doocs · acbin · Mar 21, 2024 · Mar 21, 2024
diff --git a/solution/3000-3099/3060.User Activities within Time Bounds/README.md b/solution/3000-3099/3060.User Activities within Time Bounds/README.md
@@ -1,4 +1,4 @@
-# [3060. User Activities within Time Bounds](https://leetcode.cn/problems/user-activities-within-time-bounds)
+# [3060. 时间范围内的用户活动](https://leetcode.cn/problems/user-activities-within-time-bounds)
 
 [English Version](/solution/3000-3099/3060.User%20Activities%20within%20Time%20Bounds/README_EN.md)
 
@@ -8,7 +8,7 @@
 
 <!-- 这里写题目描述 -->
 
-<p>Table: <code>Sessions</code></p>
+<p>表：<code>Sessions</code></p>
 
 <pre>
 +---------------+----------+
@@ -20,23 +20,24 @@
 | session_id    | int      |
 | session_type  | enum     |
 +---------------+----------+
-session_id is column of unique values for this table.
-session_type is an ENUM (category) type of (Viewer, Streamer).
-This table contains user id, session start, session end, session id and session type.
+session_id 是这张表中有不同值的列。
+session_type 是 (Viewer, Streamer) 的一个 ENUM (category) 类型。
+这张表包含 user id, session start, session end, session id 和 session 类型。
 </pre>
 
-<p>Write a solution to find the the <strong>users</strong> who have had <strong>at least one</strong> <strong>consecutive session</strong> of the <strong>same</strong> type (either &#39;<strong>Viewer</strong>&#39; or &#39;<strong>Streamer</strong>&#39;) with a <strong>maximum</strong> gap of <code>12</code> hours <strong>between</strong> sessions.</p>
+<p>编写一个解决方案，以查找 <strong>至少有一个相同</strong> 类型的 <strong>连续会话</strong>（无论是“<strong>Viewer</strong>”还是“<strong>Streamer</strong>”）的 <strong>用户</strong>，会话 <strong>之间</strong> 的 <strong>最大</strong> 间隔为 <code>12</code> 小时。</p>
 
-<p>Return <em>the result table ordered by </em><code>user_id</code><em> in <b>ascending</b> order.</em></p>
+<p>返回结果表，以<em>&nbsp;</em><code>user_id</code><em>&nbsp;<strong>升序</strong> 排序。</em></p>
 
-<p>The result format is in the following example.</p>
+<p>结果格式如下所述。</p>
 
 <p>&nbsp;</p>
-<p><strong class="example">Example:</strong></p>
+
+<p><strong class="example">示例：</strong></p>
 
 <pre>
-<strong>Input:</strong> 
-Sessions table:
+<strong>输入:</strong> 
+Sessions 表:
 +---------+---------------------+---------------------+------------+--------------+
 | user_id | session_start       | session_end         | session_id | session_type | 
 +---------+---------------------+---------------------+------------+--------------+
@@ -52,18 +53,18 @@ Sessions table:
 | 103     | 2023-11-02 20:00:00 | 2023-11-02 23:00:00 | 10         | Viewer       | 
 | 103     | 2023-11-03 09:00:00 | 2023-11-03 10:00:00 | 11         | Viewer       | 
 +---------+---------------------+---------------------+------------+--------------+
-<strong>Output:</strong> 
+<strong>输出:</strong> 
 +---------+
 | user_id |
 +---------+
 | 102     |
 | 103     |
 +---------+
-<strong>Explanation:</strong>
-- User ID 101 will not be included in the final output as they do not have any consecutive sessions of the same session type.
-- User ID 102 will be included in the final output as they had two viewer sessions with session IDs 3 and 4, respectively, and the time gap between them was less than 12 hours.
-- User ID 103 participated in two viewer sessions with a gap of less than 12 hours between them, identified by session IDs 10 and 11. Therefore, user 103 will be included in the final output.
-Output table is ordered by user_id in increasing order.
+<strong>解释:</strong>
+- 用户 ID 101 将不会包含在最终输出中，因为他们没有相同会话类型的连续回话。
+- 用户 ID 102 将会包含在最终输出中，因为他们分别有两个 session ID 为 3 和 4 的 viewer 会话，并且时间间隔在 12 小时内。
+- 用户 ID 103 参与了两次 viewer 会话，间隔不到 12 小时，会话 ID 为 10 和 11。因此，用户 103 将会被包含在最终输出中。
+输出表根据 user_id 升序排列。
 </pre>
 
