链表做题总结（六）[234]

234. 回文链表

思路：快指针走到末尾，慢指针刚好到中间。其中慢指针将前半部分反转。然后比较
注意：如何在利用快慢指针找中点的同时反转链表，注意两者互不影响就可以实现
代码：

class Solution:
    def isPalindrome(self, head: ListNode) -> bool:
        # 边界判断
        if not head or not head.next:
            return True
        
        # 找中点的同时对左半链表进行反转
        pre=head
        prepre=None
        fast, slow = head, head

        while fast and fast.next:
            pre = slow
            slow = slow.next
            fast = fast.next.next
            pre.next = prepre
            prepre = pre

        # 奇数个节点时
        if fast:
            slow = slow.next
        

        # 左右两链表进行比较

        while pre and slow:
            if pre.val != slow.val:
                return False
            pre=pre.next
            slow = slow.next
        return True