= queries.length, dictionary.length &lt;= 100</code></li>
+	<li><code>n == queries[i].length == dictionary[j].length</code></li>
+	<li><code>1 &lt;= n &lt;= 100</code></li>
+	<li>All <code>queries[i]</code> and <code>dictionary[j]</code> are composed of lowercase English letters.</li>
+</ul>