 ## 解法

diff --git a/solution/3000-3099/3063.Linked List Frequency/README.md b/solution/3000-3099/3063.Linked List Frequency/README.md
@@ -8,7 +8,7 @@
 
 <!-- 这里写题目描述 -->
 
-<p>给定包含 <code>k</code> 个&nbsp;<strong>不同&nbsp;</strong>元素的链表的&nbsp;<code>head</code>&nbsp;节点，创建一个长度为&nbsp;<code>k</code>&nbsp;的链表，包含给定链表中每个 <strong>不同元素</strong> 以 <strong>任何顺序</strong> 出现的 <span data-keyword="frequency-linkedlist">频率</span>&nbsp;。返回这个链表的头节点。</p>
+<p>给定包含 <code>k</code> 个&nbsp;<strong>不同&nbsp;</strong>元素的链表的&nbsp;<code>head</code>&nbsp;节点，创建一个长度为&nbsp;<code>k</code>&nbsp;的链表，以 <strong>任何顺序</strong> 返回链表中所有 <strong>不同元素</strong> 出现的 <strong>频率</strong>。返回这个链表的头节点。</p>
 
 <p>&nbsp;</p>
 

diff --git a/solution/3000-3099/3068.Find the Maximum Sum of Node Values/README.md b/solution/3000-3099/3068.Find the Maximum Sum of Node Values/README.md
@@ -10,7 +10,7 @@
 
 <p>给你一棵 <code>n</code>&nbsp;个节点的 <strong>无向</strong>&nbsp;树，节点从 <code>0</code>&nbsp;到 <code>n - 1</code>&nbsp;编号。树以长度为 <code>n - 1</code>&nbsp;下标从 <strong>0</strong>&nbsp;开始的二维整数数组 <code>edges</code>&nbsp;的形式给你，其中&nbsp;<code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code>&nbsp;表示树中节点&nbsp;<code>u<sub>i</sub></code>&nbsp;和&nbsp;<code>v<sub>i</sub></code>&nbsp;之间有一条边。同时给你一个 <strong>正</strong>&nbsp;整数&nbsp;<code>k</code>&nbsp;和一个长度为 <code>n</code>&nbsp;下标从&nbsp;<strong>0</strong>&nbsp;开始的&nbsp;<strong>非负</strong>&nbsp;整数数组&nbsp;<code>nums</code>&nbsp;，其中&nbsp;<code>nums[i]</code>&nbsp;表示节点 <code>i</code>&nbsp;的 <strong>价值</strong>&nbsp;。</p>
 
-<p>日增哥哥想 <strong>最大化</strong>&nbsp;树中所有节点价值之和。为了实现这一目标，日增哥哥可以执行以下操作 <strong>任意</strong>&nbsp;次（<strong>包括</strong><strong>&nbsp;0 次</strong>）：</p>
+<p>Alice&nbsp;想 <strong>最大化</strong>&nbsp;树中所有节点价值之和。为了实现这一目标，Alice 可以执行以下操作 <strong>任意</strong>&nbsp;次（<strong>包括</strong><strong>&nbsp;0 次</strong>）：</p>
 
 <ul>
 	<li>选择连接节点&nbsp;<code>u</code>&nbsp;和&nbsp;<code>v</code>&nbsp;的边&nbsp;<code>[u, v]</code>&nbsp;，并将它们的值更新为：
@@ -23,7 +23,7 @@
 
 </ul>
 
-<p>请你返回日增哥哥通过执行以上操作 <strong>任意次</strong>&nbsp;后，可以得到所有节点 <strong>价值之和</strong>&nbsp;的 <strong>最大值</strong>&nbsp;。</p>
+<p>请你返回 Alice 通过执行以上操作 <strong>任意次</strong>&nbsp;后，可以得到所有节点 <strong>价值之和</strong>&nbsp;的 <strong>最大值</strong>&nbsp;。</p>
 
