feat: update lc problems (#3763)

Author: yanglbme

lcp/LCP 12. 小张刷题计划/README.md

@@ -224,7 +224,7 @@ class Solution {
     func minTime(_ time: [Int], _ m: Int) -> Int {
         var left = 0
         var right = time.reduce(0, +)
-        
+
         while left < right {
             let mid = (left + right) / 2
             if check(mid, time, m) {
@@ -235,12 +235,12 @@ class Solution {
         }
         return left
     }
-    
+
     private func check(_ t: Int, _ time: [Int], _ m: Int) -> Bool {
         var sum = 0
         var maxTime = 0
         var days = 1
-        
+
         for x in time {
             sum += x
             maxTime = max(maxTime, x)

solution/0300-0399/0384.Shuffle an Array/README.md

@@ -3,6 +3,7 @@ comments: true
 difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/0300-0399/0384.Shuffle%20an%20Array/README.md
 tags:
+    - 设计
     - 数组
     - 数学
     - 随机化

solution/0300-0399/0384.Shuffle an Array/README_EN.md

@@ -3,6 +3,7 @@ comments: true
 difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/0300-0399/0384.Shuffle%20an%20Array/README_EN.md
 tags:
+    - Design
     - Array
     - Math
     - Randomized

solution/0400-0499/0490.The Maze/README.md

@@ -28,14 +28,14 @@ tags:
 
 <p>你可以 <strong>假定迷宫的边缘都是墙壁</strong>（参考示例）。</p>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>示例 1：</strong></p>
 <img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0490.The%20Maze/images/maze1-1-grid.jpg" style="width: 573px; height: 573px;" />
 <pre>
 <strong>输入：</strong>maze = [[0,0,1,0,0],[0,0,0,0,0],[0,0,0,1,0],[1,1,0,1,1],[0,0,0,0,0]], start = [0,4], destination = [4,4]
 <strong>输出：</strong>true
-<strong>解释：</strong>一种可能的路径是 : 左 -> 下 -> 左 -> 下 -> 右 -> 下 -> 右。
+<strong>解释：</strong>一种可能的路径是 : 左 -&gt; 下 -&gt; 左 -&gt; 下 -&gt; 右 -&gt; 下 -&gt; 右。
 </pre>
 
 <p><strong>示例 2：</strong></p>
@@ -53,19 +53,19 @@ tags:
 <strong>输出：</strong>false
 </pre>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>
 
 <ul>
 	<li><code>m == maze.length</code></li>
 	<li><code>n == maze[i].length</code></li>
-	<li><code>1 <= m, n <= 100</code></li>
+	<li><code>1 &lt;= m, n &lt;= 100</code></li>
 	<li><code>maze[i][j]</code> is <code>0</code> or <code>1</code>.</li>
 	<li><code>start.length == 2</code></li>
 	<li><code>destination.length == 2</code></li>
-	<li><code>0 <= start<sub>row</sub>, destination<sub>row</sub> <= m</code></li>
-	<li><code>0 <= start<sub>col</sub>, destination<sub>col</sub> <= n</code></li>
+	<li><code>0 &lt;= start<sub>row</sub>, destination<sub>row</sub> &lt;&nbsp;m</code></li>
+	<li><code>0 &lt;= start<sub>col</sub>, destination<sub>col</sub> &lt;&nbsp;n</code></li>
 	<li>球和目的地都在空地上，且初始时它们不在同一位置</li>
 	<li>迷宫 <strong>至少包括 2 块空地</strong></li>
 </ul>

solution/0400-0499/0490.The Maze/README_EN.md

@@ -59,8 +59,8 @@ tags:
 	<li><code>maze[i][j]</code> is <code>0</code> or <code>1</code>.</li>
 	<li><code>start.length == 2</code></li>
 	<li><code>destination.length == 2</code></li>
-	<li><code>0 &lt;= start<sub>row</sub>, destination<sub>row</sub> &lt;= m</code></li>
-	<li><code>0 &lt;= start<sub>col</sub>, destination<sub>col</sub> &lt;= n</code></li>
+	<li><code>0 &lt;= start<sub>row</sub>, destination<sub>row</sub> &lt; m</code></li>
+	<li><code>0 &lt;= start<sub>col</sub>, destination<sub>col</sub> &lt; n</code></li>
 	<li>Both the ball and the destination exist in an empty space, and they will not be in the same position initially.</li>
 	<li>The maze contains <strong>at least 2 empty spaces</strong>.</li>
 </ul>

solution/0500-0599/0500.Keyboard Row/README.md

@@ -20,6 +20,8 @@ tags:
 
