234. Palindrome Linked List

先用stack把它装起来，计算count， 然后出栈， 后半部分和前半部分比较

# 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 not head or not head.next:
            return True
        stack = []
        count = 0
        p = head
        while p:
            stack.append(p)
            p = p.next
            count += 1
        count = count / 2
        while count:
            if head.val != stack.pop().val:
                return False
            head = head.next
            count -= 1
        return True
    

一边找到中点一边把前面反转了，然后比较前半部分和后半部分

# 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 not head or not head.next:
            return True
        slow = fast = head
        pre = None
        while fast and fast.next:
            fast = fast.next.next
            curr = slow.next
            slow.next = pre
            pre = slow
            slow = curr
        if fast:
            slow = slow.next
        while pre and slow:
            if pre.val != slow.val:
                return False
            pre = pre.next
            slow = slow.next
        return True