chore: update lc problems (doocs#1739)

* chore: update lc problems
* chore: optimised images with calibre/image-actions

---------

Co-authored-by: Doocs Bot <doocs-bot@outlook.com>

Author: yanglbme

solution/0100-0199/0123.Best Time to Buy and Sell Stock III/README.md

@@ -72,7 +72,7 @@
 
 最后返回 `f4` 即可。
 
-时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组 `prices` 的长度。
+时间复杂度 $O(n)$，其中 $n$ 为数组 `prices` 的长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -183,8 +183,7 @@ func max(a, b int) int {
 public class Solution {
     public int MaxProfit(int[] prices) {
         int f1 = -prices[0], f2 = 0, f3 = -prices[0], f4 = 0;
-        for (int i = 1; i < prices.Length; ++i)
-        {
+        for (int i = 1; i < prices.Length; ++i) {
             f1 = Math.Max(f1, -prices[i]);
             f2 = Math.Max(f2, f1 + prices[i]);
             f3 = Math.Max(f3, f2 - prices[i]);
@@ -199,10 +198,7 @@ public class Solution {
 
 ```ts
 function maxProfit(prices: number[]): number {
-    let f1 = -prices[0],
-        f2 = 0,
-        f3 = -prices[0],
-        f4 = 0;
+    let [f1, f2, f3, f4] = [-prices[0], 0, -prices[0], 0];
     for (let i = 1; i < prices.length; ++i) {
         f1 = Math.max(f1, -prices[i]);
         f2 = Math.max(f2, f1 + prices[i]);

solution/0100-0199/0123.Best Time to Buy and Sell Stock III/README_EN.md

@@ -149,8 +149,7 @@ func max(a, b int) int {
 public class Solution {
     public int MaxProfit(int[] prices) {
         int f1 = -prices[0], f2 = 0, f3 = -prices[0], f4 = 0;
-        for (int i = 1; i < prices.Length; ++i)
-        {
+        for (int i = 1; i < prices.Length; ++i) {
             f1 = Math.Max(f1, -prices[i]);
             f2 = Math.Max(f2, f1 + prices[i]);
             f3 = Math.Max(f3, f2 - prices[i]);
@@ -165,10 +164,7 @@ public class Solution {
 
 ```ts
 function maxProfit(prices: number[]): number {
-    let f1 = -prices[0],
-        f2 = 0,
-        f3 = -prices[0],
-        f4 = 0;
+    let [f1, f2, f3, f4] = [-prices[0], 0, -prices[0], 0];
     for (let i = 1; i < prices.length; ++i) {
         f1 = Math.max(f1, -prices[i]);
         f2 = Math.max(f2, f1 + prices[i]);

solution/0100-0199/0123.Best Time to Buy and Sell Stock III/Solution.cs

@@ -1,8 +1,7 @@
 public class Solution {
     public int MaxProfit(int[] prices) {
         int f1 = -prices[0], f2 = 0, f3 = -prices[0], f4 = 0;
-        for (int i = 1; i < prices.Length; ++i)
-        {
+        for (int i = 1; i < prices.Length; ++i) {
             f1 = Math.Max(f1, -prices[i]);
             f2 = Math.Max(f2, f1 + prices[i]);
             f3 = Math.Max(f3, f2 - prices[i]);

solution/0100-0199/0123.Best Time to Buy and Sell Stock III/Solution.ts

@@ -1,8 +1,5 @@
 function maxProfit(prices: number[]): number {
-    let f1 = -prices[0],
-        f2 = 0,
-        f3 = -prices[0],
-        f4 = 0;
+    let [f1, f2, f3, f4] = [-prices[0], 0, -prices[0], 0];
     for (let i = 1; i < prices.length; ++i) {
         f1 = Math.max(f1, -prices[i]);
         f2 = Math.max(f2, f1 + prices[i]);

solution/0400-0499/0489.Robot Room Cleaner/README.md

@@ -14,7 +14,8 @@
 
 <p>请利用提供的4个API编写让机器人清理整个房间的算法。</p>
 
-<pre>interface Robot {
+<pre>
+interface Robot {
 &nbsp; // 若下一个方格为空，则返回true，并移动至该方格
 &nbsp; // 若下一个方格为障碍物，则返回false，并停留在原地
 &nbsp; boolean move();
@@ -31,7 +32,10 @@
 
