feat: add python and java solutions to lcci question

添加《程序员面试金典》题解：面试题 02.06. 回文链表

Author: yanglbme

lcci/02.06.Palindrome Linked List/README.md

@@ -26,20 +26,83 @@
 
 ## 解法
 <!-- 这里可写通用的实现逻辑 -->
-
+先利用快慢指针找到链表中点，之后将后半部分链表利用头插法逆序，再比对前后两段链表得出结果。
 
 ### Python3
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def isPalindrome(self, head: ListNode) -> bool:
+        if head is None or head.next is None:
+            return True
+        slow, fast = head, head.next
+        while fast and fast.next:
+            fast = fast.next.next
+            slow = slow.next
+        cur = slow.next
+        slow.next = None
+        while cur:
+            t = cur.next
+            cur.next = slow.next
+            slow.next = cur
+            cur = t
+        t = slow.next
+        while t:
+            if head.val != t.val:
+                return False
+            t = t.next
+            head = head.next
+        return True
 ```
 
 ### Java
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public boolean isPalindrome(ListNode head) {
+        if (head == null || head.next == null) {
+            return true;
+        }
+        ListNode slow = head, fast = head.next;
+        while (fast != null && fast.next != null) {
+            slow = slow.next;
+            fast = fast.next.next;
+        }
+        ListNode cur = slow.next;
+        slow.next = null;
+        while (cur != null) {
+            ListNode t = cur.next;
+            cur.next = slow.next;
+            slow.next = cur;
+            cur = t;
+        }
+        ListNode t = slow.next;
+        while (t != null) {
+            if (head.val != t.val) {
+                return false;
+            }
+            head = head.next;
+            t = t.next;
+        }
+        return true;
+    }
+}
 ```
 
 ### ...

lcci/02.06.Palindrome Linked List/README_EN.md

@@ -1,47 +1,129 @@
-# [02.06. Palindrome Linked List](https://leetcode-cn.com/problems/palindrome-linked-list-lcci)
-
-## Description
-<p>Implement a function to check if a linked list is a palindrome.</p>
-
-<p>&nbsp;</p>
-
-<p><strong>Example 1: </strong></p>
-
-<pre>
-<strong>Input:  </strong>1-&gt;2
-<strong>Output: </strong> false 
-</pre>
-
-<p><strong>Example 2: </strong></p>
-
-<pre>
-<strong>Input:  </strong>1-&gt;2-&gt;2-&gt;1
-<strong>Output: </strong> true 
-</pre>
-
-<p>&nbsp;</p>
-
-<p><b>Follow up:</b><br />
-Could you do it in O(n) time and O(1) space?</p>
-
-
-
-## Solutions
-
-
-### Python3
-
-```python
-
-```
-
-### Java
-
-```java
-
-```
-
-### ...
-```
-
-```
+# [02.06. Palindrome Linked List](https://leetcode-cn.com/problems/palindrome-linked-list-lcci)
+
+## Description
+<p>Implement a function to check if a linked list is a palindrome.</p>
+
+
+
+<p>&nbsp;</p>
+
+
+
+<p><strong>Example 1: </strong></p>
+
+
+
+<pre>
+
+<strong>Input:  </strong>1-&gt;2
+
+<strong>Output: </strong> false 
+
+</pre>
+
+
+
+<p><strong>Example 2: </strong></p>
+
+
+
+<pre>
+
+<strong>Input:  </strong>1-&gt;2-&gt;2-&gt;1
+
+<strong>Output: </strong> true 
+
+</pre>
+
+
+
+<p>&nbsp;</p>
+
+
+
+<p><b>Follow up:</b><br />
+
+Could you do it in O(n) time and O(1) space?</p>
+
+## Solutions
+
+
+### Python3
+
+```python
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def isPalindrome(self, head: ListNode) -> bool:
+        if head is None or head.next is None:
+            return True
+        slow, fast = head, head.next
+        while fast and fast.next:
+            fast = fast.next.next
+            slow = slow.next
+        cur = slow.next
+        slow.next = None
+        while cur:
+            t = cur.next
+            cur.next = slow.next
+            slow.next = cur
+            cur = t
+        t = slow.next
+        while t:
+            if head.val != t.val:
+                return False
+            t = t.next
+            head = head.next
+        return True
+```
+
+### Java
+
+```java
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public boolean isPalindrome(ListNode head) {
+        if (head == null || head.next == null) {
+            return true;
+        }
+        ListNode slow = head, fast = head.next;
+        while (fast != null && fast.next != null) {
+            slow = slow.next;
+            fast = fast.next.next;
+        }
+        ListNode cur = slow.next;
+        slow.next = null;
+        while (cur != null) {
+            ListNode t = cur.next;
+            cur.next = slow.next;
+            slow.next = cur;
+            cur = t;
+        }
+        ListNode t = slow.next;
+        while (t != null) {
+            if (head.val != t.val) {
+                return false;
+            }
+            head = head.next;
+            t = t.next;
+        }
+        return true;
+    }
+}
+```
+
+### ...
+```
+
+```

lcci/02.06.Palindrome Linked List/Solution.java

@@ -0,0 +1,37 @@
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public boolean isPalindrome(ListNode head) {
+        if (head == null || head.next == null) {
+            return true;
+        }
+        ListNode slow = head, fast = head.next;
+        while (fast != null && fast.next != null) {
+            slow = slow.next;
+            fast = fast.next.next;
+        }
+        ListNode cur = slow.next;
+        slow.next = null;
+        while (cur != null) {
+            ListNode t = cur.next;
+            cur.next = slow.next;
+            slow.next = cur;
+            cur = t;
+        }
+        ListNode t = slow.next;
+        while (t != null) {
+            if (head.val != t.val) {
+                return false;
+            }
+            head = head.next;
+            t = t.next;
+        }
+        return true;
+    }
+}

lcci/02.06.Palindrome Linked List/Solution.py

@@ -0,0 +1,28 @@
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def isPalindrome(self, head: ListNode) -> bool:
+        if head is None or head.next is None:
+            return True
+        slow, fast = head, head.next
+        while fast and fast.next:
+            fast = fast.next.next
+            slow = slow.next
+        cur = slow.next
+        slow.next = None
+        while cur:
+            t = cur.next
+            cur.next = slow.next
+            slow.next = cur
+            cur = t
+        t = slow.next
+        while t:
+            if head.val != t.val:
+                return False
+            t = t.next
+            head = head.next
+        return True