docs: update

doocs · yanglbme · Dec 7, 2023 · Dec 7, 2023 · Dec 7, 2023 · Dec 7, 2023
commit 35120d381a3f57b44e78988724d7d9671f5fc7e3
diff --git a/solution/0500-0599/0557.Reverse Words in a String III/README.md b/solution/0500-0599/0557.Reverse Words in a String III/README.md
@@ -20,8 +20,8 @@
 <p><strong>示例 2:</strong></p>
 
 <pre>
-<strong>输入：</strong> s = "God Ding"
-<strong>输出：</strong>"doG gniD"
+<strong>输入：</strong> s = "Mr Ding"
+<strong>输出：</strong>"rM gniD"
 </pre>
 
 <p>&nbsp;</p>

diff --git a/solution/0500-0599/0557.Reverse Words in a String III/README_EN.md b/solution/0500-0599/0557.Reverse Words in a String III/README_EN.md
@@ -8,12 +8,19 @@
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
-<pre><strong>Input:</strong> s = "Let's take LeetCode contest"
-<strong>Output:</strong> "s'teL ekat edoCteeL tsetnoc"
-</pre><p><strong class="example">Example 2:</strong></p>
-<pre><strong>Input:</strong> s = "God Ding"
-<strong>Output:</strong> "doG gniD"
+
+<pre>
+<strong>Input:</strong> s = &quot;Let&#39;s take LeetCode contest&quot;
+<strong>Output:</strong> &quot;s&#39;teL ekat edoCteeL tsetnoc&quot;
 </pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;Mr Ding&quot;
+<strong>Output:</strong> &quot;rM gniD&quot;
+</pre>
+
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>
 

diff --git a/solution/1000-1099/1084.Sales Analysis III/README.md b/solution/1000-1099/1084.Sales Analysis III/README.md
@@ -76,10 +76,10 @@ Product table:
 | 1           | S8           |
 +-------------+--------------+
 <strong>解释:</strong>
-id为1的产品仅在2019年春季销售。
-id为2的产品在2019年春季销售，但也在2019年春季之后销售。
-id 3的产品在2019年春季之后销售。
-我们只退回产品1，因为它是2019年春季才销售的产品。</pre>
+id 为 1 的产品仅在 2019 年春季销售。
+id 为 2 的产品在 2019 年春季销售，但也在 2019 年春季之后销售。
+id 为 3 的产品在 2019 年春季之后销售。
+我们只返回 id 为 1 的产品，因为它是 2019 年春季才销售的产品。</pre>
 
 ## 解法
 

diff --git a/solution/1500-1599/1522.Diameter of N-Ary Tree/README.md b/solution/1500-1599/1522.Diameter of N-Ary Tree/README.md
@@ -6,7 +6,7 @@
 
 <!-- 这里写题目描述 -->
 
-<p>给定一棵 N 叉树的根节点&nbsp;<code>root</code>&nbsp;，计算这棵树的直径长度。</p>
+<p>给定一棵 <code>N 叉树</code> 的根节点&nbsp;<code>root</code>&nbsp;，计算这棵树的直径长度。</p>
 
 <p>N 叉树的直径指的是树中任意两个节点间路径中<strong> 最长 </strong>路径的长度。这条路径可能经过根节点，也可能不经过根节点。</p>
 

diff --git a/solution/1500-1599/1522.Diameter of N-Ary Tree/README_EN.md b/solution/1500-1599/1522.Diameter of N-Ary Tree/README_EN.md
@@ -4,7 +4,7 @@
 
 ## Description
 
-<p>Given a <code>root</code> of an <a href="https://leetcode.com/articles/introduction-to-n-ary-trees/" target="_blank">N-ary tree</a>, you need to compute the length of the diameter of the tree.</p>
+<p>Given a <code>root</code> of an <code>N-ary tree</code>, you need to compute the length of the diameter of the tree.</p>
 
 <p>The diameter of an N-ary tree is the length of the <strong>longest</strong> path between any two nodes in the tree. This path may or may not pass through the root.</p>
 

diff --git a/solution/2400-2499/2431.Maximize Total Tastiness of Purchased Fruits/README.md b/solution/2400-2499/2431.Maximize Total Tastiness of Purchased Fruits/README.md
@@ -26,8 +26,8 @@
 <p><strong>注意:</strong></p>
 
