feat: add new lc problems

Author: yanglbme

solution/2200-2299/2239.Find Closest Number to Zero/README.md

@@ -0,0 +1,76 @@
+# [2239. 找到最接近 0 的数字](https://leetcode-cn.com/problems/find-closest-number-to-zero)
+
+[English Version](/solution/2200-2299/2239.Find%20Closest%20Number%20to%20Zero/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个长度为 <code>n</code>&nbsp;的整数数组&nbsp;<code>nums</code>&nbsp;，请你返回 <code>nums</code>&nbsp;中最 <strong>接近</strong>&nbsp;<code>0</code>&nbsp;的数字。如果有多个答案，请你返回它们中的 <strong>最大值</strong>&nbsp;。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><b>输入：</b>nums = [-4,-2,1,4,8]
+<b>输出：</b>1
+<strong>解释：</strong>
+-4 到 0 的距离为 |-4| = 4 。
+-2 到 0 的距离为 |-2| = 2 。
+1 到 0 的距离为 |1| = 1 。
+4 到 0 的距离为 |4| = 4 。
+8 到 0 的距离为 |8| = 8 。
+所以，数组中距离 0 最近的数字为 1 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><b>输入：</b>nums = [2,-1,1]
+<b>输出：</b>1
+<b>解释：</b>1 和 -1 都是距离 0 最近的数字，所以返回较大值 1 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 1000</code></li>
+	<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 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
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2200-2299/2239.Find Closest Number to Zero/README_EN.md

@@ -0,0 +1,67 @@
+# [2239. Find Closest Number to Zero](https://leetcode.com/problems/find-closest-number-to-zero)
+
+[中文文档](/solution/2200-2299/2239.Find%20Closest%20Number%20to%20Zero/README.md)
+
+## Description
+
+<p>Given an integer array <code>nums</code> of size <code>n</code>, return <em>the number with the value <strong>closest</strong> to </em><code>0</code><em> in </em><code>nums</code>. If there are multiple answers, return <em>the number with the <strong>largest</strong> value</em>.</p>
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [-4,-2,1,4,8]
+<strong>Output:</strong> 1
+<strong>Explanation:</strong>
+The distance from -4 to 0 is |-4| = 4.
+The distance from -2 to 0 is |-2| = 2.
+The distance from 1 to 0 is |1| = 1.
+The distance from 4 to 0 is |4| = 4.
+The distance from 8 to 0 is |8| = 8.
+Thus, the closest number to 0 in the array is 1.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [2,-1,1]
+<strong>Output:</strong> 1
+<strong>Explanation:</strong> 1 and -1 are both the closest numbers to 0, so 1 being larger is returned.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 1000</code></li>
+	<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2200-2299/2240.Number of Ways to Buy Pens and Pencils/README.md

@@ -0,0 +1,75 @@
+# [2240. 买钢笔和铅笔的方案数](https://leetcode-cn.com/problems/number-of-ways-to-buy-pens-and-pencils)
+
+[English Version](/solution/2200-2299/2240.Number%20of%20Ways%20to%20Buy%20Pens%20and%20Pencils/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个整数&nbsp;<code>total</code>&nbsp;，表示你拥有的总钱数。同时给你两个整数&nbsp;<code>cost1</code> 和&nbsp;<code>cost2</code>&nbsp;，分别表示一支钢笔和一支铅笔的价格。你可以花费你部分或者全部的钱，去买任意数目的两种笔。</p>
+
+<p>请你返回购买钢笔和铅笔的&nbsp;<strong>不同方案数目</strong>&nbsp;。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><b>输入：</b>total = 20, cost1 = 10, cost2 = 5
+<b>输出：</b>9
+<b>解释：</b>一支钢笔的价格为 10 ，一支铅笔的价格为 5 。
+- 如果你买 0 支钢笔，那么你可以买 0 ，1 ，2 ，3 或者 4 支铅笔。
+- 如果你买 1 支钢笔，那么你可以买 0 ，1 或者 2 支铅笔。
+- 如果你买 2 支钢笔，那么你没法买任何铅笔。
+所以买钢笔和铅笔的总方案数为 5 + 3 + 1 = 9 种。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><b>输入：</b>total = 5, cost1 = 10, cost2 = 10
+<b>输出：</b>1
+<b>解释：</b>钢笔和铅笔的价格都为 10 ，都比拥有的钱数多，所以你没法购买任何文具。所以只有 1 种方案：买 0 支钢笔和 0 支铅笔。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= total, cost1, cost2 &lt;= 10<sup>6</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
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2200-2299/2240.Number of Ways to Buy Pens and Pencils/README_EN.md

@@ -0,0 +1,67 @@
+# [2240. Number of Ways to Buy Pens and Pencils](https://leetcode.com/problems/number-of-ways-to-buy-pens-and-pencils)
+
+[中文文档](/solution/2200-2299/2240.Number%20of%20Ways%20to%20Buy%20Pens%20and%20Pencils/README.md)
+
+## Description
+
+<p>You are given an integer <code>total</code> indicating the amount of money you have. You are also given two integers <code>cost1</code> and <code>cost2</code> indicating the price of a pen and pencil respectively. You can spend <strong>part or all</strong> of your money to buy multiple quantities (or none) of each kind of writing utensil.</p>
+
+<p>Return <em>the <strong>number of distinct ways</strong> you can buy some number of pens and pencils.</em></p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> total = 20, cost1 = 10, cost2 = 5
+<strong>Output:</strong> 9
+<strong>Explanation:</strong> The price of a pen is 10 and the price of a pencil is 5.
+- If you buy 0 pens, you can buy 0, 1, 2, 3, or 4 pencils.
+- If you buy 1 pen, you can buy 0, 1, or 2 pencils.
+- If you buy 2 pens, you cannot buy any pencils.
+The total number of ways to buy pens and pencils is 5 + 3 + 1 = 9.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> total = 5, cost1 = 10, cost2 = 10
+<strong>Output:</strong> 1
+<strong>Explanation:</strong> The price of both pens and pencils are 10, which cost more than total, so you cannot buy any writing utensils. Therefore, there is only 1 way: buy 0 pens and 0 pencils.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= total, cost1, cost2 &lt;= 10<sup>6</sup></code></li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2200-2299/2241.Design an ATM Machine/README.md

