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yanglbmeidoocs
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chore: update lc problems (doocs#1712)
* 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/0000-0099/0026.Remove Duplicates from Sorted Array/README.md‎

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<!-- 这里写题目描述 -->
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<p>给你一个 <strong>升序排列</strong> 的数组 <code>nums</code> ,请你<strong><a href="http://baike.baidu.com/item/%E5%8E%9F%E5%9C%B0%E7%AE%97%E6%B3%95" target="_blank"> 原地</a></strong> 删除重复出现的元素,使每个元素 <strong>只出现一次</strong> ,返回删除后数组的新长度。元素的 <strong>相对顺序</strong> 应该保持 <strong>一致</strong> 。然后返回 <code>nums</code> 中唯一元素的个数。</p>
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<p>给你一个 <strong>非严格递增排列</strong> 的数组 <code>nums</code> ,请你<strong><a href="http://baike.baidu.com/item/%E5%8E%9F%E5%9C%B0%E7%AE%97%E6%B3%95" target="_blank"> 原地</a></strong> 删除重复出现的元素,使每个元素 <strong>只出现一次</strong> ,返回删除后数组的新长度。元素的 <strong>相对顺序</strong> 应该保持 <strong>一致</strong> 。然后返回 <code>nums</code> 中唯一元素的个数。</p>
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<p>考虑 <code>nums</code> 的唯一元素的数量为 <code>k</code> ,你需要做以下事情确保你的题解可以被通过:</p>
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<ul>
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<li><code>1 &lt;= nums.length &lt;= 3 * 10<sup>4</sup></code></li>
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<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
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<li><code>nums</code> 已按 <strong>升序</strong>排列</li>
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<li><code>nums</code> 已按 <strong>非严格递增</strong>&nbsp;排列</li>
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</ul>
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## 解法

‎solution/0100-0199/0138.Copy List with Random Pointer/README.md‎

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# [138. 复制带随机指针的链表](https://leetcode.cn/problems/copy-list-with-random-pointer)
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# [138. 随机链表的复制](https://leetcode.cn/problems/copy-list-with-random-pointer)
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[English Version](/solution/0100-0199/0138.Copy%20List%20with%20Random%20Pointer/README_EN.md)
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<li><code>Node.random</code>&nbsp;为&nbsp;<code>null</code> 或指向链表中的节点。</li>
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</ul>
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<p>注意:本题与主站 138 题相同:<a href="https://leetcode.cn/problems/copy-list-with-random-pointer/" rel="noopener noreferrer" target="_blank">https://leetcode.cn/problems/copy-list-with-random-pointer/</a></p>
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<p>&nbsp;</p>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->

‎solution/0200-0299/0238.Product of Array Except Self/README.md‎

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<p>题目数据 <strong>保证</strong> 数组&nbsp;<code>nums</code>之中任意元素的全部前缀元素和后缀的乘积都在&nbsp; <strong>32 位</strong> 整数范围内。</p>
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<p>请<strong>不要使用除法,</strong>且在&nbsp;<code>O(<em>n</em>)</code> 时间复杂度内完成此题。</p>
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<p>请&nbsp;<strong>不要使用除法,</strong>且在&nbsp;<code>O(<em>n</em>)</code> 时间复杂度内完成此题。</p>
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<p>&nbsp;</p>
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<p>&nbsp;</p>
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<p><strong>进阶:</strong>你可以在 <code>O(1)</code>&nbsp;的额外空间复杂度内完成这个题目吗?( 出于对空间复杂度分析的目的,输出数组<strong>不被视为</strong>额外空间。)</p>
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<p><strong>进阶:</strong>你可以在 <code>O(1)</code>&nbsp;的额外空间复杂度内完成这个题目吗?( 出于对空间复杂度分析的目的,输出数组&nbsp;<strong>不被视为&nbsp;</strong>额外空间。)</p>
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## 解法
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‎solution/0200-0299/0238.Product of Array Except Self/README_EN.md‎

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</ul>
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<p>&nbsp;</p>
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<p><strong>Follow up:</strong>&nbsp;Can you solve the problem in <code>O(1)&nbsp;</code>extra&nbsp;space complexity? (The output array <strong>does not</strong> count as extra space for space complexity analysis.)</p>
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<p><strong>Follow up:</strong>&nbsp;Can you solve the problem in <code>O(1)</code>&nbsp;extra&nbsp;space complexity? (The output array <strong>does not</strong> count as extra space for space complexity analysis.)</p>
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## Solutions
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‎solution/0400-0499/0460.LFU Cache/README.md‎

