@@ -218,59 +218,41 @@ class Solution {
218218``` python
219219# 数组模拟
220220class Solution :
221- def isPalindrome (self head : ListNode) -> bool :
222- length = 0
223- tmp = head
224- while tmp: # 求链表长度
225- length += 1
226- tmp = tmp.next
227- 228- result = [0 ] * length
229- tmp = head
230- index = 0
231- while tmp: # 链表元素加入数组
232- result[index] = tmp.val
233- index += 1
234- tmp = tmp.next
235- 236- i, j = 0 , length - 1
237- while i < j: # 判断回文
238- if result[i] != result[j]:
221+ def isPalindrome (self head : Optional[ListNode]) -> bool :
222+ list = []
223+ while head:
224+ list .append(head.val)
225+ head= head.next
226+ l,r= 0 , len (list )- 1
227+ while l<= r:
228+ if list[l]!= list[r]:
239229 return False
240- i += 1
241- j -= 1
242- return True
243-
230+ l += 1
231+ r -= 1
232+ return True
233+ 244234# 反转后半部分链表
245235class Solution :
246- def isPalindrome (self head : ListNode) -> bool :
247- if head == None or head.next == None :
248- return True
249- slow, fast = head, head
236+ def isPalindrome (self head : Optional[ ListNode] ) -> bool :
237+ fast = slow = head
238+ 239+ # find mid point which including (first) mid point into the first half linked list
250240 while fast and fast.next:
251- pre = slow
252- slow = slow.next
253241 fast = fast.next.next
254- 255- pre.next = None # 分割链表
256- cur1 = head # 前半部分
257- 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
258- while cur1:
259- if cur1.val != cur2.val:
242+ slow = slow.next
243+ node = None
244+ 245+ # reverse second half linked list
246+ while slow:
247+ slow.next, slow, node = node, slow.next, slow
248+ 249+ # compare reversed and original half; must maintain reversed linked list is shorter than 1st half
250+ while node:
251+ if node.val != head.val:
260252 return False
261- cur1 = cur1 .next
262- cur2 = cur2 .next
253+ node = node .next
254+ head = head .next
263255 return True
264- 265- def reverseList (self head : ListNode) -> ListNode:
266- cur = head
267- pre = None
268- while (cur!= None ):
269- temp = cur.next # 保存一下cur的下一个节点
270- cur.next = pre # 反转
271- pre = cur
272- cur = temp
273- return pre
274256```
275257
276258## Go
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