Palindrome Linked List 解答

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。

 # Definition for singly-linked list.
 # class ListNode(object):
 #     def __init__(self, x):
 #         self.val = x
 #         self.next = None

 class Solution(object):
     def isPalindrome(self, head):
         """
         :type head: ListNode
         :rtype: bool
         """
         if head is None or head.next is None:
             return True
         slow = head
         quick = head
         while quick.next is not None:
             quick = quick.next
             if quick.next is not None:
                 quick = quick.next
                 slow = slow.next
         # Reverse sub list between slow.next and quick
         cur = slow.next
         slow.next = None
         then = cur.next
         cur.next = None
         while then is not None:
             tmp = then.next
             then.next = cur
             cur = then
             then = tmp
         second_head = cur
         # Compare first sub list and second sub list
         cur1 = head
         cur2 = second_head
         while cur1 is not None and cur2 is not None:
             if cur1.val != cur2.val:
                 return False
             cur1 = cur1.next
             cur2 = cur2.next
         return True