Question
Given a singly linked list, determine if it is a palindrome.
Follow up:
Could you do it in O(n) time and O(1) space?
Solution
这一题思路并不难。要满足follow-up的要求，我们用到了快慢指针。
1. 用快慢指针得到前后两半list，这里有个技巧是quick先判断有无next，slow再走。这样就保证slow永远指向后半部分的前一个结点
2. Reverse 后半部分的list。三指针方法
3. 比较前半链表和反转后的后半链表
思路虽不难，但是要做到bug-free还是有难度。关键在于对以上每个子问题都熟悉。
1 # Definition for singly-linked list. 2 # class ListNode(object): 3 # def __init__(self, x): 4 # self.val = x 5 # self.next = None 6 7 class Solution(object): 8 def isPalindrome(self, head): 9 """ 10 :type head: ListNode 11 :rtype: bool 12 """ 13 if head is None or head.next is None: 14 return True 15 slow = head 16 quick = head 17 while quick.next is not None: 18 quick = quick.next 19 if quick.next is not None: 20 quick = quick.next 21 slow = slow.next 22 # Reverse sub list between slow.next and quick 23 cur = slow.next 24 slow.next = None 25 then = cur.next 26 cur.next = None 27 while then is not None: 28 tmp = then.next 29 then.next = cur 30 cur = then 31 then = tmp 32 second_head = cur 33 # Compare first sub list and second sub list 34 cur1 = head 35 cur2 = second_head 36 while cur1 is not None and cur2 is not None: 37 if cur1.val != cur2.val: 38 return False 39 cur1 = cur1.next 40 cur2 = cur2.next 41 return True