 <p><strong>示例:</strong></p>
 
-<pre><strong>输入:</strong>
+<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0489.Robot%20Room%20Cleaner/images/1695782910-iCEGqJ-image.png" style="width: 644px; height: 405px;" /></strong></p>
+
+<pre>
+<strong>输入:</strong>
 room = [
   [1,1,1,1,1,0,1,1],
   [1,1,1,1,1,0,1,1],
@@ -50,7 +54,7 @@ col = 3
 <p><strong>注意:</strong></p>
 
 <ol>
-	<li>输入只用于初始化房间和机器人的位置。你需要&ldquo;盲解&rdquo;这个问题。换而言之，你必须在对房间和机器人位置一无所知的情况下，只使用4个给出的API解决问题。&nbsp;</li>
+	<li>输入只用于初始化房间和机器人的位置。你需要“盲解”这个问题。换而言之，你必须在对房间和机器人位置一无所知的情况下，只使用4个给出的API解决问题。&nbsp;</li>
 	<li>扫地机器人的初始位置一定是空地。</li>
 	<li>扫地机器人的初始方向向上。</li>
 	<li>所有可抵达的格子都是相连的，亦即所有标记为1的格子机器人都可以抵达。</li>

solution/0400-0499/0489.Robot Room Cleaner/images/1695782910-iCEGqJ-image.png

@@ -4,7 +4,7 @@
 
 ## Description
 
-<p>Given an integer array <code>nums</code> and two integers <code>k</code> and <code>p</code>, return <em>the number of <strong>distinct subarrays,</strong> which have <strong>at most</strong></em> <code>k</code> <em>elements </em>that are&nbsp;<em>&nbsp;divisible by</em> <code>p</code>.</p>
+<p>Given an integer array <code>nums</code> and two integers <code>k</code> and <code>p</code>, return <em>the number of <strong>distinct subarrays,</strong> which have <strong>at most</strong></em> <code>k</code> <em>elements </em>that are <em>divisible by</em> <code>p</code>.</p>
 
 <p>Two arrays <code>nums1</code> and <code>nums2</code> are said to be <strong>distinct</strong> if:</p>
 

solution/2200-2299/2261.K Divisible Elements Subarrays/README_EN.md

@@ -8,15 +8,15 @@
 
 <p>现给定一个整数 <code>n</code> 和一个 <code>CategoryHandler</code> 类的对象 <code>categoryHandler</code> 。</p>
 
-<p>有 <code>n</code> 个元素，编号从 <code>0</code> 到 <code>n - 1</code>。每个元素都有一个类别，你的任务是找出唯一类别的数量。</p>
+<p>有 <code>n&nbsp;</code> 个元素，编号从 <code>0</code> 到 <code>n - 1</code>。每个元素都有一个类别，你的任务是找出唯一类别的数量。</p>
 
 <p><code>CategoryHandler</code> 类包含以下方法，可能对你有帮助：</p>
 
 <ul>
 	<li><code>boolean haveSameCategory(integer a, integer b)</code>：如果 <code>a</code> 和 <code>b</code> 属于相同的类别，则返回 <code>true</code>，否则返回 <code>false</code>。同时，如果 <code>a</code> 或 <code>b</code> 不是有效的数字（即大于等于 <code>n</code> 或小于 <code>0</code>），它也会返回 <code>false</code>。</li>
 </ul>
 
-<p>返回唯一类别的数量。</p>
+<p>返回&nbsp;<em>唯一类别的数量</em>。</p>
 
 <p>&nbsp;</p>
 

solution/2700-2799/2782.Number of Unique Categories/README.md

@@ -6,7 +6,7 @@
 
 <p>You are given an integer <code>n</code> and an object <code>categoryHandler</code> of class <code>CategoryHandler</code>.</p>
 
-<p>There are <code>n</code>elements, numbered from <code>0</code> to <code>n - 1</code>. Each element has a category, and your task is to find the number of unique categories.</p>
+<p>There are <code>n&nbsp;</code>elements, numbered from <code>0</code> to <code>n - 1</code>. Each element has a category, and your task is to find the number of unique categories.</p>
 
 <p>The class <code>CategoryHandler</code> contains the following function, which may help you:</p>
 

solution/2700-2799/2782.Number of Unique Categories/README_EN.md

@@ -1,4 +1,4 @@
-# [2872. 可以被 K 整除连通块的最大数目](https://leetcode.com/problems/maximum-number-of-k-divisible-components/)
+# [2872. 可以被 K 整除连通块的最大数目](https://leetcode.cn/problems/maximum-number-of-k-divisible-components/)
 
 [English Version](/solution/2800-2899/2872.Maximum%20Number%20of%20K-Divisible%20Components/README_EN.md)