LeetCode#234 Palindrome Linked List

Problem Definition:

　　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?

思路：为了达到O(1)的空间复杂度，可以把单链表的后半部分反转。

代码：

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

class Solution:
    # @param {ListNode} head
    # @return {boolean}
    def isPalindrome(self, head):
        if head==None:
            return True
        tail=head
        tailNext=tail.next
        length=1
        while tailNext!=None:
            tail=tailNext
            tailNext=tailNext.next
            length+=1
        if length==1:
            return True
        half=length/2
       
        newHead=head
        n=0
        while n<half:
            newHead=newHead.next
            n+=1
        if length%2==1: #odd number
            newHead=newHead.next
        
        #reverse
        while newHead!=tail:
            newHeadNext=newHead.next
            newHead.next=tail.next
            tail.next=newHead
            newHead=newHeadNext
        while newHead!=None:
            if head.val!=newHead.val:
                return False
            head=head.next
            newHead=newHead.next
        return True