feat: add python and java solutions to lcci question

添加《程序员面试金典》题解：面试题 01.06. 字符串压缩

Author: yanglbme

lcci/01.06.Compress String/README.md

@@ -28,20 +28,51 @@
 
 ## 解法
 <!-- 这里可写通用的实现逻辑 -->
-
+双指针遍历字符串求解。
 
 ### Python3
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def compressString(self, S: str) -> str:
+        if len(S) < 2:
+            return S
+        p, q = 0, 1
+        res = ''
+        while q < len(S):
+            if S[p] != S[q]:
+                res += (S[p] + str(q - p))
+                p = q
+            q += 1
+        res += (S[p] + str(q - p))
+        return res if len(res) < len(S) else S
 ```
 
 ### Java
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public String compressString(String S) {
+        if (S == null || S.length() < 2) {
+            return S;
+        }
+        char[] chars = S.toCharArray();
+        int p = 0, q = 1, n = chars.length;
+        StringBuilder sb = new StringBuilder();
+        while (q < n) {
+            if (chars[p] != chars[q]) {
+                sb.append(chars[p]).append(q - p);
+                p = q;
+            }
+            q += 1;
+        }
+        sb.append(chars[p]).append(q - p);
+        String res = sb.toString();
+        return res.length() < n ? res : S;
+    }
+}
 ```
 
 ### ...

lcci/01.06.Compress String/README_EN.md

@@ -1,50 +1,95 @@
-# [01.06. Compress String](https://leetcode-cn.com/problems/compress-string-lcci)
-
-## Description
-<p>Implement a method to perform basic string compression using the counts of repeated characters. For example, the string aabcccccaaa would become a2blc5a3. If the &quot;compressed&quot; string would not become smaller than the original string, your method should return the original string. You can assume the string has only uppercase and lowercase letters (a - z).</p>
-
-<p><strong>Example 1:</strong></p>
-
-<pre>
-<strong>Input: </strong>&quot;aabcccccaaa&quot;
-<strong>Output: </strong>&quot;a2b1c5a3&quot;
-</pre>
-
-<p><strong>Example 2:</strong></p>
-
-<pre>
-<strong>Input: </strong>&quot;abbccd&quot;
-<strong>Output: </strong>&quot;abbccd&quot;
-<strong>Explanation: </strong>
-The compressed string is &quot;a1b2c2d1&quot;, which is longer than the original string.
-</pre>
-
-<p>&nbsp;</p>
-
-<p><strong>Note:</strong></p>
-
-<ol>
-	<li><code>0 &lt;= S.length &lt;= 50000</code></li>
-</ol>
-
-
-
-## Solutions
-
-
-### Python3
-
-```python
-
-```
-
-### Java
-
-```java
-
-```
-
-### ...
-```
-
-```
+# [01.06. Compress String](https://leetcode-cn.com/problems/compress-string-lcci)
+
+## Description
+<p>Implement a method to perform basic string compression using the counts of repeated characters. For example, the string aabcccccaaa would become a2blc5a3. If the &quot;compressed&quot; string would not become smaller than the original string, your method should return the original string. You can assume the string has only uppercase and lowercase letters (a - z).</p>
+
+
+
+<p><strong>Example 1:</strong></p>
+
+
+
+<pre>
+
+<strong>Input: </strong>&quot;aabcccccaaa&quot;
+
+<strong>Output: </strong>&quot;a2b1c5a3&quot;
+
+</pre>
+
+
+
+<p><strong>Example 2:</strong></p>
+
+
+
+<pre>
+
+<strong>Input: </strong>&quot;abbccd&quot;
+
+<strong>Output: </strong>&quot;abbccd&quot;
+
+<strong>Explanation: </strong>
+
+The compressed string is &quot;a1b2c2d1&quot;, which is longer than the original string.
+
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>Note:</strong></p>
+
+- `0 <= S.length <= 50000`
+
+
+
+## Solutions
+
+
+### Python3
+
+```python
+class Solution:
+    def compressString(self, S: str) -> str:
+        if len(S) < 2:
+            return S
+        p, q = 0, 1
+        res = ''
+        while q < len(S):
+            if S[p] != S[q]:
+                res += (S[p] + str(q - p))
+                p = q
+            q += 1
+        res += (S[p] + str(q - p))
+        return res if len(res) < len(S) else S
+```
+
+### Java
+
+```java
+class Solution {
+    public String compressString(String S) {
+        if (S == null || S.length() < 2) {
+            return S;
+        }
+        char[] chars = S.toCharArray();
+        int p = 0, q = 1, n = chars.length;
+        StringBuilder sb = new StringBuilder();
+        while (q < n) {
+            if (chars[p] != chars[q]) {
+                sb.append(chars[p]).append(q - p);
+                p = q;
+            }
+            q += 1;
+        }
+        sb.append(chars[p]).append(q - p);
+        String res = sb.toString();
+        return res.length() < n ? res : S;
+    }
+}
+```
+
+### ...
+```
+
+```

lcci/01.06.Compress String/Solution.java

@@ -0,0 +1,20 @@
+class Solution {
+    public String compressString(String S) {
+        if (S == null || S.length() < 2) {
+            return S;
+        }
+        char[] chars = S.toCharArray();
+        int p = 0, q = 1, n = chars.length;
+        StringBuilder sb = new StringBuilder();
+        while (q < n) {
+            if (chars[p] != chars[q]) {
+                sb.append(chars[p]).append(q - p);
+                p = q;
+            }
+            q += 1;
+        }
+        sb.append(chars[p]).append(q - p);
+        String res = sb.toString();
+        return res.length() < n ? res : S;
+    }
+}

lcci/01.06.Compress String/Solution.py

@@ -0,0 +1,13 @@
+class Solution:
+    def compressString(self, S: str) -> str:
+        if len(S) < 2:
+            return S
+        p, q = 0, 1
+        res = ''
+        while q < len(S):
+            if S[p] != S[q]:
+                res += (S[p] + str(q - p))
+                p = q
+            q += 1
+        res += (S[p] + str(q - p))
+        return res if len(res) < len(S) else S