feat: update lc problems

Author: yanglbme

lcci/03.03.Stack of Plates/README.md

@@ -1,6 +1,9 @@
 # [面试题 03.03. 堆盘子](https://leetcode.cn/problems/stack-of-plates-lcci)
+
 [English Version](/lcci/03.03.Stack%20of%20Plates/README_EN.md)
+
 ## 题目描述
+
 <!-- 这里写题目描述 -->
 <p>堆盘子。设想有一堆盘子，堆太高可能会倒下来。因此，在现实生活中，盘子堆到一定高度时，我们就会另外堆一堆盘子。请实现数据结构<code>SetOfStacks</code>，模拟这种行为。<code>SetOfStacks</code>应该由多个栈组成，并且在前一个栈填满时新建一个栈。此外，<code>SetOfStacks.push()</code>和<code>SetOfStacks.pop()</code>应该与普通栈的操作方法相同（也就是说，pop()返回的值，应该跟只有一个栈时的情况一样）。 进阶：实现一个<code>popAt(int index)</code>方法，根据指定的子栈，执行pop操作。</p>
 <p>当某个栈为空时，应当删除该栈。当栈中没有元素或不存在该栈时，<code>pop</code>，<code>popAt</code>&nbsp;应返回 -1.</p>
@@ -18,20 +21,29 @@
 <strong> 输出</strong>：
 [null, null, null, null, 2, 1, 3]
 </pre>
+
 ## 解法
 <!-- 这里可写通用的实现逻辑 -->
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 <!-- tabs:start -->
+
 ### **Python3**
+
 <!-- 这里可写当前语言的特殊实现逻辑 -->
+
 ```python
 
 ```
 ### **Java**
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 <!-- 这里可写当前语言的特殊实现逻辑 -->
+
 ```java
 
 ```
+
 ### **TypeScript**
+
 ```ts
 class StackOfPlates {
     private cap: number;
@@ -84,8 +96,11 @@ class StackOfPlates {
  * var param_3 = obj.popAt(index)
  */
 ```
+
 ### **...**
+
 ```
 
 ```
+
 <!-- tabs:end -->

lcci/03.03.Stack of Plates/README_EN.md

@@ -1,49 +1,60 @@
 # [03.03. Stack of Plates](https://leetcode.cn/problems/stack-of-plates-lcci)
+
 [中文文档](/lcci/03.03.Stack%20of%20Plates/README.md)
+
 ## Description
+
 <p>Imagine a (literal) stack of plates. If the stack gets too high, it might topple. Therefore, in real life, we would likely start a new stack when the previous stack exceeds some threshold. Implement a data structure <code>SetOfStacks</code> that mimics this.&nbsp;<code>SetOfStacks</code> should be composed of several stacks and should create a new stack once the previous one exceeds capacity. <code>SetOfStacks.push()</code> and <code>SetOfStacks.pop()</code> should behave identically to a single stack (that is, <code>pop()</code> should return the same values as it would if there were just a single stack). Follow Up: Implement a function <code>popAt(int index)</code> which performs a pop operation on a specific sub-stack.</p>
 <p>You should delete the sub-stack when it becomes empty. <code>pop</code>, <code>popAt</code> should return -1 when there&#39;s no element to pop.</p>
 <p><strong>Example1:</strong></p>
 <pre>
 
-<strong> Input</strong>: 
+<strong> Input</strong>:
 
 [&quot;StackOfPlates&quot;, &quot;push&quot;, &quot;push&quot;, &quot;popAt&quot;, &quot;pop&quot;, &quot;pop&quot;]
 
 [[1], [1], [2], [1], [], []]
 
-<strong> Output</strong>: 
+<strong> Output</strong>:
 
 [null, null, null, 2, 1, -1]
 
-<strong> Explanation</strong>: 
+<strong> Explanation</strong>:
 
 </pre>
 <p><strong>Example2:</strong></p>
 <pre>
 
-<strong> Input</strong>: 
+<strong> Input</strong>:
 
 [&quot;StackOfPlates&quot;, &quot;push&quot;, &quot;push&quot;, &quot;push&quot;, &quot;popAt&quot;, &quot;popAt&quot;, &quot;popAt&quot;]
 
 [[2], [1], [2], [3], [0], [0], [0]]
 
-<strong> Output</strong>: 
+<strong> Output</strong>:
 
 [null, null, null, null, 2, 1, 3]
 
 </pre>
+
 ## Solutions
+
 <!-- tabs:start -->
+
 ### **Python3**
+
 ```python
 
 ```
+
 ### **Java**
+
 ```java
 
 ```
+
 ### **TypeScript**
+
 ```ts
 class StackOfPlates {
     private cap: number;
@@ -96,8 +107,11 @@ class StackOfPlates {
  * var param_3 = obj.popAt(index)
  */
 ```
+
 ### **...**
+
 ```
 
 ```
+
 <!-- tabs:end -->