 <p>给你一个字符串数组 <code>words</code> ，只返回可以使用在 <strong>美式键盘</strong> 同一行的字母打印出来的单词。键盘如下图所示。</p>
 
+<p><strong>请注意</strong>，字符串&nbsp;<strong>不区分大小写</strong>，相同字母的大小写形式都被视为在同一行<strong>。</strong></p>
+
 <p><strong>美式键盘</strong> 中：</p>
 
 <ul>
@@ -30,36 +32,43 @@ tags:
 
 <p><img alt="American keyboard" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0500.Keyboard%20Row/images/keyboard.png" style="width: 100%; max-width: 600px" /></p>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>示例 1：</strong></p>
 
-<pre>
-<strong>输入：</strong>words = ["Hello","Alaska","Dad","Peace"]
-<strong>输出：</strong>["Alaska","Dad"]
-</pre>
+<div class="example-block">
+<p><strong>输入：</strong><span class="example-io">words = ["Hello","Alaska","Dad","Peace"]</span></p>
+
+<p><b>输出：</b><span class="example-io">["Alaska","Dad"]</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>由于不区分大小写，<code>"a"</code> 和&nbsp;<code>"A"</code> 都在美式键盘的第二行。</p>
+</div>
 
 <p><strong>示例 2：</strong></p>
 
-<pre>
-<strong>输入：</strong>words = ["omk"]
-<strong>输出：</strong>[]
-</pre>
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>words = ["omk"]</span></p>
+
+<p><span class="example-io"><b>输出：</b>[]</span></p>
+</div>
+
+<p><strong class="example">示例 3：</strong></p>
 
-<p><strong>示例 3：</strong></p>
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b></span><span class="example-io">words = ["adsdf","sfd"]</span></p>
 
-<pre>
-<strong>输入：</strong>words = ["adsdf","sfd"]
-<strong>输出：</strong>["adsdf","sfd"]
-</pre>
+<p><span class="example-io"><b>输出：</b></span><span class="example-io">["adsdf","sfd"]</span></p>
+</div>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>1 <= words.length <= 20</code></li>
-	<li><code>1 <= words[i].length <= 100</code></li>
+	<li><code>1 &lt;= words.length &lt;= 20</code></li>
+	<li><code>1 &lt;= words[i].length &lt;= 100</code></li>
 	<li><code>words[i]</code> 由英文字母（小写和大写字母）组成</li>
 </ul>
 

solution/0500-0599/0500.Keyboard Row/README_EN.md

@@ -20,6 +20,8 @@ tags:
 
 <p>Given an array of strings <code>words</code>, return <em>the words that can be typed using letters of the alphabet on only one row of American keyboard like the image below</em>.</p>
 
+<p><strong>Note</strong> that the strings are <strong>case-insensitive</strong>, both lowercased and uppercased of the same letter are treated as if they are at the same row.</p>
+
 <p>In the <strong>American keyboard</strong>:</p>
 
 <ul>
@@ -31,24 +33,31 @@ tags:
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
 
-<pre>
-<strong>Input:</strong> words = [&quot;Hello&quot;,&quot;Alaska&quot;,&quot;Dad&quot;,&quot;Peace&quot;]
-<strong>Output:</strong> [&quot;Alaska&quot;,&quot;Dad&quot;]
-</pre>
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">words = [&quot;Hello&quot;,&quot;Alaska&quot;,&quot;Dad&quot;,&quot;Peace&quot;]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[&quot;Alaska&quot;,&quot;Dad&quot;]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>Both <code>&quot;a&quot;</code> and <code>&quot;A&quot;</code> are in the 2nd row of the American keyboard due to case insensitivity.</p>
+</div>
 
 <p><strong class="example">Example 2:</strong></p>
 
-<pre>
-<strong>Input:</strong> words = [&quot;omk&quot;]
-<strong>Output:</strong> []
-</pre>
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">words = [&quot;omk&quot;]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[]</span></p>
+</div>
 
 <p><strong class="example">Example 3:</strong></p>
 
-<pre>
-<strong>Input:</strong> words = [&quot;adsdf&quot;,&quot;sfd&quot;]
-<strong>Output:</strong> [&quot;adsdf&quot;,&quot;sfd&quot;]
-</pre>
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">words = [&quot;adsdf&quot;,&quot;sfd&quot;]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[&quot;adsdf&quot;,&quot;sfd&quot;]</span></p>
+</div>
 
