这道题如果不要求space O(1) 代码写起来就相当简单。 但是要求 O(1),唯一的方法就是reverse 前半部分linked List 在做比较， 稍微注意长度为奇偶时一点小小的区别。写起来还是很费劲的

class Solution:    # @param {ListNode} head    # @return {boolean}    def isPalindrome(self, head):        dummy = ListNode(0)        dummy.next = head        p1, p2, length = dummy, dummy,0        while p2.next != None:            p1 = p1.next            p2 = p2.next            length += 1            if p2.next == None:                break            p2 = p2.next            length += 1        p2 = p1.next        p1 = self.reverse(head, p1)        if length % 2 == 1:            p1 = p1.next        while p2 != None:            if p1.val != p2.val:                return False            p1, p2 = p1.next, p2.next        return True    def reverse(self, head, p1):        if head == p1:            return head        if head == None:            return        net = head.next        head.next = None        rest = self.reverse(net,p1)        net.next = head        return rest