Easy
https://leetcode.com/problems/palindrome-linked-list/
Given a singly linked list, determine if it is a palindrome.
Example 1:
Input: 1->2
Output: false
Follow up:Could you do it in O(n) time and O(1) space?
方法一：遍历链表，把值依次存在数组中，然后对数组判断是否对称。时间 O(n)，空间 O(n)
1 class Solution(object): 2 def isPalindrome(self, head): 3 """ 4 :type head: ListNode 5 :rtype: bool 6 """ 7 if not head: 8 return True 9 num = [] 10 while head: 11 num.append(head.val) 12 head = head.next 13 i, j = 0, len(num)-1 14 while i<j: 15 if num[i]!=num[j]: 16 return False 17 i += 1 18 j -= 1 19 return True
方法二：先用快慢指针找到链表中点，然后逆转链表后半段，比较前半段和逆转后的后半段，若相同则对称。时间 O(n)，空间 O(1)
1 class Solution(object): 2 def isPalindrome(self, head): 3 """ 4 :type head: ListNode 5 :rtype: bool 6 """ 7 if not head: 8 return True 9 mid = findMid(head) 10 mid.next = reverseLinkedList(mid.next) 11 p,q = head, mid.next 12 while q and p.val==q.val: 13 p = p.next 14 q = q.next 15 return False if q else True 16 17 def findMid(head): 18 slow = head 19 fast = head.next 20 while fast and fast.next: 21 fast = fast.next.next 22 slow = slow.next 23 return slow 24 25 def reverseLinkedList(head): 26 pre = None 27 while head: 28 temp = head.next 29 head.next = pre 30 pre = head 31 head = temp 32 return pre
* 链表中点查找
* 逆转链表
原文：https://www.cnblogs.com/alison-f/p/11386440.html