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>

solution/0700-0799/0777.Swap Adjacent in LR String/README.md

@@ -9,24 +9,24 @@ tags:
 
 <!-- problem:start -->
 
-# [777. 在LR字符串中交换相邻字符](https://leetcode.cn/problems/swap-adjacent-in-lr-string)
+# [777. 在 LR 字符串中交换相邻字符](https://leetcode.cn/problems/swap-adjacent-in-lr-string)
 
 [English Version](/solution/0700-0799/0777.Swap%20Adjacent%20in%20LR%20String/README_EN.md)
 
 ## 题目描述
 
 <!-- description:start -->
 
-<p>在一个由 <code>'L'</code> , <code>'R'</code> 和 <code>'X'</code> 三个字符组成的字符串（例如<code>"RXXLRXRXL"</code>）中进行移动操作。一次移动操作指用一个&nbsp;<code>"LX"</code>&nbsp;替换一个&nbsp;<code>"XL"</code>，或者用一个&nbsp;<code>"XR"</code>&nbsp;替换一个&nbsp;<code>"RX"</code>。现给定起始字符串&nbsp;<code>start</code>&nbsp;和结束字符串&nbsp;<code>end</code>，请编写代码，当且仅当存在一系列移动操作使得&nbsp;<code>start</code>&nbsp;可以转换成&nbsp;<code>end</code>&nbsp;时， 返回&nbsp;<code>True</code>。</p>
+<p>在一个由 <code>'L'</code> , <code>'R'</code> 和 <code>'X'</code> 三个字符组成的字符串（例如<code>"RXXLRXRXL"</code>）中进行移动操作。一次移动操作指用一个&nbsp;<code>"LX"</code>&nbsp;替换一个&nbsp;<code>"XL"</code>，或者用一个&nbsp;<code>"XR"</code>&nbsp;替换一个&nbsp;<code>"RX"</code>。现给定起始字符串&nbsp;<code>start</code>&nbsp;和结束字符串&nbsp;<code>result</code>，请编写代码，当且仅当存在一系列移动操作使得&nbsp;<code>start</code>&nbsp;可以转换成&nbsp;<code>result</code>&nbsp;时， 返回&nbsp;<code>True</code>。</p>
 
 <p>&nbsp;</p>
 
 <p><strong class="example">示例 1：</strong></p>
 
 <pre>
-<strong>输入：</strong>start = "RXXLRXRXL", end = "XRLXXRRLX"
+<strong>输入：</strong>start = "RXXLRXRXL", result = "XRLXXRRLX"
 <strong>输出：</strong>true
-<strong>解释：</strong>通过以下步骤我们可以将 start 转化为 end：
+<strong>解释：</strong>通过以下步骤我们可以将 start 转化为 result：
 RXXLRXRXL -&gt;
 XRXLRXRXL -&gt;
 XRLXRXRXL -&gt;
@@ -37,7 +37,7 @@ XRLXXRRLX
 <p><strong class="example">示例 2：</strong></p>
 
 <pre>
-<strong>输入：</strong>start = "X", end = "L"
+<strong>输入：</strong>start = "X", result = "L"
 <strong>输出：</strong>false
 </pre>
 
@@ -47,8 +47,8 @@ XRLXXRRLX
 
 <ul>
 	<li><code>1 &lt;= start.length&nbsp;&lt;= 10<sup>4</sup></code></li>
-	<li><code>start.length == end.length</code></li>
-	<li><code>start</code> 和&nbsp;<code>end</code>&nbsp;都只包含&nbsp;<code>'L'</code>, <code>'R'</code>&nbsp;或&nbsp;<code>'X'</code>。</li>
+	<li><code>start.length == result.length</code></li>
+	<li><code>start</code> 和&nbsp;<code>result</code>&nbsp;都只包含&nbsp;<code>'L'</code>, <code>'R'</code>&nbsp;或&nbsp;<code>'X'</code>。</li>
 </ul>
 
 <!-- description:end -->

solution/0900-0999/0909.Snakes and Ladders/README.md

@@ -36,13 +36,13 @@ tags:
 
