chore: update lc problems

Author: yanglbme

solution/0000-0099/0001.Two Sum/README.md

@@ -174,7 +174,7 @@ public class Solution {
             {
                 m.Add(v, i);
             }
-            
+
         }
         return null;
     }

solution/0000-0099/0001.Two Sum/README_EN.md

@@ -135,7 +135,7 @@ public class Solution {
             {
                 m.Add(v, i);
             }
-            
+
         }
         return null;
     }

solution/0100-0199/0146.LRU Cache/README_EN.md

@@ -14,7 +14,7 @@
 	<li><code>void put(int key, int value)</code>&nbsp;Update the value of the <code>key</code> if the <code>key</code> exists. Otherwise, add the <code>key-value</code> pair to the cache. If the number of keys exceeds the <code>capacity</code> from this operation, <strong>evict</strong> the least recently used key.</li>
 </ul>
 
-<p>The functions&nbsp;<code data-stringify-type="code">get</code>&nbsp;and&nbsp;<code data-stringify-type="code">put</code>&nbsp;must each run in <code>O(1)</code> average time complexity.</p>
+<p>The functions <code data-stringify-type="code">get</code> and <code data-stringify-type="code">put</code>&nbsp;must each run in <code>O(1)</code> average time complexity.</p>
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
@@ -46,7 +46,7 @@ lRUCache.get(4);    // return 4
 	<li><code>1 &lt;= capacity &lt;= 3000</code></li>
 	<li><code>0 &lt;= key &lt;= 10<sup>4</sup></code></li>
 	<li><code>0 &lt;= value &lt;= 10<sup>5</sup></code></li>
-	<li>At most 2<code>&nbsp;* 10<sup>5</sup></code>&nbsp;calls will be made to <code>get</code> and <code>put</code>.</li>
+	<li>At most <code>2 * 10<sup>5</sup></code> calls will be made to <code>get</code> and <code>put</code>.</li>
 </ul>
 
 ## Solutions

solution/0100-0199/0159.Longest Substring with At Most Two Distinct Characters/README.md

@@ -6,22 +6,35 @@
 
 <!-- 这里写题目描述 -->
 
-<p>给定一个字符串<strong><em> s</em></strong> ，找出&nbsp;<strong>至多&nbsp;</strong>包含两个不同字符的最长子串 <strong><em>t</em> </strong>，并返回该子串的长度。</p>
+给你一个字符串 <code>s</code> ，请你找出&nbsp;<strong>至多&nbsp;</strong>包含 <strong>两个不同字符</strong> 的最长子串，并返回该子串的长度。
 
-<p><strong>示例 1:</strong></p>
+<p>&nbsp;</p>
 
-<pre><strong>输入:</strong> &quot;eceba&quot;
-<strong>输出: </strong>3
-<strong>解释: <em>t</em></strong> 是 &quot;ece&quot;，长度为3。
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "eceba"
+<strong>输出：</strong>3
+<strong>解释：</strong>满足题目要求的子串是 "ece" ，长度为 3 。
 </pre>
 
-<p><strong>示例 2:</strong></p>
+<p><strong>示例 2：</strong></p>
 
-<pre><strong>输入:</strong> &quot;ccaabbb&quot;
-<strong>输出: </strong>5
-<strong>解释: <em>t</em></strong><em> </em>是 &quot;aabbb&quot;，长度为5。
+<pre>
+<strong>输入：</strong>s = "ccaabbb"
+<strong>输出：</strong>5
+<strong>解释：</strong>满足题目要求的子串是 "aabbb" ，长度为 5 。
 </pre>
 