 <p>&nbsp;</p>
 
@@ -34,7 +34,7 @@
 <pre>
 <b>输入：</b>nums = [1,2,1], k = 3, edges = [[0,1],[0,2]]
 <b>输出：</b>6
-<b>解释：</b>日增哥哥可以通过一次操作得到最大价值和 6 ：
+<b>解释：</b>Alice 可以通过一次操作得到最大价值和 6 ：
 - 选择边 [0,2] 。nums[0] 和 nums[2] 都变为：1 XOR 3 = 2 ，数组 nums 变为：[1,2,1] -&gt; [2,2,2] 。
 所有节点价值之和为 2 + 2 + 2 = 6 。
 6 是可以得到最大的价值之和。
@@ -47,7 +47,7 @@
 <pre>
 <b>输入：</b>nums = [2,3], k = 7, edges = [[0,1]]
 <b>输出：</b>9
-<b>解释：</b>日增哥哥可以通过一次操作得到最大和 9 ：
+<b>解释：</b>Alice 可以通过一次操作得到最大和 9 ：
 - 选择边 [0,1] 。nums[0] 变为：2 XOR 7 = 5 ，nums[1] 变为：3 XOR 7 = 4 ，数组 nums 变为：[2,3] -&gt; [5,4] 。
 所有节点价值之和为 5 + 4 = 9 。
 9 是可以得到最大的价值之和。
@@ -60,7 +60,7 @@
 <pre>
 <b>输入：</b>nums = [7,7,7,7,7,7], k = 3, edges = [[0,1],[0,2],[0,3],[0,4],[0,5]]
 <b>输出：</b>42
-<b>解释：</b>日增哥哥不需要执行任何操作，就可以得到最大价值之和 42 。
+<b>解释：</b>Alice 不需要执行任何操作，就可以得到最大价值之和 42 。
 </pre>
 
 <p>&nbsp;</p>

diff --git a/solution/3000-3099/3078.Match Alphanumerical Pattern in Matrix I/README.md b/solution/3000-3099/3078.Match Alphanumerical Pattern in Matrix I/README.md
@@ -2,7 +2,7 @@
 
 [English Version](/solution/3000-3099/3078.Match%20Alphanumerical%20Pattern%20in%20Matrix%20I/README_EN.md)
 
-<!-- tags: -->
+<!-- tags:数组,哈希表,字符串,矩阵 -->
 
 ## 题目描述
 
@@ -15,8 +15,8 @@
 <p>如果我们能用一些数字（每个 <strong>不同</strong> 的字母对应 <strong>不同</strong> 的数字）替换&nbsp;<code>pattern</code>&nbsp;中包含的字母使得结果矩阵与整数矩阵&nbsp;<code>part</code>&nbsp;相同，我们称整数矩阵&nbsp;<code>part</code>&nbsp;与&nbsp;<code>pattern</code>&nbsp;匹配。换句话说，</p>
 
 <ul>
-	<li>矩阵具有相同的维数。</li>
-	<li>如果&nbsp;<code>pattern[r][c]</code>&nbsp;是一个数字，那么&nbsp;<code>part[r][c]</code>&nbsp;一定 <strong>也</strong> 是数字。</li>
+	<li>这两个矩阵具有相同的维数。</li>
+	<li>如果&nbsp;<code>pattern[r][c]</code>&nbsp;是一个数字，那么&nbsp;<code>part[r][c]</code>&nbsp;必须是&nbsp;<strong>相同的</strong> 数字。</li>
 	<li>如果&nbsp;<code>pattern[r][c]</code>&nbsp;是一个字母&nbsp;<code>x</code>：
 	<ul>
 		<li>对于每个&nbsp;<code>pattern[i][j] == x</code>，<code>part[i][j]</code>&nbsp;一定与 <code>part[r][c]</code>&nbsp;<strong>相同</strong>。</li>
@@ -25,7 +25,7 @@
 	</li>
 </ul>
 
-<p>返回一个长度为<em>&nbsp;</em><code>2</code>&nbsp;的数组，包含匹配&nbsp;<code>pattern</code>&nbsp;的&nbsp;<code>board</code>&nbsp;的子矩阵左上角的行号和列号。如果有一个以上这样的子矩阵，返回行号更小的子矩阵。如果依然相同，则返回列号更小的子矩阵。如果没有符合的答案，返回&nbsp;<code>[-1, -1]</code>。</p>
+<p>返回一个长度为<em>&nbsp;</em><code>2</code>&nbsp;的数组，包含匹配&nbsp;<code>pattern</code>&nbsp;的&nbsp;<code>board</code>&nbsp;的子矩阵左上角的行号和列号。如果有多个这样的子矩阵，返回行号更小的子矩阵。如果依然有多个，则返回列号更小的子矩阵。如果没有符合的答案，返回&nbsp;<code>[-1, -1]</code>。</p>
 
