feat: update lc problems (doocs#4097)

Author: yanglbme

solution/0200-0299/0254.Factor Combinations/README.md

@@ -20,7 +20,8 @@ tags:
 
 <p>例如：</p>
 
-<pre>8 = 2 x 2 x 2;
+<pre>
+8 = 2 x 2 x 2;
   = 2 x 4.</pre>
 
 <p>请实现一个函数，该函数接收一个整数 <em>n</em>&nbsp;并返回该整数所有的因子组合。</p>
@@ -34,18 +35,21 @@ tags:
 
 <p><strong>示例 1：</strong></p>
 
-<pre><strong>输入: </strong><code>1</code>
+<pre>
+<strong>输入: </strong><code>1</code>
 <strong>输出: </strong>[]
 </pre>
 
 <p><strong>示例 2：</strong></p>
 
-<pre><strong>输入: </strong><code>37</code>
+<pre>
+<strong>输入: </strong><code>37</code>
 <strong>输出: </strong>[]</pre>
 
 <p><strong>示例 3：</strong></p>
 
-<pre><strong>输入: </strong><code>12</code>
+<pre>
+<strong>输入: </strong><code>12</code>
 <strong>输出:</strong>
 [
   [2, 6],
@@ -55,7 +59,8 @@ tags:
 
 <p><strong>示例 4: </strong></p>
 
-<pre><strong>输入: </strong><code>32</code>
+<pre>
+<strong>输入: </strong><code>32</code>
 <strong>输出:</strong>
 [
   [2, 16],
@@ -67,6 +72,14 @@ tags:
 ]
 </pre>
 
+<p>&nbsp;</p>
+
+<p><b>提示：</b></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 10<sup>7</sup></code></li>
+</ul>
+
 <!-- description:end -->
 
 ## 解法

solution/0200-0299/0262.Trips and Users/README_EN.md

@@ -55,7 +55,7 @@ banned is an ENUM (category) type of ('Yes', 'No').
 
 <p>The <strong>cancellation rate</strong> is computed by dividing the number of canceled (by client or driver) requests with unbanned users by the total number of requests with unbanned users on that day.</p>
 
-<p>Write a solution to find the <strong>cancellation rate</strong> of requests with unbanned users (<strong>both client and driver must not be banned</strong>) each day between <code>&quot;2013-10-01&quot;</code> and <code>&quot;2013-10-03&quot;</code>. Round <code>Cancellation Rate</code> to <strong>two decimal</strong> points.</p>
+<p>Write a solution to find the <strong>cancellation rate</strong> of requests with unbanned users (<strong>both client and driver must not be banned</strong>) each day between <code>&quot;2013-10-01&quot;</code> and <code>&quot;2013-10-03&quot;</code> with <strong>at least</strong> one trip. Round <code>Cancellation Rate</code> to <strong>two decimal</strong> points.</p>
 
 <p>Return the result table in <strong>any order</strong>.</p>
 

solution/0700-0799/0720.Longest Word in Dictionary/README.md

@@ -41,7 +41,7 @@ tags:
 <pre>
 <strong>输入：</strong>words = ["a", "banana", "app", "appl", "ap", "apply", "apple"]
 <strong>输出：</strong>"apple"
-<strong>解释：</strong>"apply" 和 "apple" 都能由词典中的单词组成。但是 "apple" 的字典序小于 "apply"
+<strong>解释：</strong>"apply" 和 "apple" 都能由词典中的单词组成。但是 "apple" 的字典序小于 "apply" 
 </pre>
 
 <p>&nbsp;</p>

solution/0700-0799/0722.Remove Comments/README.md

@@ -59,20 +59,20 @@ tags:
 <strong>解释:</strong> 示例代码可以编排成这样:
 /*Test program */
 int main()
-{
-  // variable declaration
+{ 
+  // variable declaration 
 int a, b, c;
 /* This is a test
-   multiline
-   comment for
+   multiline  
+   comment for 
    testing */
 a = b + c;
 }
 第 1 行和第 6-9 行的字符串 /* 表示块注释。第 4 行的字符串 // 表示行注释。
-编排后:
+编排后: 
 int main()
-{
-
+{ 
+  
 int a, b, c;
 a = b + c;
 }</pre>

solution/0700-0799/0722.Remove Comments/README_EN.md

@@ -58,20 +58,20 @@ tags:
 <strong>Explanation:</strong> The line by line code is visualized as below:
 /*Test program */
 int main()
-{
-  // variable declaration
+{ 
+  // variable declaration 
 int a, b, c;
 /* This is a test
-   multiline
-   comment for
+   multiline  
+   comment for 
    testing */
 a = b + c;
 }
 The string /* denotes a block comment, including line 1 and lines 6-9. The string // denotes line 4 as comments.
 The line by line output code is visualized as below:
 int main()
-{
-
+{ 
+  
 int a, b, c;
 a = b + c;
 }

solution/0700-0799/0727.Minimum Window Subsequence/README_EN.md

@@ -28,7 +28,7 @@ tags:
 <pre>
 <strong>Input:</strong> s1 = &quot;abcdebdde&quot;, s2 = &quot;bde&quot;
 <strong>Output:</strong> &quot;bcde&quot;
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 &quot;bcde&quot; is the answer because it occurs before &quot;bdde&quot; which has the same length.
 &quot;deb&quot; is not a smaller window because the elements of s2 in the window must occur in order.
 </pre>

