【leetcode】剑指 Offer II 027. 回文链表（python）

# 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: ListNode) -> bool:
        nums = []
        p = head
        while p:
            nums.append(p.val)
            p = p.next
        n = len(nums)
        for i in range(n//2):
            if nums[i] != nums[n - i - 1]:
                return False 
        return True

# 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: ListNode) -> bool:
        # 空间复杂度为O(1)解法：
        # 1.用快慢指针找到中间位置，
        # 2.将后半部分翻转链表(若为奇数个结点，则将中间结点算为前半部分)
        # 3.双指针p1,p2，遍历
        if not head or not head.next:
            return True

        # 1.用快慢指针找到中间位置
        slow, fast = head, head
        while fast.next and fast.next.next:
            slow = slow.next
            fast = fast.next.next

        # 2.迭代翻转后半部分链表,若为奇数个结点，则将中间结点算为前半部分
        pre = slow.next
        cur = pre.next
        pre.next = None
        while cur:
            tmp = cur.next
            cur.next = pre
            pre = cur
            cur = tmp

        p1 = head  # 3.双指针p1,p2，遍历
        p2 = pre
        while p1 and p2:
            if p1.val != p2.val:
                return False
            p1 = p1.next
            p2 = p2.next
        return True