doocs · acbin · Jul 3, 2024 · Jul 3, 2024
diff --git a/lcof2/剑指 Offer II 065. 最短的单词编码/README.md b/lcof2/剑指 Offer II 065. 最短的单词编码/README.md
@@ -228,7 +228,7 @@ class Trie {
 class Solution {
     func minimumLengthEncoding(_ words: [String]) -> Int {
         let root = Trie()
-        
+
         for word in words {
             var current = root
             for char in word.reversed() {
@@ -239,25 +239,25 @@ class Solution {
                 current = current.children[index]!
             }
         }
-        
+
         return dfs(root, 1)
     }
-    
+
     private func dfs(_ current: Trie, _ length: Int) -> Int {
         var isLeaf = true
         var result = 0
-        
+
         for child in current.children {
             if let child = child {
                 isLeaf = false
                 result += dfs(child, length + 1)
             }
         }
-        
+
         if isLeaf {
             result += length
         }
-        
+
         return result
     }
 }

diff --git a/solution/0200-0299/0251.Flatten 2D Vector/README.md b/solution/0200-0299/0251.Flatten 2D Vector/README.md
@@ -19,34 +19,50 @@ tags:
 
 <!-- description:start -->
 
-<p>请设计并实现一个能够展开二维向量的迭代器。该迭代器需要支持 <code>next</code> 和 <code>hasNext</code> 两种操作。</p>
+<p>请设计并实现一个能够展开二维向量的迭代器。该迭代器需要支持&nbsp;<code>next</code> 和&nbsp;<code>hasNext</code>&nbsp;两种操作。</p>
 
-<p> </p>
+<p>实现&nbsp;<code>Vector2D</code>&nbsp;类：</p>
 
-<p><strong>示例：</strong></p>
+<ul>
+	<li><code>Vector2D(int[][] vec)</code>&nbsp;使用二维向量&nbsp;<code>vec</code>&nbsp;初始化对象</li>
+	<li><code>next()</code>&nbsp;从二维向量返回下一个元素并将指针移动到下一个位置。你可以假设对&nbsp;<code>next</code>&nbsp;的所有调用都是合法的。</li>
+	<li><code>hasNext()</code>&nbsp;当向量中还有元素返回&nbsp;<code>true</code>，否则返回 <code>false</code>。</li>
+</ul>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
 
 <pre>
-Vector2D iterator = new Vector2D([[1,2],[3],[4]]);
-
-iterator.next(); // 返回 1
-iterator.next(); // 返回 2
-iterator.next(); // 返回 3
-iterator.hasNext(); // 返回 true
-iterator.hasNext(); // 返回 true
-iterator.next(); // 返回 4
-iterator.hasNext(); // 返回 false
+<strong>输入：</strong>
+["Vector2D", "next", "next", "next", "hasNext", "hasNext", "next", "hasNext"]
+[[[[1, 2], [3], [4]]], [], [], [], [], [], [], []]
+<strong>输出：</strong>
+[null, 1, 2, 3, true, true, 4, false]
+
+<strong>解释：</strong>
+Vector2D vector2D = new Vector2D([[1, 2], [3], [4]]);
+vector2D.next();    // return 1
+vector2D.next();    // return 2
+vector2D.next();    // return 3
+vector2D.hasNext(); // return True
+vector2D.hasNext(); // return True
+vector2D.next();    // return 4
+vector2D.hasNext(); // return False
 </pre>
 
-<p> </p>
+<p>&nbsp;</p>
 
-<p><strong>注意：</strong></p>
+<p><b>提示：</b></p>
 
-<ol>
-	<li>请记得 <strong>重置 </strong>在 Vector2D 中声明的类变量（静态变量），因为类变量会 <strong>在多个测试用例中保持不变</strong>，影响判题准确。请 <a href="https://support.leetcode.cn/hc/kb/section/1071534/" target="_blank">查阅</a> 这里。</li>
-	<li>你可以假定 <code>next()</code> 的调用总是合法的，即当 <code>next()</code> 被调用时，二维向量总是存在至少一个后续元素。</li>
-</ol>
+<ul>
+	<li><code>0 &lt;= vec.length &lt;= 200</code></li>
+	<li><code>0 &lt;= vec[i].length &lt;= 500</code></li>
+	<li><code>-500 &lt;= vec[i][j] &lt;= 500</code></li>
+	<li>最多调用&nbsp;<code>next</code> 和&nbsp;<code>hasNext</code>&nbsp;<code>10<sup>5</sup></code>&nbsp;次。</li>
+</ul>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>进阶：</strong>尝试在代码中仅使用 <a href="http://www.cplusplus.com/reference/iterator/iterator/">C++ 提供的迭代器</a> 或 <a href="https://docs.oracle.com/javase/7/docs/api/java/util/Iterator.html">Java 提供的迭代器</a>。</p>
 

diff --git a/solution/0200-0299/0271.Encode and Decode Strings/README.md b/solution/0200-0299/0271.Encode and Decode Strings/README.md
@@ -22,41 +22,78 @@ tags:
 
