feat: add weekly contest 437 & biweekly contest 150 (doocs#4069)

Author: yanglbme

solution/0000-0099/0003.Longest Substring Without Repeating Characters/README_EN.md

@@ -18,7 +18,7 @@ tags:
 
 <!-- description:start -->
 
-<p>Given a string <code>s</code>, find the length of the <strong>longest</strong> <span data-keyword="substring-nonempty"><strong>substring</strong></span> without repeating characters.</p>
+<p>Given a string <code>s</code>, find the length of the <strong>longest</strong> <span data-keyword="substring-nonempty"><strong>substring</strong></span> without duplicate characters.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>

solution/0700-0799/0763.Partition Labels/README_EN.md

@@ -19,7 +19,7 @@ tags:
 
 <!-- description:start -->
 
-<p>You are given a string <code>s</code>. We want to partition the string into as many parts as possible so that each letter appears in at most one part.</p>
+<p>You are given a string <code>s</code>. We want to partition the string into as many parts as possible so that each letter appears in at most one part. For example, the string <code>&quot;ababcc&quot;</code> can be partitioned into <code>[&quot;abab&quot;, &quot;cc&quot;]</code>, but partitions such as <code>[&quot;aba&quot;, &quot;bcc&quot;]</code> or <code>[&quot;ab&quot;, &quot;ab&quot;, &quot;cc&quot;]</code> are invalid.</p>
 
 <p>Note that the partition is done so that after concatenating all the parts in order, the resultant string should be <code>s</code>.</p>
 

solution/0900-0999/0980.Unique Paths III/README.md

@@ -46,7 +46,7 @@ tags:
 
 <pre><strong>输入：</strong>[[1,0,0,0],[0,0,0,0],[0,0,0,2]]
 <strong>输出：</strong>4
-<strong>解释：</strong>我们有以下四条路径：
+<strong>解释：</strong>我们有以下四条路径： 
 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3)
 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3)
 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3)

solution/0900-0999/0980.Unique Paths III/README_EN.md

@@ -36,7 +36,7 @@ tags:
 <pre>
 <strong>Input:</strong> grid = [[1,0,0,0],[0,0,0,0],[0,0,2,-1]]
 <strong>Output:</strong> 2
-<strong>Explanation:</strong> We have the following two paths:
+<strong>Explanation:</strong> We have the following two paths: 
 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2)
 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2)
 </pre>
@@ -46,7 +46,7 @@ tags:
 <pre>
 <strong>Input:</strong> grid = [[1,0,0,0],[0,0,0,0],[0,0,0,2]]
 <strong>Output:</strong> 4
-<strong>Explanation:</strong> We have the following four paths:
+<strong>Explanation:</strong> We have the following four paths: 
 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3)
 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3)
 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3)

solution/0900-0999/0981.Time Based Key-Value Store/README.md

@@ -41,10 +41,10 @@ tags:
 
 <strong>解释：</strong>
 TimeMap timeMap = new TimeMap();
-timeMap.set("foo", "bar", 1);  // 存储键 "foo" 和值 "bar" ，时间戳 timestamp = 1 &nbsp;
+timeMap.set("foo", "bar", 1);  // 存储键 "foo" 和值 "bar" ，时间戳 timestamp = 1 &nbsp; 
 timeMap.get("foo", 1);         // 返回 "bar"
 timeMap.get("foo", 3);         // 返回 "bar", 因为在时间戳 3 和时间戳 2 处没有对应 "foo" 的值，所以唯一的值位于时间戳 1 处（即 "bar"） 。
-timeMap.set("foo", "bar2", 4); // 存储键 "foo" 和值 "bar2" ，时间戳 timestamp = 4&nbsp;
+timeMap.set("foo", "bar2", 4); // 存储键 "foo" 和值 "bar2" ，时间戳 timestamp = 4&nbsp; 
 timeMap.get("foo", 4);         // 返回 "bar2"
 timeMap.get("foo", 5);         // 返回 "bar2"
 </pre>

solution/1200-1299/1261.Find Elements in a Contaminated Binary Tree/README.md

@@ -27,8 +27,12 @@ tags:
 