 <ul>
-	<li>每种水果最多只能购买一次。</li>
-	<li>一些水果你最多只能用一次折价券。</li>
+	<li>每个水果最多只能购买一次。</li>
+	<li>一个水果你最多只能用一次折价券。</li>
 </ul>
 
 <p>&nbsp;</p>

diff --git a/...2400-2499/2436.Minimum Split Into Subarrays With GCD Greater Than One/README.md b/...2400-2499/2436.Minimum Split Into Subarrays With GCD Greater Than One/README.md
@@ -8,7 +8,7 @@
 
 <p>给定一个由正整数组成的数组 <code>nums</code>。</p>
 
-<p>将数组拆分为&nbsp;<strong>一个或多个&nbsp;</strong>不相连的子数组，如下所示:</p>
+<p>将数组拆分为&nbsp;<strong>一个或多个&nbsp;</strong>互相不覆盖的子数组，如下所示:</p>
 
 <ul>
 	<li>数组中的每个元素都&nbsp;<strong>只属于一个&nbsp;</strong>子数组，并且</li>

diff --git a/solution/2600-2699/2661.First Completely Painted Row or Column/README.md b/solution/2600-2699/2661.First Completely Painted Row or Column/README.md
@@ -10,7 +10,7 @@
 
 <p>从下标 <code>0</code> 开始遍历 <code>arr</code> 中的每个下标 <code>i</code> ，并将包含整数 <code>arr[i]</code> 的 <code>mat</code> 单元格涂色。</p>
 
-<p>请你找出 <code>arr</code> 中在 <code>mat</code> 的某一行或某一列上都被涂色且下标最小的元素，并返回其下标 <code>i</code> 。</p>
+<p>请你找出 <code>arr</code> 中第一个使得&nbsp;<code>mat</code> 的某一行或某一列都被涂色的元素，并返回其下标 <code>i</code> 。</p>
 
 <p>&nbsp;</p>
 

diff --git a/solution/2700-2799/2715.Timeout Cancellation/README_EN.md b/solution/2700-2799/2715.Timeout Cancellation/README_EN.md
@@ -6,53 +6,66 @@
 
 <p>Given a function <code>fn</code>, an array of&nbsp;arguments&nbsp;<code>args</code>, and a timeout&nbsp;<code>t</code>&nbsp;in milliseconds, return a cancel function <code>cancelFn</code>.</p>
 
-<p>After a delay of&nbsp;<code>t</code>,&nbsp;<code>fn</code>&nbsp;should be called with <code>args</code> passed as parameters <strong>unless</strong> <code>cancelFn</code> was invoked before the delay of <code>t</code> milliseconds elapses, specifically at <code>cancelT</code>&nbsp;ms.&nbsp;In that case,&nbsp;<code>fn</code> should never be called.</p>
+<p>After a delay of <code>cancelT</code>, the returned cancel function <code>cancelFn</code> will be invoked.</p>
+
+<pre>
+setTimeout(cancelFn, cancelT)
+</pre>
+
+<p>Initially, the execution of the function <code>fn</code> should be delayed by <code>t</code> milliseconds.</p>
+
+<p>If, before the delay of <code>t</code> milliseconds, the function <code>cancelFn</code> is invoked, it should cancel the delayed execution of <code>fn</code>. Otherwise, if <code>cancelFn</code> is not invoked within the specified delay <code>t</code>, <code>fn</code> should be executed with the provided <code>args</code> as arguments.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
 
 <pre>
-<strong>Input:</strong> fn = (x) =&gt; x * 5, args = [2], t = 20, cancelT = 50
+<strong>Input:</strong> fn = (x) =&gt; x * 5, args = [2], t = 20
 <strong>Output:</strong> [{&quot;time&quot;: 20, &quot;returned&quot;: 10}]
 <strong>Explanation:</strong> 
-const result = []
+const cancelT = 50;
+const result = [];
 
-const fn = (x) =&gt; x * 5
+const fn = (x) =&gt; x * 5;
 
-const start = performance.now()&nbsp;
+const start = performance.now();
 
 const log = (...argsArr) =&gt; {
     const diff = Math.floor(performance.now() - start);
-    result.push({&quot;time&quot;: diff, &quot;returned&quot;: fn(...argsArr)})
+    result.push({&quot;time&quot;: diff, &quot;returned&quot;: fn(...argsArr)});
 }
  &nbsp; &nbsp;&nbsp;
 const cancel = cancellable(log, [2], 20);
 