 <p>1 号机（发送方）有如下函数：</p>
 
-<pre>string encode(vector&lt;string&gt; strs) {
+<pre>
+string encode(vector&lt;string&gt; strs) {
   // ... your code
   return encoded_string;
 }</pre>
 
 <p>2 号机（接收方）有如下函数：</p>
 
-<pre>vector&lt;string&gt; decode(string s) {
+<pre>
+vector&lt;string&gt; decode(string s) {
   //... your code
   return strs;
 }
 </pre>
 
 <p>1 号机（发送方）执行：</p>
 
-<pre>string encoded_string = encode(strs);
+<pre>
+string encoded_string = encode(strs);
 </pre>
 
 <p>2 号机（接收方）执行：</p>
 
-<pre>vector&lt;string&gt; strs2 = decode(encoded_string);
+<pre>
+vector&lt;string&gt; strs2 = decode(encoded_string);
 </pre>
 
 <p>此时，2 号机（接收方）的 <code>strs2</code>&nbsp;需要和 1 号机（发送方）的 <code>strs</code> 相同。</p>
 
 <p>请你来实现这个&nbsp;<code>encode</code> 和&nbsp;<code>decode</code> 方法。</p>
 
-<p><strong>注意：</strong></p>
+<p>不允许使用任何序列化方法解决这个问题（例如 <code>eval</code>）。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<pre>
+<b>输入：</b>dummy_input = ["Hello","World"]
+<b>输出：</b>["Hello","World"]
+<strong>解释：</strong>
+1 号机：
+Codec encoder = new Codec();
+String msg = encoder.encode(strs);
+Machine 1 ---msg---&gt; Machine 2
+
+2 号机：
+Codec decoder = new Codec();
+String[] strs = decoder.decode(msg);
+</pre>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<pre>
+<b>输入：</b>dummy_input = [""]
+<b>输出：</b>[""]
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
 
 <ul>
-	<li>因为字符串可能会包含 256 个合法&nbsp;ascii 字符中的任何字符，所以您的算法必须要能够处理任何可能会出现的字符。</li>
-	<li>请勿使用 &ldquo;类成员&rdquo;、&ldquo;全局变量&rdquo; 或 &ldquo;静态变量&rdquo; 来存储这些状态，您的编码和解码算法应该是非状态依赖的。</li>
-	<li>请不要依赖任何方法库，例如 <code>eval</code>&nbsp;又或者是&nbsp;<code>serialize</code>&nbsp;之类的方法。本题的宗旨是需要您自己实现 &ldquo;编码&rdquo; 和 &ldquo;解码&rdquo; 算法。</li>
+	<li><code>1 &lt;= strs.length &lt;= 200</code></li>
+	<li><code>0 &lt;= strs[i].length &lt;= 200</code></li>
+	<li><code>strs[i]</code>&nbsp;包含 256 个有效 ASCII 字符中的任何可能字符。</li>
 </ul>
 
+<p>&nbsp;</p>
+
+<p><strong>进阶：</strong>你能编写一个通用算法来处理任何可能的字符集吗？</p>
+
 <!-- description:end -->
 
 ## 解法

diff --git a/solution/0500-0599/0553.Optimal Division/README.md b/solution/0500-0599/0553.Optimal Division/README.md
@@ -18,7 +18,7 @@ tags:
 
 <!-- description:start -->
 
-<p>给定一正整数数组<strong> </strong><code>nums</code><strong>，</strong><code>nums</code> 中的相邻整数将进行浮点除法。例如，&nbsp;[2,3,4] -&gt; 2 / 3 / 4 。</p>
+<p>给定一正整数数组<strong> </strong><code>nums</code><strong>，</strong><code>nums</code> 中的相邻整数将进行浮点除法。</p>
 
 <ul>
 	<li>例如，<code>nums = [2,3,4]</code>，我们将求表达式的值&nbsp;<code>"2/3/4"</code>。</li>

diff --git a/solution/0700-0799/0720.Longest Word in Dictionary/README.md b/solution/0700-0799/0720.Longest Word in Dictionary/README.md
@@ -24,6 +24,8 @@ tags:
 
 <p>若其中有多个可行的答案，则返回答案中字典序最小的单词。若无答案，则返回空字符串。</p>
 
+<p>请注意，单词应该从左到右构建，每个额外的字符都添加到前一个单词的结尾。</p>
+
 <p>&nbsp;</p>
 
 <p><strong>示例 1：</strong></p>

diff --git a/solution/1200-1299/1236.Web Crawler/README.md b/solution/1200-1299/1236.Web Crawler/README.md
@@ -19,7 +19,7 @@ tags:
 
