palindrome_linked_list
palindrome_linked_list.py 源码
# 回文链表
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
class Solution:
# 使用一个数组存储
def isPalindrome_1(self, head: ListNode) -> bool:
data = []
while head:
data.append(head.val)
head = head.next
return data == data[::-1]
# 原地遍历，快慢指针
def isPalindrome_2(self, head: ListNode) -> bool:
# 找到中间节点
slow, fast = head, head
while fast and fast.next:
slow, fast = slow.next, fast.next.next
mid = slow
if fast:
mid = slow.next
# 反转mid->tail的节点
prev, back = None, mid
while back:
back.next, prev, back = prev, back, back.next
# 依次遍历并比较
while prev and head != slow:
if prev.val != head.val:
return False
prev, head = prev.next, head.next
return True
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