 <ol>
 	<li><code>root.val == 0</code></li>
-	<li>如果 <code>treeNode.val == x</code> 且&nbsp;<code>treeNode.left != null</code>，那么&nbsp;<code>treeNode.left.val == 2 * x + 1</code></li>
-	<li>如果 <code>treeNode.val == x</code> 且 <code>treeNode.right != null</code>，那么&nbsp;<code>treeNode.right.val == 2 * x + 2</code></li>
+	<li>对于任意 <code>treeNode</code>：
+	<ol type="a">
+		<li>如果 <code>treeNode.val</code> 为&nbsp;<code>x</code> 且&nbsp;<code>treeNode.left != null</code>，那么&nbsp;<code>treeNode.left.val == 2 * x + 1</code></li>
+		<li>如果 <code>treeNode.val</code> 为&nbsp;<code>x</code> 且 <code>treeNode.right != null</code>，那么&nbsp;<code>treeNode.right.val == 2 * x + 2</code></li>
+	</ol>
+	</li>
 </ol>
 
 <p>现在这个二叉树受到「污染」，所有的&nbsp;<code>treeNode.val</code>&nbsp;都变成了&nbsp;<code>-1</code>。</p>
@@ -42,12 +46,13 @@ tags:
 
 <p>&nbsp;</p>
 
-<p><strong>示例 1：</strong></p>
+<p><strong class="example">示例 1：</strong></p>
 
-<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1200-1299/1261.Find%20Elements%20in%20a%20Contaminated%20Binary%20Tree/images/untitled-diagram-4-1.jpg" style="height: 119px; width: 320px;"></strong></p>
+<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1200-1299/1261.Find%20Elements%20in%20a%20Contaminated%20Binary%20Tree/images/untitled-diagram-4-1.jpg" style="height: 119px; width: 320px;" /></strong></p>
 
-<pre><strong>输入：</strong>
-[&quot;FindElements&quot;,&quot;find&quot;,&quot;find&quot;]
+<pre>
+<strong>输入：</strong>
+["FindElements","find","find"]
 [[[-1,null,-1]],[1],[2]]
 <strong>输出：</strong>
 [null,false,true]
@@ -56,12 +61,13 @@ FindElements findElements = new FindElements([-1,null,-1]);
 findElements.find(1); // return False 
 findElements.find(2); // return True </pre>
 
-<p><strong>示例 2：</strong></p>
+<p><strong class="example">示例 2：</strong></p>
 
-<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1200-1299/1261.Find%20Elements%20in%20a%20Contaminated%20Binary%20Tree/images/untitled-diagram-4.jpg" style="height: 198px; width: 400px;"></strong></p>
+<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1200-1299/1261.Find%20Elements%20in%20a%20Contaminated%20Binary%20Tree/images/untitled-diagram-4.jpg" style="height: 198px; width: 400px;" /></strong></p>
 
-<pre><strong>输入：</strong>
-[&quot;FindElements&quot;,&quot;find&quot;,&quot;find&quot;,&quot;find&quot;]
+<pre>
+<strong>输入：</strong>
+["FindElements","find","find","find"]
 [[[-1,-1,-1,-1,-1]],[1],[3],[5]]
 <strong>输出：</strong>
 [null,true,true,false]
@@ -71,12 +77,13 @@ findElements.find(1); // return True
 findElements.find(3); // return True
 findElements.find(5); // return False</pre>
 
-<p><strong>示例 3：</strong></p>
+<p><strong class="example">示例 3：</strong></p>
 
-<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1200-1299/1261.Find%20Elements%20in%20a%20Contaminated%20Binary%20Tree/images/untitled-diagram-4-1-1.jpg" style="height: 274px; width: 306px;"></strong></p>
+<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1200-1299/1261.Find%20Elements%20in%20a%20Contaminated%20Binary%20Tree/images/untitled-diagram-4-1-1.jpg" style="height: 274px; width: 306px;" /></strong></p>
 