-const maxT = Math.max(t, 50)
+const maxT = Math.max(t, 50);
 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;
-setTimeout(cancel, cancelT)
+setTimeout(cancel, cancelT);
 
 setTimeout(() =&gt; {
- &nbsp; &nbsp; console.log(result) // [{&quot;time&quot;:20,&quot;returned&quot;:10}]
-}, 65)
+ &nbsp; &nbsp; console.log(result); // [{&quot;time&quot;:20,&quot;returned&quot;:10}]
+}, 65);
 
 The cancellation was scheduled to occur after a delay of cancelT (50ms), which happened after the execution of fn(2) at 20ms.
 </pre>
 
 <p><strong class="example">Example 2:</strong></p>
 
 <pre>
-<strong>Input:</strong> fn = (x) =&gt; x**2, args = [2], t = 100, cancelT = 50 
+<strong>Input:</strong> fn = (x) =&gt; x**2, args = [2], t = 100
 <strong>Output:</strong> []
-<strong>Explanation:</strong> The cancellation was scheduled to occur after a delay of cancelT (50ms), which happened before the execution of fn(2) at 100ms, resulting in fn(2) never being called.
+<strong>Explanation:</strong> 
+const cancelT = 50;
+The cancellation was scheduled to occur after a delay of cancelT (50ms), which happened before the execution of fn(2) at 100ms, resulting in fn(2) never being called.
 </pre>
 
 <p><strong class="example">Example 3:</strong></p>
 
 <pre>
-<strong>Input:</strong> fn = (x1, x2) =&gt; x1 * x2, args = [2,4], t = 30, cancelT = 100
+<strong>Input:</strong> fn = (x1, x2) =&gt; x1 * x2, args = [2,4], t = 30
 <strong>Output:</strong> [{&quot;time&quot;: 30, &quot;returned&quot;: 8}]
-<strong>Explanation: </strong>The cancellation was scheduled to occur after a delay of cancelT (100ms), which happened after the execution of fn(2,4) at 30ms.
+<strong>Explanation: 
+</strong>const cancelT = 100;
+The cancellation was scheduled to occur after a delay of cancelT (100ms), which happened after the execution of fn(2,4) at 30ms.
 </pre>
 
 <p>&nbsp;</p>

diff --git a/...-2999/2948.Make Lexicographically Smallest Array by Swapping Elements/README.md b/...-2999/2948.Make Lexicographically Smallest Array by Swapping Elements/README.md
@@ -12,7 +12,7 @@
 
 <p>返回执行任意次操作后能得到的 <strong>字典序最小的数组</strong><em> </em>。</p>
 
-<p>如果在数组 <code>a</code> 和数组 <code>b</code> 第一个不同的位置上，数组 <code>a</code> 中的对应字符比数组 <code>b</code> 中的对应字符的字典序更小，则认为数组 <code>a</code> 就比数组 <code>b</code> 字典序更小。例如，数组 <code>[2,10,3]</code> 比数组 <code>[10,2,3]</code> 字典序更小，下标 <code>0</code> 处是两个数组第一个不同的位置，且 <code>2 &lt; 10</code> 。</p>
+<p>如果在数组 <code>a</code> 和数组 <code>b</code> 第一个不同的位置上，数组 <code>a</code> 中的对应元素比数组 <code>b</code> 中的对应元素的字典序更小，则认为数组 <code>a</code> 就比数组 <code>b</code> 字典序更小。例如，数组 <code>[2,10,3]</code> 比数组 <code>[10,2,3]</code> 字典序更小，下标 <code>0</code> 处是两个数组第一个不同的位置，且 <code>2 &lt; 10</code> 。</p>
 
