234. 回文链表 - 力扣（LeetCode）

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def isPalindrome(self, head: Optional[ListNode]) -> bool:
        if not head or not head.next:
            return True

        # 使用快慢指针找到链表中点
        slow, fast = head, head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next

        # 反转后半部分链表
        prev = None
        while slow:
            temp = slow.next
            slow.next = prev
            prev = slow
            slow = temp

        # 比较前半部分和后半部分
        left, right = head, prev
        while right:  # 只需比较后半部分长度
            if left.val != right.val:
                return False
            left = left.next
            right = right.next

        return True