fix: update

Author: yanglbme

solution/2600-2699/2674.Split a Circular Linked List/README.md

@@ -46,7 +46,7 @@ tags:
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>list</code> 中的节点数范围为 <code>[2, 105]</code></li>
+	<li><code>list</code> 中的节点数范围为 <code>[2, 10<sup>5</sup>]</code></li>
 	<li><code>0 &lt;= Node.val &lt;= 10<sup>9</sup></code></li>
 	<li><code>LastNode.next = FirstNode</code> ，其中 <code>LastNode</code> 是链表的最后一个节点，<code>FirstNode</code> 是第一个节点。</li>
 </ul>

solution/3400-3499/3442.Maximum Difference Between Even and Odd Frequency I/README.md

@@ -71,7 +71,11 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3442.Ma
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：计数
+
+我们可以用一个哈希表或数组 $\textit{cnt}$ 记录字符串 $s$ 中每个字符的出现次数。然后遍历 $\textit{cnt}$，找出出现奇数次的字符的最大频次 $a$ 和出现偶数次的字符的最小频次 $b$，最后返回 $a - b$ 即可。
+
+时间复杂度 $O(n)$，其中 $n$ 是字符串 $s$ 的长度。空间复杂度 $O(|\Sigma|)$，其中 $\Sigma$ 是字符集，本题中 $|\Sigma| = 26$。
 
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solution/3400-3499/3442.Maximum Difference Between Even and Odd Frequency I/README_EN.md

@@ -69,7 +69,11 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3442.Ma
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Counting
+
+We can use a hash table or an array $\textit{cnt}$ to record the occurrences of each character in the string $s$. Then, we traverse $\textit{cnt}$ to find the maximum frequency $a$ of characters that appear an odd number of times and the minimum frequency $b$ of characters that appear an even number of times. Finally, we return $a - b$.
+
+The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(|\Sigma|)$, where $\Sigma$ is the character set. In this problem, $|\Sigma| = 26$.
 
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