feat: add solutions to lc problems: No.0521,0522

* No.0521.Longest Uncommon Subsequence
* No.0522.Longest Uncommon Subsequence II

Author: yanglbme

solution/0500-0599/0521.Longest Uncommon Subsequence I/README.md

@@ -6,42 +6,49 @@
 
 <!-- 这里写题目描述 -->
 
-<p>给你两个字符串，请你从这两个字符串中找出最长的特殊序列。</p>
+<p>给你两个字符串&nbsp;<code>a</code>&nbsp;和&nbsp;<code>b</code>，请返回 <em>这两个字符串中 <strong>最长的特殊序列</strong>&nbsp;</em> 。如果不存在，则返回 <code>-1</code>&nbsp;。</p>
 
-<p>「最长特殊序列」定义如下：该序列为某字符串独有的最长子序列（即不能是其他字符串的子序列）。</p>
+<p><strong>「最长特殊序列」</strong>&nbsp;定义如下：该序列为&nbsp;<strong>某字符串独有的最长子序列（即不能是其他字符串的子序列）</strong>&nbsp;。</p>
 
-<p><strong>子序列</strong> 可以通过删去字符串中的某些字符实现，但不能改变剩余字符的相对顺序。空序列为所有字符串的子序列，任何字符串为其自身的子序列。</p>
+<p>字符串&nbsp;<code>s</code>&nbsp;的子序列是在从&nbsp;<code>s</code>&nbsp;中删除任意数量的字符后可以获得的字符串。</p>
 
-<p>输入为两个字符串，输出最长特殊序列的长度。如果不存在，则返回 -1。</p>
+<ul>
+	<li>例如，<code>“abc”</code>&nbsp;是 <code>“aebdc”</code> 的子序列，因为您可以删除 <code>“aebdc”</code> 中的下划线字符来得到 <code>“abc”</code> 。 <code>“aebdc”</code> 的子序列还包括 <code>“aebdc”</code> 、 <code>“aeb”</code> 和 <code>“”</code> (空字符串)。</li>
+</ul>
 
 <p>&nbsp;</p>
 
 <p><strong>示例 1：</strong></p>
 
-<pre><strong>输入:</strong> &quot;aba&quot;, &quot;cdc&quot;
+<pre>
+<strong>输入:</strong> a = "aba", b = "cdc"
 <strong>输出:</strong> 3
-<strong>解释:</strong> 最长特殊序列可为 &quot;aba&quot; (或 &quot;cdc&quot;)，两者均为自身的子序列且不是对方的子序列。</pre>
+<strong>解释:</strong> 最长特殊序列可为 "aba" (或 "cdc")，两者均为自身的子序列且不是对方的子序列。</pre>
 
 <p><strong>示例 2：</strong></p>
 
-<pre><strong>输入：</strong>a = &quot;aaa&quot;, b = &quot;bbb&quot;
+<pre>
+<strong>输入：</strong>a = "aaa", b = "bbb"
 <strong>输出：</strong>3
+<strong>解释:</strong> 最长特殊序列是“aaa”和“bbb”。
 </pre>
 
 <p><strong>示例 3：</strong></p>
 
-<pre><strong>输入：</strong>a = &quot;aaa&quot;, b = &quot;aaa&quot;
+<pre>
+<strong>输入：</strong>a = "aaa", b = "aaa"
 <strong>输出：</strong>-1
+<strong>解释:</strong> 字符串a的每个子序列也是字符串b的每个子序列。同样，字符串b的每个子序列也是字符串a的子序列。
 </pre>
 
 <p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>
 
-<ol>
-	<li>两个字符串长度均处于区间&nbsp;<code>[1 - 100]</code> 。</li>
-	<li>字符串中的字符仅含有&nbsp;<code>&#39;a&#39;~&#39;z&#39;</code> 。</li>
-</ol>
+<ul>
+	<li><code>1 &lt;= a.length, b.length &lt;= 100</code></li>
+	<li><code>a</code>&nbsp;和&nbsp;<code>b</code>&nbsp;由小写英文字母组成</li>
+</ul>
 
 ## 解法
 
@@ -55,7 +62,7 @@
 
 而特殊序列则是求**非子序列**，此时列举 `a` 的子序列 `"abc"`，`b` 拿不出来，那这就是一个成功的非子序列。
 
-如此，在 `a != b` 时，谁最长谁就是 *最长的特殊序列*
+如此，在 `a != b` 时，谁最长谁就是 _最长的特殊序列_
 
 <!-- tabs:start -->
 
@@ -65,12 +72,7 @@
 
 ```python
 class Solution:
-    def findLUSlength(self, a, b):
-        """
-        :type a: str
-        :type b: str
-        :rtype: int
-        """
+    def findLUSlength(self, a: str, b: str) -> int:
         return -1 if a == b else max(len(a), len(b))
 ```
 
@@ -81,9 +83,7 @@ class Solution:
 ```java
 class Solution {
     public int findLUSlength(String a, String b) {
-        if (a.equals(b))
-            return -1;
-        return Math.max(a.length(), b.length());
+        return a.equals(b) ? -1 : Math.max(a.length(), b.length());
     }
 }
 ```
@@ -101,6 +101,31 @@ impl Solution {
 }
 ```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int findLUSlength(string a, string b) {
+        return a == b ? -1 : max(a.size(), b.size());
+    }
+};
+```
+
+### **Go**
+
+```go
+func findLUSlength(a string, b string) int {
+	if a == b {
+		return -1
+	}
+	if len(a) > len(b) {
+		return len(a)
+	}
+	return len(b)
+}
+```
+
 ### **...**
 