 <p>如果 <code>board[r][c] != -1</code>&nbsp;，位于&nbsp;<code>r</code> 行 <code>c</code> 列的棋盘格中可能存在 “蛇” 或 “梯子”。那个蛇或梯子的目的地将会是 <code>board[r][c]</code>。编号为 <code>1</code> 和 <code>n<sup>2</sup></code> 的方格不是任何蛇或梯子的起点。</p>
 
-<p>注意，玩家在每回合的前进过程中最多只能爬过蛇或梯子一次：就算目的地是另一条蛇或梯子的起点，玩家也 <strong>不能</strong> 继续移动。</p>
+<p>注意，玩家在每次掷骰的前进过程中最多只能爬过蛇或梯子一次：就算目的地是另一条蛇或梯子的起点，玩家也 <strong>不能</strong> 继续移动。</p>
 
 <ul>
 	<li>举个例子，假设棋盘是 <code>[[-1,4],[-1,3]]</code> ，第一次移动，玩家的目标方格是 <code>2</code> 。那么这个玩家将会顺着梯子到达方格 <code>3</code> ，但 <strong>不能</strong> 顺着方格 <code>3</code> 上的梯子前往方格 <code>4</code> 。（简单来说，类似飞行棋，玩家掷出骰子点数后移动对应格数，遇到单向的路径（即梯子或蛇）可以直接跳到路径的终点，但如果多个路径首尾相连，也不能连续跳多个路径）</li>
 </ul>
 
-<p>返回达到编号为&nbsp;<code>n<sup>2</sup></code> 的方格所需的最少移动次数，如果不可能，则返回 <code>-1</code>。</p>
+<p>返回达到编号为&nbsp;<code>n<sup>2</sup></code> 的方格所需的最少掷骰次数，如果不可能，则返回 <code>-1</code>。</p>
 
 <p>&nbsp;</p>
 

solution/1000-1099/1095.Find in Mountain Array/README.md

@@ -53,14 +53,14 @@ tags:
 <p><strong>示例 1：</strong></p>
 
 <pre>
-<strong>输入：</strong>array = [1,2,3,4,5,3,1], target = 3
+<strong>输入：</strong>mountainArr = [1,2,3,4,5,3,1], target = 3
 <strong>输出：</strong>2
 <strong>解释：</strong>3 在数组中出现了两次，下标分别为 2 和 5，我们返回最小的下标 2。</pre>
 
 <p><strong>示例 2：</strong></p>
 
 <pre>
-<strong>输入：</strong>array = [0,1,2,4,2,1], target = 3
+<strong>输入：</strong>mountainArr = [0,1,2,4,2,1], target = 3
 <strong>输出：</strong>-1
 <strong>解释：</strong>3 在数组中没有出现，返回 -1。
 </pre>
@@ -70,9 +70,9 @@ tags:
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>3 &lt;= mountain_arr.length() &lt;= 10<sup>4</sup></code></li>
+	<li><code>3 &lt;= mountainArr.length() &lt;= 10<sup>4</sup></code></li>
 	<li><code>0 &lt;= target &lt;= 10<sup>9</sup></code></li>
-	<li><code>0 &lt;= mountain_arr.get(index) &lt;=&nbsp;10<sup>9</sup></code></li>
+	<li><code>0 &lt;= mountainArr.get(index) &lt;=&nbsp;10<sup>9</sup></code></li>
 </ul>
 
 <!-- description:end -->

solution/1000-1099/1095.Find in Mountain Array/README_EN.md

@@ -49,14 +49,14 @@ tags:
 <p><strong class="example">Example 1:</strong></p>
 
 <pre>
-<strong>Input:</strong> array = [1,2,3,4,5,3,1], target = 3
+<strong>Input:</strong> mountainArr = [1,2,3,4,5,3,1], target = 3
 <strong>Output:</strong> 2
 <strong>Explanation:</strong> 3 exists in the array, at index=2 and index=5. Return the minimum index, which is 2.</pre>
 
 <p><strong class="example">Example 2:</strong></p>
 
 <pre>
-<strong>Input:</strong> array = [0,1,2,4,2,1], target = 3
+<strong>Input:</strong> mountainArr = [0,1,2,4,2,1], target = 3
 <strong>Output:</strong> -1
 <strong>Explanation:</strong> 3 does not exist in <code>the array,</code> so we return -1.
 </pre>
@@ -65,9 +65,9 @@ tags:
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>3 &lt;= mountain_arr.length() &lt;= 10<sup>4</sup></code></li>
+	<li><code>3 &lt;= mountainArr.length() &lt;= 10<sup>4</sup></code></li>
 	<li><code>0 &lt;= target &lt;= 10<sup>9</sup></code></li>
-	<li><code>0 &lt;= mountain_arr.get(index) &lt;= 10<sup>9</sup></code></li>
+	<li><code>0 &lt;= mountainArr.get(index) &lt;= 10<sup>9</sup></code></li>
 </ul>
 
