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 -->

solution/1700-1799/1718.Construct the Lexicographically Largest Valid Sequence/README_EN.md

@@ -19,7 +19,7 @@ tags:
 
 <!-- description:start -->
 
-<p>Given an integer <code>n</code>, find a sequence that satisfies all of the following:</p>
+<p>Given an integer <code>n</code>, find a sequence with elements in the range <code>[1, n]</code> that satisfies all of the following:</p>
 
 <ul>
 	<li>The integer <code>1</code> occurs once in the sequence.</li>

solution/1700-1799/1730.Shortest Path to Get Food/README.md

@@ -55,6 +55,12 @@ tags:
 <strong>输出:</strong> 6
 <strong>解释:</strong> 这里有多个食物。拿到下边的食物仅需走 6 步。</pre>
 
+<p><strong class="example">示例 4：</strong></p>
+
+<pre>
+<b>输入：</b>grid = [["X","X","X","X","X","X","X","X"],["X","*","O","X","O","#","O","X"],["X","O","O","X","O","O","X","X"],["X","O","O","O","O","#","O","X"],["O","O","O","O","O","O","O","O"]]
+<b>输出：</b>5</pre>
+
 <p>&nbsp;</p>
 
 <p><b>提示：</b></p>

solution/1700-1799/1752.Check if Array Is Sorted and Rotated/README.md

@@ -24,7 +24,7 @@ tags:
 
 <p>源数组中可能存在 <strong>重复项</strong> 。</p>
 
-<p><strong>注意：</strong>我们称数组 <code>A</code> 在轮转 <code>x</code> 个位置后得到长度相同的数组 <code>B</code> ，当它们满足 <code>A[i] == B[(i+x) % A.length]</code> ，其中 <code>%</code> 为取余运算。</p>
+<p><strong>注意：</strong>数组 <code>A</code> 在轮转 <code>x</code> 个位置后得到长度相同的数组 <code>B</code> ，使得对于每一个有效的下标&nbsp;<code>i</code>，满足&nbsp;<code>B[i] == A[(i+x) % A.length]</code>。</p>
 
 <p>&nbsp;</p>
 

solution/1800-1899/1800.Maximum Ascending Subarray Sum/README.md

@@ -18,13 +18,11 @@ tags:
 
 <!-- description:start -->
 
-<p>给你一个正整数组成的数组 <code>nums</code> ，返回 <code>nums</code> 中一个 <strong>升序 </strong>子数组的最大可能元素和。</p>
+<p>给你一个正整数组成的数组 <code>nums</code> ，返回 <code>nums</code> 中一个 <span data-keyword="strictly-increasing-array">严格递增子数组</span> 的最大可能元素和。</p>
 
 <p>子数组是数组中的一个连续数字序列。</p>
 
-<p>已知子数组 <code>[nums<sub>l</sub>, nums<sub>l+1</sub>, ..., nums<sub>r-1</sub>, nums<sub>r</sub>]</code> ，若对所有 <code>i</code>（<code>l <= i < r</code>），<code>nums<sub>i </sub> < nums<sub>i+1</sub></code> 都成立，则称这一子数组为 <strong>升序</strong> 子数组。注意，大小为 <code>1</code> 的子数组也视作 <strong>升序</strong> 子数组。</p>
-
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>示例 1：</strong></p>
 
@@ -50,20 +48,13 @@ tags:
 <strong>解释：</strong>[10,11,12] 是元素和最大的升序子数组，最大元素和为 33 。 
 </pre>
 
-<p><strong>示例 4：</strong></p>
-
-<pre>
-<strong>输入：</strong>nums = [100,10,1]
-<strong>输出：</strong>100
-</pre>
-
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>1 <= nums.length <= 100</code></li>
-	<li><code>1 <= nums[i] <= 100</code></li>
+	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
 </ul>
 
 <!-- description:end -->

solution/1800-1899/1852.Distinct Numbers in Each Subarray/README.md

@@ -18,9 +18,11 @@ tags:
 
 <!-- description:start -->
 
-<p>给你一个整数数组&nbsp;<code>nums</code>与一个整数 <code>k</code>，请你构造一个长度 <code>n-k+1</code> 的数组 <code>ans</code>，这个数组第<code>i</code>个元素 <code>ans[i]</code> 是每个长度为k的子数组 <code>nums[i:i+k-1] = [nums[i], nums[i+1], ..., nums[i+k-1]]</code>中数字的种类数。</p>
+<p>给你一个长度为&nbsp;<code>n</code>&nbsp;的整数数组&nbsp;<code>nums</code>&nbsp;与一个整数 <code>k</code>。你的任务是找到&nbsp;<code>nums</code>&nbsp;<strong>所有</strong>&nbsp;长度为&nbsp;<code>k</code>&nbsp;的子数组中&nbsp;<strong>不同</strong>&nbsp;元素的数量。</p>
 
-<p>返回这个数组 <code>ans</code>。</p>
+<p>返回一个数组 <code>ans</code>，其中&nbsp;<code>ans[i]</code>&nbsp;是对于每个索引&nbsp;<code>0 &lt;= i &lt; n - k</code>，<code>nums[i..(i + k - 1)]</code>&nbsp;中不同元素的数量。</p>
+
+<p>&nbsp;</p>
 