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>s</code> 由英文字母组成</li>
+</ul>
+
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->

solution/0200-0299/0215.Kth Largest Element in an Array/README.md

@@ -10,6 +10,8 @@
 
 <p>请注意，你需要找的是数组排序后的第 <code>k</code> 个最大的元素，而不是第 <code>k</code> 个不同的元素。</p>
 
+<p>你必须设计并实现时间复杂度为 <code>O(n)</code> 的算法解决此问题。</p>
+
 <p>&nbsp;</p>
 
 <p><strong>示例 1:</strong></p>

solution/0300-0399/0304.Range Sum Query 2D - Immutable/README_EN.md

@@ -17,6 +17,8 @@
 	<li><code>int sumRegion(int row1, int col1, int row2, int col2)</code> Returns the <strong>sum</strong> of the elements of <code>matrix</code> inside the rectangle defined by its <strong>upper left corner</strong> <code>(row1, col1)</code> and <strong>lower right corner</strong> <code>(row2, col2)</code>.</li>
 </ul>
 
+<p>You must design an algorithm where <code>sumRegion</code> works on <code>O(1)</code> time complexity.</p>
+
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
 <img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0300-0399/0304.Range%20Sum%20Query%202D%20-%20Immutable/images/sum-grid.jpg" style="width: 415px; height: 415px;" />

solution/0300-0399/0340.Longest Substring with At Most K Distinct Characters/README.md

@@ -6,32 +6,32 @@
 
 <!-- 这里写题目描述 -->
 
-<p>给定一个字符串<strong><em> <code>s</code></em></strong> ，找出 <strong>至多 </strong>包含<em> <code>k</code></em> 个不同字符的最长子串 <strong><em>T</em></strong>。</p>
+<p>给你一个字符串 <code>s</code> 和一个整数 <code>k</code> ，请你找出&nbsp;<strong>至多&nbsp;</strong>包含<em> <code>k</code></em> 个 <strong>不同</strong> 字符的最长子串，并返回该子串的长度。</p>
 
-<p> </p>
+<p>&nbsp;</p>
 
-<p><strong>示例 1:</strong></p>
+<p><strong>示例 1：</strong></p>
 
 <pre>
-<strong>输入: </strong>s = "eceba", k = 2
-<strong>输出: </strong>3
-<strong>解释: </strong>则<strong> </strong>T 为 "ece"，所以长度为 3。</pre>
+<strong>输入：</strong>s = "eceba", k = 2
+<strong>输出：</strong>3
+<strong>解释：</strong>满足题目要求的子串是 "ece" ，长度为 3 。</pre>
 
-<p><strong>示例 2:</strong></p>
+<p><strong>示例 2：</strong></p>
 
 <pre>
-<strong>输入: </strong>s = "aa", k = 1
-<strong>输出: </strong>2
-<strong>解释: </strong>则 T 为 "aa"，所以长度为 2。
+<strong>输入：</strong>s = "aa", k = 1
+<strong>输出：</strong>2
+<strong>解释：</strong>满足题目要求的子串是 "aa" ，长度为 2 。
 </pre>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>1 <= s.length <= 5 * 10<sup>4</sup></code></li>
-	<li><code>0 <= k <= 50</code></li>
+	<li><code>1 &lt;= s.length &lt;= 5 * 10<sup>4</sup></code></li>
+	<li><code>0 &lt;= k &lt;= 50</code></li>
 </ul>
 
 ## 解法

solution/0600-0699/0626.Exchange Seats/README_EN.md

@@ -11,7 +11,7 @@
 | Column Name | Type    |
 +-------------+---------+
 | id          | int     |
-| name        | varchar |
+| student     | varchar |
 +-------------+---------+
 id is the primary key column for this table.
 Each row of this table indicates the name and the ID of a student.

solution/0600-0699/0636.Exclusive Time of Functions/README.md

@@ -176,7 +176,7 @@ function exclusiveTime(n: number, logs: string[]): number[] {
             }
         }
     }
-    
+
     return res;
 }
 ```

solution/0600-0699/0636.Exclusive Time of Functions/README_EN.md

@@ -150,7 +150,7 @@ function exclusiveTime(n: number, logs: string[]): number[] {
             }
         }
     }
-    
+
     return res;
 }
 ```