 <p>&nbsp;</p>
 

diff --git a/solution/README.md b/solution/README.md
@@ -1294,7 +1294,7 @@
 |  1281  |  [整数的各位积和之差](/solution/1200-1299/1281.Subtract%20the%20Product%20and%20Sum%20of%20Digits%20of%20an%20Integer/README.md)  |  `数学`  |  简单  |  第 166 场周赛  |
 |  1282  |  [用户分组](/solution/1200-1299/1282.Group%20the%20People%20Given%20the%20Group%20Size%20They%20Belong%20To/README.md)  |  `数组`,`哈希表`  |  中等  |  第 166 场周赛  |
 |  1283  |  [使结果不超过阈值的最小除数](/solution/1200-1299/1283.Find%20the%20Smallest%20Divisor%20Given%20a%20Threshold/README.md)  |  `数组`,`二分查找`  |  中等  |  第 166 场周赛  |
-|  1284  |  [转化为全零矩阵的最少反转次数](/solution/1200-1299/1284.Minimum%20Number%20of%20Flips%20to%20Convert%20Binary%20Matrix%20to%20Zero%20Matrix/README.md)  |  `位运算`,`广度优先搜索`,`数组`,`矩阵`  |  困难  |  第 166 场周赛  |
+|  1284  |  [转化为全零矩阵的最少反转次数](/solution/1200-1299/1284.Minimum%20Number%20of%20Flips%20to%20Convert%20Binary%20Matrix%20to%20Zero%20Matrix/README.md)  |  `位运算`,`广度优先搜索`,`数组`,`哈希表`,`矩阵`  |  困难  |  第 166 场周赛  |
 |  1285  |  [找到连续区间的开始和结束数字](/solution/1200-1299/1285.Find%20the%20Start%20and%20End%20Number%20of%20Continuous%20Ranges/README.md)  |  `数据库`  |  中等  |  🔒  |
 |  1286  |  [字母组合迭代器](/solution/1200-1299/1286.Iterator%20for%20Combination/README.md)  |  `设计`,`字符串`,`回溯`,`迭代器`  |  中等  |  第 15 场双周赛  |
 |  1287  |  [有序数组中出现次数超过25%的元素](/solution/1200-1299/1287.Element%20Appearing%20More%20Than%2025%25%20In%20Sorted%20Array/README.md)  |  `数组`  |  简单  |  第 15 场双周赛  |
@@ -1895,7 +1895,7 @@
 |  1882  |  [使用服务器处理任务](/solution/1800-1899/1882.Process%20Tasks%20Using%20Servers/README.md)  |  `数组`,`堆（优先队列）`  |  中等  |  第 243 场周赛  |
 |  1883  |  [准时抵达会议现场的最小跳过休息次数](/solution/1800-1899/1883.Minimum%20Skips%20to%20Arrive%20at%20Meeting%20On%20Time/README.md)  |  `数组`,`动态规划`  |  困难  |  第 243 场周赛  |
 |  1884  |  [鸡蛋掉落-两枚鸡蛋](/solution/1800-1899/1884.Egg%20Drop%20With%202%20Eggs%20and%20N%20Floors/README.md)  |  `数学`,`动态规划`  |  中等  |    |
-|  1885  |  [统计数对](/solution/1800-1899/1885.Count%20Pairs%20in%20Two%20Arrays/README.md)  |  `数组`,`二分查找`,`排序`  |  中等  |  🔒  |
+|  1885  |  [统计数对](/solution/1800-1899/1885.Count%20Pairs%20in%20Two%20Arrays/README.md)  |  `数组`,`双指针`,`二分查找`,`排序`  |  中等  |  🔒  |
 |  1886  |  [判断矩阵经轮转后是否一致](/solution/1800-1899/1886.Determine%20Whether%20Matrix%20Can%20Be%20Obtained%20By%20Rotation/README.md)  |  `数组`,`矩阵`  |  简单  |  第 244 场周赛  |
 |  1887  |  [使数组元素相等的减少操作次数](/solution/1800-1899/1887.Reduction%20Operations%20to%20Make%20the%20Array%20Elements%20Equal/README.md)  |  `数组`,`排序`  |  中等  |  第 244 场周赛  |
 |  1888  |  [使二进制字符串字符交替的最少反转次数](/solution/1800-1899/1888.Minimum%20Number%20of%20Flips%20to%20Make%20the%20Binary%20String%20Alternating/README.md)  |  `贪心`,`字符串`,`动态规划`,`滑动窗口`  |  中等  |  第 244 场周赛  |
@@ -2593,7 +2593,7 @@
 |  2580  |  [统计将重叠区间合并成组的方案数](/solution/2500-2599/2580.Count%20Ways%20to%20Group%20Overlapping%20Ranges/README.md)  |  `数组`,`排序`  |  中等  |  第 99 场双周赛  |