 <!-- description:start -->
 
-<p>给定一个链接&nbsp;<code>startUrl</code> 和一个接口&nbsp;<code>HtmlParser</code>&nbsp;，请你实现一个网络爬虫，以实现爬取同&nbsp;<code>startUrl</code>&nbsp;拥有相同&nbsp;<strong>域名标签&nbsp;</strong>的全部链接。</p>
+<p>给定一个链接&nbsp;<code>startUrl</code> 和一个接口&nbsp;<code>HtmlParser</code>&nbsp;，请你实现一个网络爬虫，以实现爬取同&nbsp;<code>startUrl</code>&nbsp;拥有相同&nbsp;<strong>主机名&nbsp;</strong>的全部链接。</p>
 
 <p>该爬虫得到的全部链接可以&nbsp;<strong>任何顺序&nbsp;</strong>返回结果。</p>
 
@@ -98,8 +98,8 @@ startUrl = "http://news.google.com"
 	<li><code>1 &lt;= urls.length &lt;= 1000</code></li>
 	<li><code>1 &lt;= urls[i].length &lt;= 300</code></li>
 	<li><code>startUrl</code>&nbsp;为&nbsp;<code>urls</code>&nbsp;中的一个。</li>
-	<li>域名标签的长为1到63个字符（包括点），只能包含从‘a’到‘z’的ASCII字母、‘0’到‘9’的数字以及连字符即减号（‘-’）。</li>
-	<li>域名标签不会以连字符即减号（‘-’）开头或结尾。</li>
+	<li>主机名的长为1到63个字符（包括点），只能包含从‘a’到‘z’的ASCII字母、‘0’到‘9’的数字以及连字符即减号（‘-’）。</li>
+	<li>主机名不会以连字符即减号（‘-’）开头或结尾。</li>
 	<li>关于域名有效性的约束可参考:&nbsp;&nbsp;<a href="https://en.wikipedia.org/wiki/Hostname#Restrictions_on_valid_hostnames">https://en.wikipedia.org/wiki/Hostname#Restrictions_on_valid_hostnames</a></li>
 	<li>你可以假定url库中不包含重复项。</li>
 </ul>

diff --git a/solution/1200-1299/1240.Tiling a Rectangle with the Fewest Squares/README.md b/solution/1200-1299/1240.Tiling a Rectangle with the Fewest Squares/README.md
@@ -18,13 +18,7 @@ tags:
 
 <!-- description:start -->
 
-<p>你是一位施工队的工长，根据设计师的要求准备为一套设计风格独特的房子进行室内装修。</p>
-
-<p>房子的客厅大小为&nbsp;<code>n</code>&nbsp;x <code>m</code>，为保持极简的风格，需要使用尽可能少的 <strong>正方形</strong> 瓷砖来铺盖地面。</p>
-
-<p>假设正方形瓷砖的规格不限，边长都是整数。</p>
-
-<p>请你帮设计师计算一下，最少需要用到多少块方形瓷砖？</p>
+<p>给定一个大小为&nbsp;<code>n</code>&nbsp;x&nbsp;<code>m</code>&nbsp;的长方形，返回贴满矩形所需的整数边正方形的最小数量。</p>
 
 <p>&nbsp;</p>
 
@@ -35,9 +29,9 @@ tags:
 <pre>
 <strong>输入：</strong>n = 2, m = 3
 <strong>输出：</strong>3
-<code><strong>解释：</strong>3</code> 块地砖就可以铺满卧室。
-<code>     2</code> 块 <code>1x1 地砖</code>
-<code>     1</code> 块 <code>2x2 地砖</code></pre>
+<code><strong>解释：</strong>需要<strong> </strong>3</code> 个正方形来覆盖长方形。
+<code>     2</code> 个 <code>1x1 的正方形</code>
+<code>     1</code> 个 <code>2x2 的正方形</code></pre>
 
 <p><strong>示例 2：</strong></p>
 

diff --git a/solution/1700-1799/1797.Design Authentication Manager/README.md b/solution/1700-1799/1797.Design Authentication Manager/README.md
@@ -7,6 +7,8 @@ source: 第 48 场双周赛 Q2
 tags:
     - 设计
     - 哈希表
+    - 链表
+    - 双向链表
 ---
 
 <!-- problem:start -->

diff --git a/solution/1700-1799/1797.Design Authentication Manager/README_EN.md b/solution/1700-1799/1797.Design Authentication Manager/README_EN.md
@@ -7,6 +7,8 @@ source: Biweekly Contest 48 Q2
 tags:
     - Design
     - Hash Table
+    - Linked List
+    - Doubly-Linked List
 ---
 