 <!-- description:end -->

solution/1200-1299/1257.Smallest Common Region/README.md

@@ -25,38 +25,48 @@ tags:
 
 <p>给你一些区域列表&nbsp;<code>regions</code> ，每个列表的第一个区域都包含这个列表内所有其他区域。</p>
 
-<p>很自然地，如果区域&nbsp;<code>X</code> 包含区域&nbsp;<code>Y</code> ，那么区域&nbsp;<code>X</code> &nbsp;比区域&nbsp;<code>Y</code> 大。</p>
+<p>很自然地，如果区域&nbsp;<code>x</code> 包含区域&nbsp;<code>y</code> ，那么区域&nbsp;<code>x</code> &nbsp;比区域&nbsp;<code>y</code> 大。同时根据定义，区域&nbsp;<code>x</code> 包含自身。</p>
 
 <p>给定两个区域&nbsp;<code>region1</code>&nbsp;和&nbsp;<code>region2</code> ，找到同时包含这两个区域的&nbsp;<strong>最小&nbsp;</strong>区域。</p>
 
-<p>如果区域列表中&nbsp;<code>r1</code>&nbsp;包含&nbsp;<code>r2</code>&nbsp;和&nbsp;<code>r3</code> ，那么数据保证&nbsp;<code>r2</code> 不会包含&nbsp;<code>r3</code>&nbsp;。</p>
+<p>如果给定区域&nbsp;<code>r1</code>，<code>r2</code>&nbsp;和&nbsp;<code>r3</code>，使得&nbsp;<code>r1</code>&nbsp;包含&nbsp;<code>r3</code>，那么数据保证&nbsp;<code>r2</code> 不会包含&nbsp;<code>r3</code>&nbsp;。</p>
 
-<p>数据同样保证最小公共区域一定存在。</p>
+<p>数据同样保证最小区域一定存在。</p>
 
 <p>&nbsp;</p>
 
 <p><strong>示例 1：</strong></p>
 
 <pre>
 <strong>输入：
-</strong>regions = [[&quot;Earth&quot;,&quot;North America&quot;,&quot;South America&quot;],
-[&quot;North America&quot;,&quot;United States&quot;,&quot;Canada&quot;],
-[&quot;United States&quot;,&quot;New York&quot;,&quot;Boston&quot;],
-[&quot;Canada&quot;,&quot;Ontario&quot;,&quot;Quebec&quot;],
-[&quot;South America&quot;,&quot;Brazil&quot;]],
-region1 = &quot;Quebec&quot;,
-region2 = &quot;New York&quot;
-<strong>输出：</strong>&quot;North America&quot;
+</strong>regions = [["Earth","North America","South America"],
+["North America","United States","Canada"],
+["United States","New York","Boston"],
+["Canada","Ontario","Quebec"],
+["South America","Brazil"]],
+region1 = "Quebec",
+region2 = "New York"
+<strong>输出：</strong>"North America"
+</pre>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<pre>
+<b>输入：</b>regions = [["Earth", "North America", "South America"],["North America", "United States", "Canada"],["United States", "New York", "Boston"],["Canada", "Ontario", "Quebec"],["South America", "Brazil"]], region1 = "Canada", region2 = "South America"
+<b>输出：</b>"Earth"
 </pre>
 
 <p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>2 &lt;= regions.length &lt;= 10^4</code></li>
+	<li><code>2 &lt;= regions.length &lt;= 10<sup>4</sup></code></li>
+	<li><code>2 &lt;= regions[i].length &lt;= 20</code></li>
+	<li><code>1 &lt;= regions[i][j].length, region1.length, region2.length &lt;= 20</code></li>
 	<li><code>region1 != region2</code></li>
-	<li>所有字符串只包含英文字母和空格，且最多只有&nbsp;20 个字母。</li>
+	<li><code>regions[i][j]</code>，<code>region1</code>&nbsp;和&nbsp;<code>region2</code> 由英语字母组成。</li>
+	<li>输入保证存在一个区域直接或间接包含所有其他区域。</li>
 </ul>
 
 <!-- description:end -->