152-234. 回文链表

如何判断一个链表是回文链表。(第一个我写的,下面我根据其他人改编的)
# Definition for singly-linked list.
class ListNode(object):
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next


class Solution(object):
    def isPalindrome1(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        if not head:
            return True

        temp_list = []

        temp = head
        temp_list.append(temp.val)

        while temp.next:
            temp = temp.next
            temp_list.append(temp.val)

        return temp_list == temp_list[::-1]

    def isPalindrome2(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        if head is None:
            return True

        # 这一部分代码很经典,通过fast和slow之间的差值计算出了什么地方时中点
        fast = head
        slow = head
        while fast.next and fast.next.next:
            fast = fast.next.next
            slow = slow.next

        # 这一个部分代码是链表反转
        cur = slow.next
        slow.next = None
        pre = None
        while cur:
            temp = cur.next
            cur.next = pre
            pre = cur
            cur = temp

        # 这一部分才是最后的比较部分
        while pre:
            if pre.val != head.val:
                return False
            pre = pre.next
            head = head.next
        return True

    def isPalindrome(self, head):
        """这是个神奇的东西,我没明白
        :type head: ListNode
        :rtype: bool
        """
        s1 = 0
        s2 = 0
        t = 1

        while head:
            s1 = s1 * 10 + head.val
            s2 = s2 + t * head.val
            t = t * 10
            head = head.next

        return s1 == s2


if __name__ == '__main__':
    s1 = Solution()

    root = ListNode(1)

    n1 = ListNode(2)
    n2 = ListNode(2)
    n3 = ListNode(1)

    n2.next = n3
    n1.next = n2
    root.next = n1

    print(s1.isPalindrome(root))