diff --git a/problems/0234.回文链表.md b/problems/0234.回文链表.md
index 6a24b1d0..b3ad899c 100644
--- a/problems/0234.回文链表.md
+++ b/problems/0234.回文链表.md
@@ -148,7 +148,62 @@ public:
 
 ## Python
 
-```python
+```python3
+#数组模拟
+class Solution:
+    def isPalindrome(self, head: ListNode) -> bool:
+        length = 0
+        tmp = head
+        while tmp: #求链表长度
+            length += 1
+            tmp = tmp.next
+        
+        result = [0] * length
+        tmp = head
+        index = 0
+        while tmp: #链表元素加入数组
+            result[index] = tmp.val
+            index += 1
+            tmp = tmp.next
+        
+        i, j = 0, length - 1
+        while i < j: # 判断回文
+            if result[i] != result[j]:
+                return False
+            i += 1
+            j -= 1
+        return True
+        
+#反转后半部分链表
+class Solution:
+    def isPalindrome(self, head: ListNode) -> bool:
+        if head == None or head.next == None:
+            return True
+        slow, fast = head, head
+        while fast and fast.next:
+            pre = slow
+            slow = slow.next
+            fast = fast.next.next
+        
+        pre.next = None # 分割链表
+        cur1 = head # 前半部分
+        cur2 = self.reverseList(slow) # 反转后半部分，总链表长度如果是奇数，cur2比cur1多一个节点
+        while cur1:
+            if cur1.val != cur2.val:
+                return False
+            cur1 = cur1.next
+            cur2 = cur2.next
+        return True
+
+    def reverseList(self, head: ListNode) -> ListNode:
+        cur = head   
+        pre = None
+        while(cur!=None):
+            temp = cur.next # 保存一下cur的下一个节点
+            cur.next = pre # 反转
+            pre = cur
+            cur = temp
+        return pre
 ```
 
 ## Go