回文链表：判定一个链表内容是否是回文结构，快慢指针法

class LinkNode:
    def __init__(self, data):
        self.val = data
        self.next = None


class Solution(object):
    def isPalindrome(self, head):
        if not head or not head.next:  # 链表为空或是只有一个元素
            return True
        slow = head
        fast = head.next    # 创建快慢指针，寻找链表中点
        while fast and fast.next:  # 快指针到达结尾为止
            fast = fast.next.next
            if fast:   # 当前快指针fast的位置为奇数位置，
                slow = slow.next  # 慢指针保持指向左侧最后一个元素
        # second要指向右侧第一个元素（偶）或是中间的右侧第一个元素（奇）
        if fast:  # 链表有奇数个元素
            second = slow.next
        else:
            second = slow.next.next 
            
        slow.next = None
        first = head
        pre = None
        while first:  # 翻转左侧链表
            tmp = first.next
            first.next = pre
            pre = first
            first = tmp

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


p1 = LinkNode(1)
p2 = LinkNode(2)
p3 = LinkNode(3)
p4 = LinkNode(2)
p5 = LinkNode(1)
p1.next = p2
p2.next = p3
p3.next = p4
p4.next = p5
print(Solution().isPalindrome(p1))