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