Given a non-empty, singly linked list with head node head, return a middle node of linked list.
If there are two middle nodes, return the second middle node.
class Solution:
def middleNode(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
return slow
1 返回的是list
2 重点在fast和slow的指针，当fast跑完，slow整好是我们想要的