 <p>&nbsp;</p>
 

diff --git a/solution/3000-3099/3078.Match Alphanumerical Pattern in Matrix I/README_EN.md b/solution/3000-3099/3078.Match Alphanumerical Pattern in Matrix I/README_EN.md
@@ -2,7 +2,7 @@
 
 [中文文档](/solution/3000-3099/3078.Match%20Alphanumerical%20Pattern%20in%20Matrix%20I/README.md)
 
-<!-- tags: -->
+<!-- tags:Array,Hash Table,String,Matrix -->
 
 ## Description
 

diff --git a/solution/3000-3099/3079.Find the Sum of Encrypted Integers/README.md b/solution/3000-3099/3079.Find the Sum of Encrypted Integers/README.md
@@ -2,7 +2,7 @@
 
 [English Version](/solution/3000-3099/3079.Find%20the%20Sum%20of%20Encrypted%20Integers/README_EN.md)
 
-<!-- tags: -->
+<!-- tags:数组,数学 -->
 
 ## 题目描述
 

diff --git a/solution/3000-3099/3079.Find the Sum of Encrypted Integers/README_EN.md b/solution/3000-3099/3079.Find the Sum of Encrypted Integers/README_EN.md
@@ -2,7 +2,7 @@
 
 [中文文档](/solution/3000-3099/3079.Find%20the%20Sum%20of%20Encrypted%20Integers/README.md)
 
-<!-- tags: -->
+<!-- tags:Array,Math -->
 
 ## Description
 

diff --git a/solution/3000-3099/3080.Mark Elements on Array by Performing Queries/README.md b/solution/3000-3099/3080.Mark Elements on Array by Performing Queries/README.md
@@ -2,7 +2,7 @@
 
 [English Version](/solution/3000-3099/3080.Mark%20Elements%20on%20Array%20by%20Performing%20Queries/README_EN.md)
 
-<!-- tags: -->
+<!-- tags:数组,哈希表,排序,模拟,堆（优先队列） -->
 
 ## 题目描述
 

diff --git a/solution/3000-3099/3080.Mark Elements on Array by Performing Queries/README_EN.md b/solution/3000-3099/3080.Mark Elements on Array by Performing Queries/README_EN.md
@@ -2,7 +2,7 @@
 
 [中文文档](/solution/3000-3099/3080.Mark%20Elements%20on%20Array%20by%20Performing%20Queries/README.md)
 
-<!-- tags: -->
+<!-- tags:Array,Hash Table,Sorting,Simulation,Heap (Priority Queue) -->
 
 ## Description
 

diff --git a/...3000-3099/3081.Replace Question Marks in String to Minimize Its Value/README.md b/...3000-3099/3081.Replace Question Marks in String to Minimize Its Value/README.md
@@ -2,7 +2,7 @@
 
 [English Version](/solution/3000-3099/3081.Replace%20Question%20Marks%20in%20String%20to%20Minimize%20Its%20Value/README_EN.md)
 
-<!-- tags: -->
+<!-- tags:贪心,哈希表,字符串,计数,排序,堆（优先队列） -->
 
 ## 题目描述
 

diff --git a/...0-3099/3081.Replace Question Marks in String to Minimize Its Value/README_EN.md b/...0-3099/3081.Replace Question Marks in String to Minimize Its Value/README_EN.md
@@ -2,7 +2,7 @@
 
 [中文文档](/solution/3000-3099/3081.Replace%20Question%20Marks%20in%20String%20to%20Minimize%20Its%20Value/README.md)
 
-<!-- tags: -->
+<!-- tags:Greedy,Hash Table,String,Counting,Sorting,Heap (Priority Queue) -->
 
 ## Description
 

diff --git a/solution/3000-3099/3082.Find the Sum of the Power of All Subsequences/README.md b/solution/3000-3099/3082.Find the Sum of the Power of All Subsequences/README.md
@@ -2,7 +2,7 @@
 