-<pre><strong>输入：</strong>
-[&quot;FindElements&quot;,&quot;find&quot;,&quot;find&quot;,&quot;find&quot;,&quot;find&quot;]
+<pre>
+<strong>输入：</strong>
+["FindElements","find","find","find","find"]
 [[[-1,null,-1,-1,null,-1]],[2],[3],[4],[5]]
 <strong>输出：</strong>
 [null,true,false,false,true]
@@ -95,9 +102,9 @@ findElements.find(5); // return True
 <ul>
 	<li><code>TreeNode.val == -1</code></li>
 	<li>二叉树的高度不超过&nbsp;<code>20</code></li>
-	<li>节点的总数在&nbsp;<code>[1,&nbsp;10^4]</code>&nbsp;之间</li>
-	<li>调用&nbsp;<code>find()</code>&nbsp;的总次数在&nbsp;<code>[1,&nbsp;10^4]</code>&nbsp;之间</li>
-	<li><code>0 &lt;= target &lt;= 10^6</code></li>
+	<li>节点的总数在&nbsp;<code>[1,&nbsp;10<sup>4</sup>]</code>&nbsp;之间</li>
+	<li>调用&nbsp;<code>find()</code>&nbsp;的总次数在&nbsp;<code>[1,&nbsp;10<sup>4</sup>]</code>&nbsp;之间</li>
+	<li><code>0 &lt;= target &lt;= 10<sup>6</sup></code></li>
 </ul>
 
 <!-- description:end -->

solution/1200-1299/1261.Find Elements in a Contaminated Binary Tree/README_EN.md

@@ -27,8 +27,12 @@ tags:
 
 <ol>
 	<li><code>root.val == 0</code></li>
-	<li>If <code>treeNode.val == x</code> and <code>treeNode.left != null</code>, then <code>treeNode.left.val == 2 * x + 1</code></li>
-	<li>If <code>treeNode.val == x</code> and <code>treeNode.right != null</code>, then <code>treeNode.right.val == 2 * x + 2</code></li>
+	<li>For any <code>treeNode</code>:
+	<ol type="a">
+		<li>If <code>treeNode.val</code> has a value <code>x</code> and <code>treeNode.left != null</code>, then <code>treeNode.left.val == 2 * x + 1</code></li>
+		<li>If <code>treeNode.val</code> has a value <code>x</code> and <code>treeNode.right != null</code>, then <code>treeNode.right.val == 2 * x + 2</code></li>
+	</ol>
+	</li>
 </ol>
 
 <p>Now the binary tree is contaminated, which means all <code>treeNode.val</code> have been changed to <code>-1</code>.</p>

solution/1300-1399/1352.Product of the Last K Numbers/README.md

@@ -22,29 +22,25 @@ tags:
 
 <!-- description:start -->
 
-<p>请你实现一个「数字乘积类」<code>ProductOfNumbers</code>，要求支持下述两种方法：</p>
+<p>设计一个算法，该算法接受一个整数流并检索该流中最后 <code>k</code> 个整数的乘积。</p>
 
-<p>1.<code>&nbsp;add(int num)</code></p>
+<p>实现&nbsp;<code>ProductOfNumbers</code>&nbsp;类：</p>
 
 <ul>
-	<li>将数字&nbsp;<code>num</code>&nbsp;添加到当前数字列表的最后面。</li>
+	<li><code>ProductOfNumbers()</code>&nbsp;用一个空的流初始化对象。</li>
+	<li><code>void add(int num)</code>&nbsp;将数字&nbsp;<code>num</code>&nbsp;添加到当前数字列表的最后面。</li>
+	<li><code>int getProduct(int k)</code>&nbsp;返回当前数字列表中，最后&nbsp;<code>k</code>&nbsp;个数字的乘积。你可以假设当前列表中始终 <strong>至少</strong> 包含 <code>k</code> 个数字。</li>
 </ul>
 
-<p>2.<code> getProduct(int k)</code></p>
-
-<ul>
-	<li>返回当前数字列表中，最后&nbsp;<code>k</code>&nbsp;个数字的乘积。</li>
-	<li>你可以假设当前列表中始终 <strong>至少</strong> 包含 <code>k</code> 个数字。</li>
-</ul>
-
-<p>题目数据保证：任何时候，任一连续数字序列的乘积都在 32-bit 整数范围内，不会溢出。</p>
+<p>题目数据保证：任何时候，任一连续数字序列的乘积都在 32 位整数范围内，不会溢出。</p>
 