 ```

solution/0500-0599/0521.Longest Uncommon Subsequence I/README_EN.md

@@ -4,13 +4,15 @@
 
 ## Description
 
-<p>Given two strings <code>a</code>&nbsp;and <code>b</code>, find the length of the&nbsp;<strong>longest uncommon subsequence</strong>&nbsp;between them.</p>
+<p>Given two strings <code>a</code> and <code>b</code>, return <em>the length of the <strong>longest uncommon subsequence</strong> between </em><code>a</code> <em>and</em> <code>b</code>. If the longest uncommon subsequence does not exist, return <code>-1</code>.</p>
 
-<p>A&nbsp;<b>subsequence</b>&nbsp;of&nbsp;a string&nbsp;<code>s</code>&nbsp;is a string that can be obtained after deleting any number of characters from <code>s</code>. For example, <code>&quot;abc&quot;</code>&nbsp;is a subsequence of <code>&quot;aebdc&quot;</code>&nbsp;because you can delete the underlined characters in&nbsp;<code>&quot;a<u>e</u>b<u>d</u>c&quot;</code>&nbsp;to get <code>&quot;abc&quot;</code>. Other subsequences of&nbsp;<code>&quot;aebdc&quot;</code>&nbsp;include&nbsp;<code>&quot;aebdc&quot;</code>,&nbsp;<code>&quot;aeb&quot;</code>,&nbsp;and&nbsp;<code>&quot;&quot;</code>&nbsp;(empty string).</p>
+<p>An <strong>uncommon subsequence</strong> between two strings is a string that is a <strong>subsequence of one but not the other</strong>.</p>
 
-<p>An&nbsp;<strong>uncommon subsequence</strong>&nbsp;between two strings&nbsp;is a string that is a <strong>subsequence of one&nbsp;but not the other</strong>.</p>
+<p>A <strong>subsequence</strong> of a string <code>s</code> is a string that can be obtained after deleting any number of characters from <code>s</code>.</p>
 
-<p>Return <em>the length of the <strong>longest uncommon subsequence</strong>&nbsp;between <code>a</code>&nbsp;and <code>b</code></em>. If the longest uncommon subsequence doesn&#39;t exist, return <code>-1</code>.</p>
+<ul>
+	<li>For example, <code>&quot;abc&quot;</code> is a subsequence of <code>&quot;aebdc&quot;</code> because you can delete the underlined characters in <code>&quot;a<u>e</u>b<u>d</u>c&quot;</code> to get <code>&quot;abc&quot;</code>. Other subsequences of <code>&quot;aebdc&quot;</code> include <code>&quot;aebdc&quot;</code>, <code>&quot;aeb&quot;</code>, and <code>&quot;&quot;</code> (empty string).</li>
+</ul>
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
@@ -54,12 +56,7 @@ Note that &quot;cdc&quot; is also a longest uncommon subsequence.
 
 ```python
 class Solution:
-    def findLUSlength(self, a, b):
-        """
-        :type a: str
-        :type b: str
-        :rtype: int
-        """
+    def findLUSlength(self, a: str, b: str) -> int:
         return -1 if a == b else max(len(a), len(b))
 ```
 
@@ -68,9 +65,7 @@ class Solution:
 ```java
 class Solution {
     public int findLUSlength(String a, String b) {
-        if (a.equals(b))
-            return -1;
-        return Math.max(a.length(), b.length());
+        return a.equals(b) ? -1 : Math.max(a.length(), b.length());
     }
 }
 ```
@@ -88,6 +83,31 @@ impl Solution {
 }
 ```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int findLUSlength(string a, string b) {
+        return a == b ? -1 : max(a.size(), b.size());
+    }
+};
+```
+
+### **Go**
+
+```go
+func findLUSlength(a string, b string) int {
+	if a == b {
+		return -1
+	}
+	if len(a) > len(b) {
+		return len(a)
+	}
+	return len(b)
+}
+```
+
 ### **...**
 
 ```

solution/0500-0599/0521.Longest Uncommon Subsequence I/Solution.cpp

@@ -0,0 +1,6 @@
+class Solution {
+public:
+    int findLUSlength(string a, string b) {
+        return a == b ? -1 : max(a.size(), b.size());
+    }
+};

solution/0500-0599/0521.Longest Uncommon Subsequence I/Solution.go

@@ -0,0 +1,9 @@
+func findLUSlength(a string, b string) int {
+	if a == b {
+		return -1
+	}
+	if len(a) > len(b) {
+		return len(a)
+	}
+	return len(b)
+}

solution/0500-0599/0521.Longest Uncommon Subsequence I/Solution.java

@@ -1,7 +1,5 @@
 class Solution {
     public int findLUSlength(String a, String b) {
-        if (a.equals(b))
-            return -1;
-        return Math.max(a.length(), b.length());
+        return a.equals(b) ? -1 : Math.max(a.length(), b.length());
     }
-}
+}

solution/0500-0599/0521.Longest Uncommon Subsequence I/Solution.py

@@ -1,8 +1,3 @@
 class Solution:
-    def findLUSlength(self, a, b):
-        """
-        :type a: str
-        :type b: str
-        :rtype: int
-        """
+    def findLUSlength(self, a: str, b: str) -> int:
         return -1 if a == b else max(len(a), len(b))