###### tags: `LeetCode`,`Python`,`Easy`
# 234. Palindrome Linked List
### **題目連結：** [**Palindrome Linked List**](https://leetcode.com/problems/palindrome-linked-list/description/?envType=daily-question&envId=2024-03-22)
### **解題方向**
* 用快慢指針找到中間點
* 反轉後半list
* 對比前半與反轉後的後半
### **完整程式碼**

# 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
        
        slow = head
        fast = head
        
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next
        
        reverse_head = self.reverse(slow)
        temp1 = head
        temp2 = reverse_head
        
        while temp2:
            if temp1.val != temp2.val:
                return False
            temp1 = temp1.next
            temp2 = temp2.next
        
        return True
    def reverse(self, head):
        if not head:
            return None
        prev = None
        curr = head
        while curr:
            next_node = curr.next
            curr.next = prev
            prev = curr
            curr = next_node
        return prev