 |  2581  |  [统计可能的树根数目](/solution/2500-2599/2581.Count%20Number%20of%20Possible%20Root%20Nodes/README.md)  |  `树`,`深度优先搜索`,`哈希表`,`动态规划`  |  困难  |  第 99 场双周赛  |
 |  2582  |  [递枕头](/solution/2500-2599/2582.Pass%20the%20Pillow/README.md)  |  `数学`,`模拟`  |  简单  |  第 335 场周赛  |
-|  2583  |  [二叉树中的第 K 大层和](/solution/2500-2599/2583.Kth%20Largest%20Sum%20in%20a%20Binary%20Tree/README.md)  |  `树`,`广度优先搜索`,`二分查找`  |  中等  |  第 335 场周赛  |
+|  2583  |  [二叉树中的第 K 大层和](/solution/2500-2599/2583.Kth%20Largest%20Sum%20in%20a%20Binary%20Tree/README.md)  |  `树`,`广度优先搜索`,`二叉树`,`排序`  |  中等  |  第 335 场周赛  |
 |  2584  |  [分割数组使乘积互质](/solution/2500-2599/2584.Split%20the%20Array%20to%20Make%20Coprime%20Products/README.md)  |  `数组`,`哈希表`,`数学`,`数论`  |  困难  |  第 335 场周赛  |
 |  2585  |  [获得分数的方法数](/solution/2500-2599/2585.Number%20of%20Ways%20to%20Earn%20Points/README.md)  |  `数组`,`动态规划`  |  困难  |  第 335 场周赛  |
 |  2586  |  [统计范围内的元音字符串数](/solution/2500-2599/2586.Count%20the%20Number%20of%20Vowel%20Strings%20in%20Range/README.md)  |  `数组`,`字符串`  |  简单  |  第 336 场周赛  |
@@ -2960,11 +2960,11 @@
 |  2947  |  [统计美丽子字符串 I](/solution/2900-2999/2947.Count%20Beautiful%20Substrings%20I/README.md)  |  `字符串`,`枚举`,`前缀和`  |  中等  |  第 373 场周赛  |
 |  2948  |  [交换得到字典序最小的数组](/solution/2900-2999/2948.Make%20Lexicographically%20Smallest%20Array%20by%20Swapping%20Elements/README.md)  |  `并查集`,`数组`,`排序`  |  中等  |  第 373 场周赛  |
 |  2949  |  [统计美丽子字符串 II](/solution/2900-2999/2949.Count%20Beautiful%20Substrings%20II/README.md)  |  `哈希表`,`数学`,`字符串`,`数论`,`前缀和`  |  困难  |  第 373 场周赛  |
-|  2950  |  [可整除子串的数量](/solution/2900-2999/2950.Number%20of%20Divisible%20Substrings/README.md)  |    |  中等  |  🔒  |
-|  2951  |  [找出峰值](/solution/2900-2999/2951.Find%20the%20Peaks/README.md)  |    |  简单  |  第 374 场周赛  |
-|  2952  |  [需要添加的硬币的最小数量](/solution/2900-2999/2952.Minimum%20Number%20of%20Coins%20to%20be%20Added/README.md)  |    |  中等  |  第 374 场周赛  |
-|  2953  |  [统计完全子字符串](/solution/2900-2999/2953.Count%20Complete%20Substrings/README.md)  |    |  中等  |  第 374 场周赛  |
-|  2954  |  [统计感冒序列的数目](/solution/2900-2999/2954.Count%20the%20Number%20of%20Infection%20Sequences/README.md)  |    |  困难  |  第 374 场周赛  |
+|  2950  |  [可整除子串的数量](/solution/2900-2999/2950.Number%20of%20Divisible%20Substrings/README.md)  |  `哈希表`,`字符串`,`计数`  |  中等  |  🔒  |
+|  2951  |  [找出峰值](/solution/2900-2999/2951.Find%20the%20Peaks/README.md)  |  `数组`,`枚举`  |  简单  |  第 374 场周赛  |
+|  2952  |  [需要添加的硬币的最小数量](/solution/2900-2999/2952.Minimum%20Number%20of%20Coins%20to%20be%20Added/README.md)  |  `贪心`,`数组`,`排序`  |  困难  |  第 374 场周赛  |
+|  2953  |  [统计完全子字符串](/solution/2900-2999/2953.Count%20Complete%20Substrings/README.md)  |  `哈希表`,`字符串`,`滑动窗口`  |  中等  |  第 374 场周赛  |
+|  2954  |  [统计感冒序列的数目](/solution/2900-2999/2954.Count%20the%20Number%20of%20Infection%20Sequences/README.md)  |  `数组`,`数学`,`组合数学`  |  困难  |  第 374 场周赛  |
 |  2955  |  [Number of Same-End Substrings](/solution/2900-2999/2955.Number%20of%20Same-End%20Substrings/README.md)  |    |  中等  |  🔒  |
 
 ## 版权