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

@@ -19,7 +19,7 @@ tags:
 
 <!-- description:start -->
 
-<p>给你一个字符串 <code>s</code> 。我们要把这个字符串划分为尽可能多的片段，同一字母最多出现在一个片段中。</p>
+<p>给你一个字符串 <code>s</code> 。我们要把这个字符串划分为尽可能多的片段，同一字母最多出现在一个片段中。例如，字符串&nbsp;<code>"ababcc"</code> 能够被分为 <code>["abab", "cc"]</code>，但类似&nbsp;<code>["aba", "bcc"]</code> 或&nbsp;<code>["ab", "ab", "cc"]</code> 的划分是非法的。</p>
 
 <p>注意，划分结果需要满足：将所有划分结果按顺序连接，得到的字符串仍然是 <code>s</code> 。</p>
 

solution/0800-0899/0801.Minimum Swaps To Make Sequences Increasing/README_EN.md

@@ -33,7 +33,7 @@ tags:
 <pre>
 <strong>Input:</strong> nums1 = [1,3,5,4], nums2 = [1,2,3,7]
 <strong>Output:</strong> 1
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 Swap nums1[3] and nums2[3]. Then the sequences are:
 nums1 = [1, 3, 5, 7] and nums2 = [1, 2, 3, 4]
 which are both strictly increasing.

solution/0800-0899/0850.Rectangle Area II/README.md

@@ -19,7 +19,7 @@ tags:
 
 <!-- description:start -->
 
-<p>给你一个轴对齐的二维数组&nbsp;<code>rectangles</code>&nbsp;。 对于&nbsp;<code>rectangle[i] = [x1, y1, x2, y2]</code>，其中（x1，y1）是矩形&nbsp;<code>i</code>&nbsp;左下角的坐标，<meta charset="UTF-8" />&nbsp;<code>(x<sub>i1</sub>, y<sub>i1</sub>)</code>&nbsp;是该矩形 <strong>左下角</strong> 的坐标，<meta charset="UTF-8" />&nbsp;<code>(x<sub>i2</sub>, y<sub>i2</sub>)</code>&nbsp;是该矩形&nbsp;<strong>右上角</strong> 的坐标。</p>
+<p>给你一个轴对齐的二维数组&nbsp;<code>rectangles</code>&nbsp;。 对于&nbsp;<code>rectangle[i] = [x1, y1, x2, y2]</code>，其中&nbsp;<code>(x<sub>i1</sub>, y<sub>i1</sub>)</code>&nbsp;是该矩形 <strong>左下角</strong> 的坐标，<meta charset="UTF-8" />&nbsp;<code>(x<sub>i2</sub>, y<sub>i2</sub>)</code>&nbsp;是该矩形&nbsp;<strong>右上角</strong> 的坐标。</p>
 
 <p>计算平面中所有&nbsp;<code>rectangles</code>&nbsp;所覆盖的 <strong>总面积 </strong>。任何被两个或多个矩形覆盖的区域应只计算 <strong>一次</strong> 。</p>
 

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

@@ -20,13 +20,11 @@ tags:
 
 <!-- description:start -->
 
-<p>给你一个字符串数组 <code>words</code> ，数组中的每个字符串都可以看作是一个单词。请你按 <strong>任意</strong> 顺序返回 <code>words</code> 中是其他单词的子字符串的所有单词。</p>
-
-<p>如果你可以删除 <code>words[j]</code>&nbsp;最左侧和/或最右侧的若干字符得到 <code>words[i]</code> ，那么字符串 <code>words[i]</code> 就是 <code>words[j]</code> 的一个子字符串。</p>
+<p>给你一个字符串数组 <code>words</code> ，数组中的每个字符串都可以看作是一个单词。请你按 <strong>任意</strong> 顺序返回 <code>words</code> 中是其他单词的 <span data-keyword="substring-nonempty">子字符串</span> 的所有单词。</p>
 
 <p>&nbsp;</p>
 
-<p><strong>示例 1：</strong></p>
+<p><strong class="example">示例 1：</strong></p>
 
 <pre>
 <strong>输入：</strong>words = ["mass","as","hero","superhero"]
@@ -35,15 +33,15 @@ tags:
 ["hero","as"] 也是有效的答案。
 </pre>
 
-<p><strong>示例 2：</strong></p>
+<p><strong class="example">示例 2：</strong></p>
 
 <pre>
 <strong>输入：</strong>words = ["leetcode","et","code"]
 <strong>输出：</strong>["et","code"]
 <strong>解释：</strong>"et" 和 "code" 都是 "leetcode" 的子字符串。
 </pre>
 
-<p><strong>示例 3：</strong></p>
+<p><strong class="example">示例 3：</strong></p>
 
 <pre>
 <strong>输入：</strong>words = ["blue","green","bu"]
@@ -58,7 +56,7 @@ tags:
 	<li><code>1 &lt;= words.length &lt;= 100</code></li>
 	<li><code>1 &lt;= words[i].length &lt;= 30</code></li>
 	<li><code>words[i]</code> 仅包含小写英文字母。</li>
-	<li>题目数据 <strong>保证</strong> 每个 <code>words[i]</code> 都是独一无二的。</li>
+	<li>题目数据 <strong>保证</strong> <code>words</code>&nbsp;的每个字符串都是独一无二的。</li>
 </ul>
 
 <!-- description:end -->