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<ul>
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<li><code>LFUCache(int capacity)</code> - 用数据结构的容量&nbsp;<code>capacity</code> 初始化对象</li>
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<li><code>int get(int key)</code>&nbsp;- 如果键&nbsp;<code>key</code> 存在于缓存中,则获取键的值,否则返回 <code>-1</code> 。</li>
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<li><code>void put(int key, int value)</code>&nbsp;- 如果键&nbsp;<code>key</code> 已存在,则变更其值;如果键不存在,请插入键值对。当缓存达到其容量&nbsp;<code>capacity</code> 时,则应该在插入新项之前,移除最不经常使用的项。在此问题中,当存在平局(即两个或更多个键具有相同使用频率)时,应该去除 <strong>最近最久未使用</strong> 的键。</li>
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<li><code>void put(int key, int value)</code>&nbsp;- 如果键&nbsp;<code>key</code> 已存在,则变更其值;如果键不存在,请插入键值对。当缓存达到其容量&nbsp;<code>capacity</code> 时,则应该在插入新项之前,移除最不经常使用的项。在此问题中,当存在平局(即两个或更多个键具有相同使用频率)时,应该去除 <strong>最久未使用</strong> 的键。</li>
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<p>为了确定最不常使用的键,可以为缓存中的每个键维护一个 <strong>使用计数器</strong> 。使用计数最小的键是最久未使用的键。</p>

‎solution/0400-0499/0475.Heaters/README.md‎

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<!-- 这里写题目描述 -->
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<p>冬季已经来临。你的任务是设计一个有固定加热半径的供暖器向所有房屋供暖。</p>
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<p>冬季已经来临。&nbsp;你的任务是设计一个有固定加热半径的供暖器向所有房屋供暖。</p>
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<p>在加热器的加热半径范围内的每个房屋都可以获得供暖。</p>
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<p>现在,给出位于一条水平线上的房屋<code>houses</code> 和供暖器<code>heaters</code> 的位置,请你找出并返回可以覆盖所有房屋的最小加热半径。</p>
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<p>现在,给出位于一条水平线上的房屋&nbsp;<code>houses</code> 和供暖器&nbsp;<code>heaters</code> 的位置,请你找出并返回可以覆盖所有房屋的最小加热半径。</p>
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<p><strong>说明</strong>:所有供暖器都遵循你的半径标准,加热的半径也一样。</p>
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<p><b>注意</b>:所有供暖器 <code>heaters</code> 都遵循你的半径标准,加热的半径也一样。</p>
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<p></p>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<pre>
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<strong>输入:</strong> houses = [1,2,3], heaters = [2]
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<strong>输出:</strong> 1
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<strong>解释:</strong> 仅在位置2上有一个供暖器。如果我们将加热半径设为1,那么所有房屋就都能得到供暖。
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<strong>解释:</strong> 仅在位置 2 上有一个供暖器。如果我们将加热半径设为 1,那么所有房屋就都能得到供暖。
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre>
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<strong>输入:</strong> houses = [1,2,3,4], heaters = [1,4]
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<strong>输出:</strong> 1
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<strong>解释:</strong> 在位置1, 4上有两个供暖器。我们需要将加热半径设为1,这样所有房屋就都能得到供暖。
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<strong>解释:</strong> 在位置 1, 4 上有两个供暖器。我们需要将加热半径设为 1,这样所有房屋就都能得到供暖。
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</pre>
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<p><strong>示例 3:</strong></p>
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<strong>输出:</strong>3
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</pre>
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<p></p>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<li><code>1 <= houses.length, heaters.length <= 3 * 10<sup>4</sup></code></li>
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<li><code>1 <= houses[i], heaters[i] <= 10<sup>9</sup></code></li>
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<li><code>1 &lt;= houses.length, heaters.length &lt;= 3 * 10<sup>4</sup></code></li>
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<li><code>1 &lt;= houses[i], heaters[i] &lt;= 10<sup>9</sup></code></li>
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## 解法

‎solution/0400-0499/0475.Heaters/README_EN.md‎

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<pre>
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<strong>Input:</strong> houses = [1,2,3,4], heaters = [1,4]
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<strong>Output:</strong> 1
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<strong>Explanation:</strong> The two heater was placed in the position 1 and 4. We need to use radius 1 standard, then all the houses can be warmed.
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<strong>Explanation:</strong> The two heaters were placed at positions 1 and 4. We need to use a radius 1 standard, then all the houses can be warmed.
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<p><strong class="example">Example 3:</strong></p>

‎solution/0500-0599/0533.Lonely Pixel II/README.md‎

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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0533.Lonely%20Pixel%20II/images/pixel2-1-grid.jpg" style="width: 493px; height: 333px;" />
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0533.Lonely%20Pixel%20II/images/1694957797-UWXAxl-image.png" style="width: 493px; height: 333px;" />
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<pre>
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<strong>输入:</strong>picture = [["W","B","W","B","B","W"],["W","B","W","B","B","W"],["W","B","W","B","B","W"],["W","W","B","W","B","W"]], target = 3
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<strong>输出:</strong>6
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</pre>
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<p><strong>示例 2:</strong></p>
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0533.Lonely%20Pixel%20II/images/pixel2-2-grid.jpg" style="width: 253px; height: 253px;" />
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0533.Lonely%20Pixel%20II/images/1694957806-FyCCMF-image.png" style="width: 253px; height: 253px;" />
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<pre>
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<strong>输入:</strong>picture = [["W","W","B"],["W","W","B"],["W","W","B"]], target = 1
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<strong>输出:</strong>0
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