 <p>&nbsp;</p>
 
@@ -30,11 +32,11 @@ tags:
 <strong>输入:</strong> nums = [1,2,3,2,2,1,3], k = 3
 <strong>输出:</strong> [3,2,2,2,3]
 <b>解释</b>：每个子数组的数字种类计算方法如下：
-- nums[0:2] = [1,2,3] 所以 ans[0] = 3
-- nums[1:3] = [2,3,2] 所以 ans[1] = 2
-- nums[2:4] = [3,2,2] 所以 ans[2] = 2
-- nums[3:5] = [2,2,1] 所以 ans[3] = 2
-- nums[4:6] = [2,1,3] 所以 ans[4] = 3
+- nums[0..2] = [1,2,3] 所以 ans[0] = 3
+- nums[1..3] = [2,3,2] 所以 ans[1] = 2
+- nums[2..4] = [3,2,2] 所以 ans[2] = 2
+- nums[3..5] = [2,2,1] 所以 ans[3] = 2
+- nums[4..6] = [2,1,3] 所以 ans[4] = 3
 </pre>
 
 <p><strong>示例&nbsp;2:</strong></p>
@@ -43,10 +45,10 @@ tags:
 <strong>输入:</strong> nums = [1,1,1,1,2,3,4], k = 4
 <strong>输出:</strong> [1,2,3,4]
 <strong>解释: </strong>每个子数组的数字种类计算方法如下：
-- nums[0:3] = [1,1,1,1] 所以 ans[0] = 1
-- nums[1:4] = [1,1,1,2] 所以 ans[1] = 2
-- nums[2:5] = [1,1,2,3] 所以 ans[2] = 3
-- nums[3:6] = [1,2,3,4] 所以 ans[3] = 4
+- nums[0..3] = [1,1,1,1] 所以 ans[0] = 1
+- nums[1..4] = [1,1,1,2] 所以 ans[1] = 2
+- nums[2..5] = [1,1,2,3] 所以 ans[2] = 3
+- nums[3..6] = [1,2,3,4] 所以 ans[3] = 4
 </pre>
 
 <p>&nbsp;</p>

solution/1900-1999/1944.Number of Visible People in a Queue/README.md

@@ -72,15 +72,15 @@ tags:
 
 我们观察发现，对于第 $i$ 个人来说，他能看到的人一定是按从左到右高度严格单调递增的。
 
-因此，我们可以倒序遍历数组 $heights$，用一个从栈顶到栈底单调递增的栈 $stk$ 记录已经遍历过的人的高度。
+因此，我们可以倒序遍历数组 $\textit{heights}$，用一个从栈顶到栈底单调递增的栈 $\textit{stk}$ 记录已经遍历过的人的高度。
 
-对于第 $i$ 个人，如果栈不为空并且栈顶元素小于 $heights[i]$，累加当前第 $i$ 个人能看到的人数，然后将栈顶元素出栈，直到栈为空或者栈顶元素大于等于 $heights[i]$。如果此时栈不为空，说明栈顶元素大于等于 $heights[i]$，那么第 $i$ 个人能看到的人数还要再加 $1$。
+对于第 $i$ 个人，如果栈不为空并且栈顶元素小于 $\textit{heights}[i]$，累加当前第 $i$ 个人能看到的人数，然后将栈顶元素出栈，直到栈为空或者栈顶元素大于等于 $\textit{heights}[i]$。如果此时栈不为空，说明栈顶元素大于等于 $\textit{heights}[i]$，那么第 $i$ 个人能看到的人数还要再加 $1$。
 
-接下来，我们将 $heights[i]$ 入栈，继续遍历下一个人。
+接下来，我们将 $\textit{heights}[i]$ 入栈，继续遍历下一个人。
 
-遍历结束后，返回答案数组 $ans$。
+遍历结束后，返回答案数组 $\textit{ans}$。
 
-时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $heights$ 的长度。
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $\textit{heights}$ 的长度。
 
 相似题目：
 

solution/1900-1999/1944.Number of Visible People in a Queue/README_EN.md

@@ -66,7 +66,19 @@ Person 5 can see no one since nobody is to the right of them.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Monotonic Stack
+
+We observe that for the $i$-th person, the people he can see must be strictly increasing in height from left to right.
+
+Therefore, we can traverse the array $\textit{heights}$ in reverse order, using a stack $\textit{stk}$ that is monotonically increasing from top to bottom to record the heights of the people we have traversed.
+
+For the $i$-th person, if the stack is not empty and the top element of the stack is less than $\textit{heights}[i]$, we increment the count of people the $i$-th person can see, then pop the top element of the stack, until the stack is empty or the top element of the stack is greater than or equal to $\textit{heights}[i]$. If the stack is not empty at this point, it means the top element of the stack is greater than or equal to $\textit{heights}[i]$, so we increment the count of people the $i$-th person can see by 1.
+
+Next, we push $\textit{heights}[i]$ onto the stack and continue to the next person.
+
+After traversing, we return the answer array $\textit{ans}$.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{heights}$.
 
 <!-- tabs:start -->
 

solution/1900-1999/1945.Sum of Digits of String After Convert/README.md

@@ -151,11 +151,15 @@ class Solution {
 public:
     int getLucky(string s, int k) {
         string t;
-        for (char c : s) t += to_string(c - 'a' + 1);
+        for (char c : s) {
+            t += to_string(c - 'a' + 1);
+        }
         s = t;
         while (k--) {
             int t = 0;
-            for (char c : s) t += c - '0';
+            for (char c : s) {
+                t += c - '0';
+            }
             s = to_string(t);
         }
         return stoi(s);