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feat: update lc problems (#4097)
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‎solution/0200-0299/0254.Factor Combinations/README.md‎

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<p>例如:</p>
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<pre>8 = 2 x 2 x 2;
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<pre>
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8 = 2 x 2 x 2;
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= 2 x 4.</pre>
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<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>
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<p><strong>示例 1:</strong></p>
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<pre><strong>输入: </strong><code>1</code>
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<pre>
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<strong>输入: </strong><code>1</code>
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<strong>输出: </strong>[]
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre><strong>输入: </strong><code>37</code>
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<pre>
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<strong>输入: </strong><code>37</code>
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<strong>输出: </strong>[]</pre>
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<p><strong>示例 3:</strong></p>
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<pre><strong>输入: </strong><code>12</code>
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<pre>
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<strong>输入: </strong><code>12</code>
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<strong>输出:</strong>
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[
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[2, 6],
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<p><strong>示例 4: </strong></p>
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<pre><strong>输入: </strong><code>32</code>
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<pre>
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<strong>输入: </strong><code>32</code>
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<strong>输出:</strong>
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[
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[2, 16],
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]
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</pre>
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<p>&nbsp;</p>
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<p><b>提示:</b></p>
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<ul>
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<li><code>1 &lt;= n &lt;= 10<sup>7</sup></code></li>
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</ul>
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<!-- description:end -->
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## 解法

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

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<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>
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<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年10月01日&quot;</code> and <code>&quot;2013年10月03日&quot;</code>. Round <code>Cancellation Rate</code> to <strong>two decimal</strong> points.</p>
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<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年10月01日&quot;</code> and <code>&quot;2013年10月03日&quot;</code> with <strong>at least</strong> one trip. Round <code>Cancellation Rate</code> to <strong>two decimal</strong> points.</p>
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<p>Return the result table in <strong>any order</strong>.</p>
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‎solution/0700-0799/0720.Longest Word in Dictionary/README.md‎

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<pre>
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<strong>输入:</strong>words = ["a", "banana", "app", "appl", "ap", "apply", "apple"]
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<strong>输出:</strong>"apple"
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<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"
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<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"
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</pre>
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<p>&nbsp;</p>

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

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<strong>解释:</strong> 示例代码可以编排成这样:
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/*Test program */
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int main()
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{
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// variable declaration
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{
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// variable declaration
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int a, b, c;
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/* This is a test
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multiline
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comment for
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multiline
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comment for
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testing */
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a = b + c;
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}
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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。
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编排后:
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编排后:
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int main()
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{
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{
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int a, b, c;
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a = b + c;
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}</pre>

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

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

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

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

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

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<!-- description:start -->
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<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>
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<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>
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<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>
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‎solution/0800-0899/0801.Minimum Swaps To Make Sequences Increasing/README_EN.md‎

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

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

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<!-- description:start -->
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<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,<metacharset="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>
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<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>
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<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>
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‎solution/1400-1499/1408.String Matching in an Array/README.md‎

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<!-- description:start -->
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<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>
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<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>
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<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>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<p><strongclass="example">示例 1:</strong></p>
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<pre>
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<strong>输入:</strong>words = ["mass","as","hero","superhero"]
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["hero","as"] 也是有效的答案。
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</pre>
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<p><strong>示例 2:</strong></p>
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<p><strongclass="example">示例 2:</strong></p>
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<pre>
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<strong>输入:</strong>words = ["leetcode","et","code"]
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<strong>输出:</strong>["et","code"]
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<strong>解释:</strong>"et" 和 "code" 都是 "leetcode" 的子字符串。
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</pre>
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<p><strong>示例 3:</strong></p>
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<p><strongclass="example">示例 3:</strong></p>
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<pre>
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<strong>输入:</strong>words = ["blue","green","bu"]
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<li><code>1 &lt;= words.length &lt;= 100</code></li>
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<li><code>1 &lt;= words[i].length &lt;= 30</code></li>
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<li><code>words[i]</code> 仅包含小写英文字母。</li>
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<li>题目数据 <strong>保证</strong> 每个 <code>words[i]</code> 都是独一无二的。</li>
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<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>
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</ul>
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<!-- description:end -->

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