 [English Version](/solution/3000-3099/3082.Find%20the%20Sum%20of%20the%20Power%20of%20All%20Subsequences/README_EN.md)
 
-<!-- tags: -->
+<!-- tags:数组,动态规划 -->
 
 ## 题目描述
 

diff --git a/solution/3000-3099/3082.Find the Sum of the Power of All Subsequences/README_EN.md b/solution/3000-3099/3082.Find the Sum of the Power of All Subsequences/README_EN.md
@@ -2,7 +2,7 @@
 
 [中文文档](/solution/3000-3099/3082.Find%20the%20Sum%20of%20the%20Power%20of%20All%20Subsequences/README.md)
 
-<!-- tags: -->
+<!-- tags:Array,Dynamic Programming -->
 
 ## Description
 

diff --git a/...n/3000-3099/3083.Existence of a Substring in a String and Its Reverse/README.md b/...n/3000-3099/3083.Existence of a Substring in a String and Its Reverse/README.md
@@ -2,7 +2,7 @@
 
 [English Version](/solution/3000-3099/3083.Existence%20of%20a%20Substring%20in%20a%20String%20and%20Its%20Reverse/README_EN.md)
 
-<!-- tags: -->
+<!-- tags:哈希表,字符串 -->
 
 ## 题目描述
 

diff --git a/...000-3099/3083.Existence of a Substring in a String and Its Reverse/README_EN.md b/...000-3099/3083.Existence of a Substring in a String and Its Reverse/README_EN.md
@@ -2,7 +2,7 @@
 
 [中文文档](/solution/3000-3099/3083.Existence%20of%20a%20Substring%20in%20a%20String%20and%20Its%20Reverse/README.md)
 
-<!-- tags: -->
+<!-- tags:Hash Table,String -->
 
 ## Description
 

diff --git a/...0-3099/3084.Count Substrings Starting and Ending with Given Character/README.md b/...0-3099/3084.Count Substrings Starting and Ending with Given Character/README.md
@@ -2,7 +2,7 @@
 
 [English Version](/solution/3000-3099/3084.Count%20Substrings%20Starting%20and%20Ending%20with%20Given%20Character/README_EN.md)
 
-<!-- tags: -->
+<!-- tags:数学,字符串,计数 -->
 
 ## 题目描述
 

diff --git a/...099/3084.Count Substrings Starting and Ending with Given Character/README_EN.md b/...099/3084.Count Substrings Starting and Ending with Given Character/README_EN.md
@@ -2,7 +2,7 @@
 
 [中文文档](/solution/3000-3099/3084.Count%20Substrings%20Starting%20and%20Ending%20with%20Given%20Character/README.md)
 
-<!-- tags: -->
+<!-- tags:Math,String,Counting -->
 
 ## Description
 

diff --git a/solution/3000-3099/3085.Minimum Deletions to Make String K-Special/README.md b/solution/3000-3099/3085.Minimum Deletions to Make String K-Special/README.md
@@ -2,7 +2,7 @@
 
 [English Version](/solution/3000-3099/3085.Minimum%20Deletions%20to%20Make%20String%20K-Special/README_EN.md)
 
-<!-- tags: -->
+<!-- tags:贪心,哈希表,字符串,计数,排序 -->
 
 ## 题目描述
 

diff --git a/solution/3000-3099/3085.Minimum Deletions to Make String K-Special/README_EN.md b/solution/3000-3099/3085.Minimum Deletions to Make String K-Special/README_EN.md
@@ -2,7 +2,7 @@
 
 [中文文档](/solution/3000-3099/3085.Minimum%20Deletions%20to%20Make%20String%20K-Special/README.md)
 
-<!-- tags: -->
+<!-- tags:Greedy,Hash Table,String,Counting,Sorting -->
 
 ## Description
 

diff --git a/solution/3000-3099/3086.Minimum Moves to Pick K Ones/README.md b/solution/3000-3099/3086.Minimum Moves to Pick K Ones/README.md
@@ -2,7 +2,7 @@
 
 [English Version](/solution/3000-3099/3086.Minimum%20Moves%20to%20Pick%20K%20Ones/README_EN.md)
 
-<!-- tags: -->
+<!-- tags:贪心,数组,前缀和,滑动窗口 -->
 
 ## 题目描述