[LeetCode] 234. Palindrome Linked List

Given a singly linked list, determine if it is a palindrome.

Example 1:

Input: 1->2
Output: false

Example 2:

Input: 1->2->2->1
Output: true

Follow up:
Could you do it in O(n) time and O(1) space?

这个题目就是先找中点，然后将中点后的list都reverse，进而分别将每个点比较，一旦不等，那么返回False，最后返回True。

T: O(n), S: O(1)

code

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = Noneclass Solution:def isPalindrome(self, head: ListNode) -> bool:if not head or not head.next: return Truedummy = ListNode(0)dummy.next = headslow, fast = dummy, dummywhile fast:if not fast.next:breakif not fast.next.next:fast.next = fast.next.nextslow = slow.nextfast = fast.next.nextpre, cur, count = slow, slow.next, 0while cur:count += 1temp = cur.nextcur.next = prepre = curcur = temp# compare two listsfor _ in range(count):if pre.val != head.val:return Falsepre = pre.nexthead = head.nextreturn True

转载于:https://www.cnblogs.com/Johnsonxiong/p/10817515.html

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