@@ -0,0 +1,93 @@
+# [2241. 设计一个 ATM 机器](https://leetcode-cn.com/problems/design-an-atm-machine)
+
+[English Version](/solution/2200-2299/2241.Design%20an%20ATM%20Machine/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>一个 ATM 机器，存有&nbsp;<code>5</code>&nbsp;种面值的钞票：<code>20</code>&nbsp;，<code>50</code>&nbsp;，<code>100</code>&nbsp;，<code>200</code>&nbsp;和&nbsp;<code>500</code>&nbsp;美元。初始时，ATM 机是空的。用户可以用它存或者取任意数目的钱。</p>
+
+<p>取款时，机器会优先取 <b>较大</b>&nbsp;数额的钱。</p>
+
+<ul>
+	<li>比方说，你想取&nbsp;<code>$300</code>&nbsp;，并且机器里有&nbsp;<code>2</code>&nbsp;张 <code>$50</code>&nbsp;的钞票，<code>1</code>&nbsp;张&nbsp;<code>$100</code>&nbsp;的钞票和<code>1</code>&nbsp;张&nbsp;<code>$200</code>&nbsp;的钞票，那么机器会取出&nbsp;<code>$100</code> 和&nbsp;<code>$200</code>&nbsp;的钞票。</li>
+	<li>但是，如果你想取&nbsp;<code>$600</code>&nbsp;，机器里有&nbsp;<code>3</code>&nbsp;张&nbsp;<code>$200</code>&nbsp;的钞票和<code>1</code>&nbsp;张&nbsp;<code>$500</code>&nbsp;的钞票，那么取款请求会被拒绝，因为机器会先取出&nbsp;<code>$500</code>&nbsp;的钞票，然后无法取出剩余的&nbsp;<code>$100</code>&nbsp;。注意，因为有&nbsp;<code>$500</code>&nbsp;钞票的存在，机器&nbsp;<strong>不能</strong>&nbsp;取&nbsp;<code>$200</code>&nbsp;的钞票。</li>
+</ul>
+
+<p>请你实现 ATM 类：</p>
+
+<ul>
+	<li><code>ATM()</code>&nbsp;初始化 ATM 对象。</li>
+	<li><code>void deposit(int[] banknotesCount)</code>&nbsp;分别存入&nbsp;<code>$20</code>&nbsp;，<code>$50</code>，<code>$100</code>，<code>$200</code>&nbsp;和&nbsp;<code>$500</code>&nbsp;钞票的数目。</li>
+	<li><code>int[] withdraw(int amount)</code>&nbsp;返回一个长度为&nbsp;<code>5</code>&nbsp;的数组，分别表示&nbsp;<code>$20</code>&nbsp;，<code>$50</code>，<code>$100</code>&nbsp;，<code>$200</code>&nbsp;和&nbsp;<code>$500</code>&nbsp;钞票的数目，并且更新 ATM 机里取款后钞票的剩余数量。如果无法取出指定数额的钱，请返回&nbsp;<code>[-1]</code>&nbsp;（这种情况下 <strong>不</strong>&nbsp;取出任何钞票）。</li>
+</ul>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>
+["ATM", "deposit", "withdraw", "deposit", "withdraw", "withdraw"]
+[[], [[0,0,1,2,1]], [600], [[0,1,0,1,1]], [600], [550]]
+<strong>输出：</strong>
+[null, null, [0,0,1,0,1], null, [-1], [0,1,0,0,1]]
+
+<strong>解释：</strong>
+ATM atm = new ATM();
+atm.deposit([0,0,1,2,1]); // 存入 1 张 $100 ，2 张 $200 和 1 张 $500 的钞票。
+atm.withdraw(600);        // 返回 [0,0,1,0,1] 。机器返回 1 张 $100 和 1 张 $500 的钞票。机器里剩余钞票的数量为 [0,0,0,2,0] 。
+atm.deposit([0,1,0,1,1]); // 存入 1 张 $50 ，1 张 $200 和 1 张 $500 的钞票。
+                          // 机器中剩余钞票数量为 [0,1,0,3,1] 。
+atm.withdraw(600);        // 返回 [-1] 。机器会尝试取出 $500 的钞票，然后无法得到剩余的 $100 ，所以取款请求会被拒绝。
+                          // 由于请求被拒绝，机器中钞票的数量不会发生改变。
+atm.withdraw(550);        // 返回 [0,1,0,0,1] ，机器会返回 1 张 $50 的钞票和 1 张 $500 的钞票。</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>banknotesCount.length == 5</code></li>
+	<li><code>0 &lt;= banknotesCount[i] &lt;= 10<sup>9</sup></code></li>
+	<li><code>1 &lt;= amount &lt;= 10<sup>9</sup></code></li>
+	<li><strong>总共</strong>&nbsp;最多有&nbsp;<code>5000</code>&nbsp;次&nbsp;<code>withdraw</code> 和&nbsp;<code>deposit</code>&nbsp;的调用。</li>
+	<li><span style="">函数 </span><code>withdraw</code> 和&nbsp;<code>deposit</code>&nbsp;至少各有 <strong>一次&nbsp;</strong>调用。</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
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->