 <p>&nbsp;</p>
 
 <p><strong>示例：</strong></p>
 
-<pre><strong>输入：</strong>
-[&quot;ProductOfNumbers&quot;,&quot;add&quot;,&quot;add&quot;,&quot;add&quot;,&quot;add&quot;,&quot;add&quot;,&quot;getProduct&quot;,&quot;getProduct&quot;,&quot;getProduct&quot;,&quot;add&quot;,&quot;getProduct&quot;]
+<pre>
+<strong>输入：</strong>
+["ProductOfNumbers","add","add","add","add","add","getProduct","getProduct","getProduct","add","getProduct"]
 [[],[3],[0],[2],[5],[4],[2],[3],[4],[8],[2]]
 
 <strong>输出：</strong>
@@ -61,19 +57,24 @@ productOfNumbers.getProduct(2); // 返回 20 。最后 2 个数字的乘积是 5
 productOfNumbers.getProduct(3); // 返回 40 。最后 3 个数字的乘积是 2 * 5 * 4 = 40
 productOfNumbers.getProduct(4); // 返回  0 。最后 4 个数字的乘积是 0 * 2 * 5 * 4 = 0
 productOfNumbers.add(8);        // [3,0,2,5,4,8]
-productOfNumbers.getProduct(2); // 返回 32 。最后 2 个数字的乘积是 4 * 8 = 32
+productOfNumbers.getProduct(2); // 返回 32 。最后 2 个数字的乘积是 4 * 8 = 32 
 </pre>
 
 <p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>add</code> 和 <code>getProduct</code>&nbsp;两种操作加起来总共不会超过&nbsp;<code>40000</code>&nbsp;次。</li>
 	<li><code>0 &lt;= num&nbsp;&lt;=&nbsp;100</code></li>
-	<li><code>1 &lt;= k &lt;= 40000</code></li>
+	<li><code>1 &lt;= k &lt;= 4 * 10<sup>4</sup></code></li>
+	<li><code>add</code> 和 <code>getProduct</code>&nbsp;最多被调用&nbsp;<code>4 * 10<sup>4</sup></code> 次。</li>
+	<li>在任何时间点流的乘积都在 32 位整数范围内。</li>
 </ul>
 
+<p>&nbsp;</p>
+
+<p><strong>进阶：</strong>您能否 <strong>同时</strong> 将 <code>GetProduct</code> 和 <code>Add</code> 的实现改为 <code>O(1)</code> 时间复杂度，而不是 <code>O(k)</code> 时间复杂度？</p>
+
 <!-- description:end -->
 
 ## 解法

solution/1300-1399/1352.Product of the Last K Numbers/README_EN.md

@@ -56,7 +56,7 @@ productOfNumbers.getProduct(2); // return 20. The product of the last 2 numbers
 productOfNumbers.getProduct(3); // return 40. The product of the last 3 numbers is 2 * 5 * 4 = 40
 productOfNumbers.getProduct(4); // return 0. The product of the last 4 numbers is 0 * 2 * 5 * 4 = 0
 productOfNumbers.add(8);        // [3,0,2,5,4,8]
-productOfNumbers.getProduct(2); // return 32. The product of the last 2 numbers is 4 * 8 = 32
+productOfNumbers.getProduct(2); // return 32. The product of the last 2 numbers is 4 * 8 = 32 
 </pre>
 
 <p>&nbsp;</p>
@@ -69,6 +69,9 @@ productOfNumbers.getProduct(2); // return 32. The product of the last 2 numbers
 	<li>The product of the stream at any point in time will fit in a <strong>32-bit</strong> integer.</li>
 </ul>
 
+<p>&nbsp;</p>
+<strong>Follow-up: </strong>Can you implement <strong>both</strong> <code>GetProduct</code> and <code>Add</code> to work in <code>O(1)</code> time complexity instead of <code>O(k)</code> time complexity?
+
 <!-- description:end -->
 
 ## Solutions

solution/1400-1499/1408.String Matching in an Array/README_EN.md

@@ -20,9 +20,7 @@ tags:
 
 <!-- description:start -->
 
-<p>Given an array of string <code>words</code>, return <em>all strings in </em><code>words</code><em> that is a <strong>substring</strong> of another word</em>. You can return the answer in <strong>any order</strong>.</p>
-
-<p>A <strong>substring</strong> is a contiguous sequence of characters within a string</p>
+<p>Given an array of string <code>words</code>, return all strings in<em> </em><code>words</code><em> </em>that are a <span data-keyword="substring-nonempty">substring</span> of another word. You can return the answer in <strong>any order</strong>.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>