leetcode 回文鍊錶 python3

class solution :def ispalindrome (self, head) :""" 判斷乙個鍊錶是否是回文的，很自然的想法就是兩個指標，乙個指標從前往後走，乙個指標從後往前走，判斷元素值是否相同，這裡要分幾個步驟來進行求解： 1、找到鍊錶長度的一半，用追趕法，乙個指標一次走兩步，乙個指標一次走一步 2、將後一半陣列轉置 3、判斷鍊錶是否是回文鍊錶 :type head: listnode :rtype: bool """slow = fast = head while fast and fast. next : slow = slow. next fast = fast. next .next node = none while slow: nxt = slow. next slow. next = node node = slow slow = nxt while node and head: if node.val != head.val: return false node = node. next head = head. next return

true

# definition for singly-linked list.
# class listnode:
# def __init__(self, x):
# self.val = x
# self.next = none
class
solution
:def
ispalindrome
(self, head)
:"""
:type head: listnode
:rtype: bool
"""nums =
while head:
head = head.
next
return nums == nums[::
-1]

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