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Commit 6b87938

Browse files
yanglbmeidoocs
andauthored
chore: update lc problems (#1739)
* chore: update lc problems * chore: optimised images with calibre/image-actions --------- Co-authored-by: Doocs Bot <doocs-bot@outlook.com>
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‎solution/0100-0199/0123.Best Time to Buy and Sell Stock III/README.md‎

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

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

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public class Solution {
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public int MaxProfit(int[] prices) {
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int f1 = -prices[0], f2 = 0, f3 = -prices[0], f4 = 0;
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for (int i = 1; i < prices.Length; ++i)
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{
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for (int i = 1; i < prices.Length; ++i) {
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f1 = Math.Max(f1, -prices[i]);
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f2 = Math.Max(f2, f1 + prices[i]);
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f3 = Math.Max(f3, f2 - prices[i]);
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```ts
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function maxProfit(prices: number[]): number {
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let f1 = -prices[0],
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f2 = 0,
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f3 = -prices[0],
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f4 = 0;
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let [f1, f2, f3, f4] = [-prices[0], 0, -prices[0], 0];
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for (let i = 1; i < prices.length; ++i) {
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f1 = Math.max(f1, -prices[i]);
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f2 = Math.max(f2, f1 + prices[i]);

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

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

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

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

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

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<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>
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<pre>interface Robot {
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<pre>
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interface Robot {
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&nbsp; // 若下一个方格为空,则返回true,并移动至该方格
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&nbsp; // 若下一个方格为障碍物,则返回false,并停留在原地
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&nbsp; boolean move();
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<p><strong>示例:</strong></p>
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<pre><strong>输入:</strong>
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<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>
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<pre>
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<strong>输入:</strong>
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room = [
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[1,1,1,1,1,0,1,1],
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[1,1,1,1,1,0,1,1],
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<p><strong>注意:</strong></p>
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<ol>
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<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>
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<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>
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<li>扫地机器人的初始位置一定是空地。</li>
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<li>扫地机器人的初始方向向上。</li>
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<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>
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‎solution/2200-2299/2261.K Divisible Elements Subarrays/README_EN.md‎

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## Description
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<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>
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<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>
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<p>Two arrays <code>nums1</code> and <code>nums2</code> are said to be <strong>distinct</strong> if:</p>
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‎solution/2700-2799/2782.Number of Unique Categories/README.md‎

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<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>
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<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>
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<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>
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<p><code>CategoryHandler</code> 类包含以下方法,可能对你有帮助:</p>
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<ul>
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<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>
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</ul>
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<p>返回唯一类别的数量。</p>
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<p>返回&nbsp;<em>唯一类别的数量</em>。</p>
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<p>&nbsp;</p>
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‎solution/2700-2799/2782.Number of Unique Categories/README_EN.md‎

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<p>You are given an integer <code>n</code> and an object <code>categoryHandler</code> of class <code>CategoryHandler</code>.</p>
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<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>
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<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>
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<p>The class <code>CategoryHandler</code> contains the following function, which may help you:</p>
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‎solution/2800-2899/2872.Maximum Number of K-Divisible Components/README.md‎

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# [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/)
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# [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/)
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[English Version](/solution/2800-2899/2872.Maximum%20Number%20of%20K-Divisible%20Components/README_EN.md)
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