feat: add new leetcode problems

Author: yanglbme

images/contributors.svg

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+# [1836. ](https://leetcode-cn.com/problems/remove-duplicates-from-an-unsorted-linked-list)
+
+[English Version](/solution/1800-1899/1836.Remove%20Duplicates%20From%20an%20Unsorted%20Linked%20List/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+None
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/1800-1899/1836.Remove Duplicates From an Unsorted Linked List/README.md

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+# [1836. Remove Duplicates From an Unsorted Linked List](https://leetcode.com/problems/remove-duplicates-from-an-unsorted-linked-list)
+
+[中文文档](/solution/1800-1899/1836.Remove%20Duplicates%20From%20an%20Unsorted%20Linked%20List/README.md)
+
+## Description
+
+<p>Given the <code>head</code> of a linked list, find all the values that appear <strong>more than once</strong> in the list and delete the nodes that have any of those values.</p>
+
+<p>Return <em>the linked list after the deletions.</em></p>
+
+<p>&nbsp;</p>
+
+<p><strong>Example 1:</strong></p>
+
+<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/1800-1899/1836.Remove%20Duplicates%20From%20an%20Unsorted%20Linked%20List/images/tmp-linked-list.jpg" style="width: 422px; height: 222px;" />
+
+<pre>
+
+<strong>Input:</strong> head = [1,2,3,2]
+
+<strong>Output:</strong> [1,3]
+
+<strong>Explanation:</strong> 2 appears twice in the linked list, so all 2&#39;s should be deleted. After deleting all 2&#39;s, we are left with [1,3].
+
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/1800-1899/1836.Remove%20Duplicates%20From%20an%20Unsorted%20Linked%20List/images/tmp-linked-list-1.jpg" style="width: 422px; height: 151px;" />
+
+<pre>
+
+<strong>Input:</strong> head = [2,1,1,2]
+
+<strong>Output:</strong> []
+
+<strong>Explanation:</strong> 2 and 1 both appear twice. All the elements should be deleted.
+
+</pre>
+
+<p><strong>Example 3:</strong></p>
+
+<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/1800-1899/1836.Remove%20Duplicates%20From%20an%20Unsorted%20Linked%20List/images/tmp-linked-list-2.jpg" style="width: 500px; height: 142px;" />
+
+<pre>
+
+<strong>Input:</strong> head = [3,2,2,1,3,2,4]
+
+<strong>Output:</strong> [1,4]
+
+<strong>Explanation: </strong>3 appears twice and 2 appears three times. After deleting all 3&#39;s and 2&#39;s, we are left with [1,4].
+
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li>The number of nodes in the list is in the range&nbsp;<code>[1, 10<sup>5</sup>]</code></li>
+	<li><code>1 &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/1800-1899/1836.Remove Duplicates From an Unsorted Linked List/README_EN.md

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+# [1837. K 进制表示下的各位数字总和](https://leetcode-cn.com/problems/sum-of-digits-in-base-k)
+
+[English Version](/solution/1800-1899/1837.Sum%20of%20Digits%20in%20Base%20K/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个整数 <code>n</code>（<code>10</code> 进制）和一个基数 <code>k</code> ，请你将 <code>n</code> 从 <code>10</code> 进制表示转换为 <code>k</code> 进制表示，计算并返回转换后各位数字的 <strong>总和</strong> 。</p>
+
+<p>转换后，各位数字应当视作是 <code>10</code> 进制数字，且它们的总和也应当按 <code>10</code> 进制表示返回。</p>
+
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>n = 34, k = 6
+<strong>输出：</strong>9
+<strong>解释：</strong>34 (10 进制) 在 6 进制下表示为 54 。5 + 4 = 9 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>n = 10, k = 10
+<strong>输出：</strong>1
+<strong>解释：</strong>n 本身就是 10 进制。 1 + 0 = 1 。
+</pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 <= n <= 100</code></li>
+	<li><code>2 <= k <= 10</code></li>
+</ul>
+
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/1800-1899/1836.Remove Duplicates From an Unsorted Linked List/images/tmp-linked-list-1.jpg

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+# [1837. Sum of Digits in Base K](https://leetcode.com/problems/sum-of-digits-in-base-k)
+
+[中文文档](/solution/1800-1899/1837.Sum%20of%20Digits%20in%20Base%20K/README.md)
+
+## Description
+
+<p>Given an integer <code>n</code> (in base <code>10</code>) and a base <code>k</code>, return <em>the <strong>sum</strong> of the digits of </em><code>n</code><em> <strong>after</strong> converting </em><code>n</code><em> from base </em><code>10</code><em> to base </em><code>k</code>.</p>
+
+<p>After converting, each digit should be interpreted as a base <code>10</code> number, and the sum should be returned in base <code>10</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 34, k = 6
+<strong>Output:</strong> 9
+<strong>Explanation: </strong>34 (base 10) expressed in base 6 is 54. 5 + 4 = 9.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 10, k = 10
+<strong>Output:</strong> 1
+<strong>Explanation: </strong>n is already in base 10. 1 + 0 = 1.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 100</code></li>
+	<li><code>2 &lt;= k &lt;= 10</code></li>
+</ul>
+
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/1800-1899/1836.Remove Duplicates From an Unsorted Linked List/images/tmp-linked-list-2.jpg

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+# [1838. 最高频元素的频数](https://leetcode-cn.com/problems/frequency-of-the-most-frequent-element)
+
+[English Version](/solution/1800-1899/1838.Frequency%20of%20the%20Most%20Frequent%20Element/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>元素的 <strong>频数</strong> 是该元素在一个数组中出现的次数。</p>
+
+<p>给你一个整数数组 <code>nums</code> 和一个整数 <code>k</code> 。在一步操作中，你可以选择 <code>nums</code> 的一个下标，并将该下标对应元素的值增加 <code>1</code> 。</p>
+
+<p>执行最多 <code>k</code> 次操作后，返回数组中最高频元素的 <strong>最大可能频数</strong> <em>。</em></p>
+
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [1,2,4], k = 5
+<strong>输出：</strong>3<strong>
+解释：</strong>对第一个元素执行 3 次递增操作，对第二个元素执 2 次递增操作，此时 nums = [4,4,4] 。
+4 是数组中最高频元素，频数是 3 。</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [1,4,8,13], k = 5
+<strong>输出：</strong>2
+<strong>解释：</strong>存在多种最优解决方案：
+- 对第一个元素执行 3 次递增操作，此时 nums = [4,4,8,13] 。4 是数组中最高频元素，频数是 2 。
+- 对第二个元素执行 4 次递增操作，此时 nums = [1,8,8,13] 。8 是数组中最高频元素，频数是 2 。
+- 对第三个元素执行 5 次递增操作，此时 nums = [1,4,13,13] 。13 是数组中最高频元素，频数是 2 。
+</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [3,9,6], k = 2
+<strong>输出：</strong>1
+</pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 <= nums.length <= 10<sup>5</sup></code></li>
+	<li><code>1 <= nums[i] <= 10<sup>5</sup></code></li>
+	<li><code>1 <= k <= 10<sup>5</sup></code></li>
+</ul>
+
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->