Update 0234.回文链表.md

For both solution of python3, there are shorter and more efficient ways to write it. 
For the #数组模拟, it can be solved more easily by convert the linked list to  a list
#反转后半部分链表, the original version define to function, isPalindrome, and reverseList. That's too complicated... No need.

problems/0234.回文链表.md

@@ -218,59 +218,41 @@ class Solution {
 ```python
 #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]:
+    def isPalindrome(self, head: Optional[ListNode]) -> bool:
+        list=[]
+        while head: 
+            list.append(head.val)
+            head=head.next
+        l,r=0, len(list)-1
+        while l<=r: 
+            if list[l]!=list[r]:
                 return False
-            i += 1
-            j -= 1
+            l+=1
+            r-=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
+    def isPalindrome(self, head: Optional[ListNode]) -> bool:
+        fast = slow = head
+
+        # find mid point which including (first) mid point into the first half linked list
         while fast and fast.next:
-            pre = slow
-            slow = slow.next
             fast = fast.next.next
+            slow = slow.next
+        node = None
 
-        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:
+        # reverse second half linked list
+        while slow:
+            slow.next, slow, node = node, slow.next, slow
+
+        # compare reversed and original half; must maintain reversed linked list is shorter than 1st half
+        while node:
+            if node.val != head.val:
                 return False
-            cur1 = cur1.next
-            cur2 = cur2.next
+            node = node.next
+            head = head.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