 <!-- problem:start -->

diff --git a/solution/1900-1999/1940.Longest Common Subsequence Between Sorted Arrays/README.md b/solution/1900-1999/1940.Longest Common Subsequence Between Sorted Arrays/README.md
@@ -18,7 +18,7 @@ tags:
 
 <!-- description:start -->
 
-<p>给定一个由整数数组组成的数组&nbsp;<code>arrays</code>，其中&nbsp;<code>arrays[i]</code>&nbsp;是 <strong>严格递增</strong> 排序的，返回一个表示 <strong>所有</strong> 数组之间的 <strong>最长公共子序列</strong> 的整数数组。</p>
+<p>给定一个由整数数组组成的数组&nbsp;<code>arrays</code>，其中&nbsp;<code>arrays[i]</code>&nbsp;是 <strong>严格递增</strong> 排序的，返回一个 <strong>所有</strong> 数组均包含的 <strong>最长公共子序列</strong> 的整数数组。</p>
 
 <p><strong>子序列</strong> 是从另一个序列派生出来的序列，删除一些元素或不删除任何元素，而不改变其余元素的顺序。</p>
 

diff --git a/solution/2800-2899/2806.Account Balance After Rounded Purchase/README_EN.md b/solution/2800-2899/2806.Account Balance After Rounded Purchase/README_EN.md
@@ -18,35 +18,57 @@ tags:
 
 <!-- description:start -->
 
-<p>Initially, you have a bank account balance of <code>100</code> dollars.</p>
+<p>Initially, you have a bank account balance of <strong>100</strong> dollars.</p>
 
-<p>You are given an integer <code>purchaseAmount</code> representing the amount you will spend on a purchase in dollars.</p>
+<p>You are given an integer <code>purchaseAmount</code> representing the amount you will spend on a purchase in dollars, in other words, its price.</p>
 
-<p>At the store where you will make the purchase, the purchase amount is rounded to the <strong>nearest multiple</strong> of <code>10</code>. In other words, you pay a <strong>non-negative</strong> amount, <code>roundedAmount</code>, such that <code>roundedAmount</code> is a multiple of <code>10</code> and <code>abs(roundedAmount - purchaseAmount)</code> is <strong>minimized</strong>.</p>
+<p>When making the purchase, first the <code>purchaseAmount</code> <strong>is rounded to the nearest multiple of 10</strong>. Let us call this value <code>roundedAmount</code>. Then, <code>roundedAmount</code> dollars are removed from your bank account.</p>
 
-<p>If there is more than one nearest multiple of <code>10</code>, the <strong>largest multiple</strong> is chosen.</p>
+<p>Return an integer denoting your final bank account balance after this purchase.</p>
 
-<p>Return <em>an integer denoting your account balance after making a purchase worth </em><code>purchaseAmount</code><em> dollars from the store.</em></p>
+<p><strong>Notes:</strong></p>
 
-<p><strong>Note:</strong> <code>0</code> is considered to be a multiple of <code>10</code> in this problem.</p>
+<ul>
+	<li>0 is considered to be a multiple of 10 in this problem.</li>
+	<li>When rounding, 5 is rounded upward (5 is rounded to 10, 15 is rounded to 20, 25 to 30, and so on).</li>
+</ul>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
 
-<pre>
-<strong>Input:</strong> purchaseAmount = 9
-<strong>Output:</strong> 90
-<strong>Explanation:</strong> In this example, the nearest multiple of 10 to 9 is 10. Hence, your account balance becomes 100 - 10 = 90.
-</pre>
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">purchaseAmount = 9</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">90</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The nearest multiple of 10 to 9 is 10. So your account balance becomes 100 - 10 = 90.</p>
+</div>
 
 <p><strong class="example">Example 2:</strong></p>
 
-<pre>
-<strong>Input:</strong> purchaseAmount = 15
-<strong>Output:</strong> 80
-<strong>Explanation:</strong> In this example, there are two nearest multiples of 10 to 15: 10 and 20. So, the larger multiple, 20, is chosen.
-Hence, your account balance becomes 100 - 20 = 80.
-</pre>
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">purchaseAmount = 15</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">80</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The nearest multiple of 10 to 15 is 20. So your account balance becomes 100 - 20 = 80.</p>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">purchaseAmount = 10</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">90</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>10 is a multiple of 10 itself. So your account balance becomes 100 - 10 = 90.</p>
+</div>
 
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>
-Original file line number
+Diff line change
@@ Expand Up / @@ -7,6 +7,8 @@ source: 第 48 场双周赛 Q2 @@
     tags:
         - 设计
         - 哈希表
+        - 链表
+        - 双向链表
     ---
     <!-- problem:start -->
@@ Expand Down @@