876. Middle of the Linked List

1. 原始题目

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.

Example 1:

Input: [1,2,3,4,5]
Output: Node 3 from this list (Serialization: [3,4,5])
The returned node has value 3.  (The judge's serialization of this node is [3,4,5]).
Note that we returned a ListNode object ans, such that:
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.

Example 2:

Input: [1,2,3,4,5,6]
Output: Node 4 from this list (Serialization: [4,5,6])
Since the list has two middle nodes with values 3 and 4, we return the second one.

2. 题目理解

就是找到中间结点，然后依次打印链表结点。方法就是利用两个快慢指针。从头开始，如果快指针每次走两步，慢指针每次走一步，那么当快指针走到末尾时，慢指针正好在中间位置。

3. 解题

 class Solution:
     def middleNode(self, head: ListNode) -> ListNode:
         pre = head   # 快指针
         cur = head   # 慢指针
         while(pre and pre.next):
             cur = cur.next
             pre = pre.next.next

         return cur

检验：

 # 新建链表1
 listnode1 = ListNode_handle(None)
 #s1 = [1,2,3,666,8,3,2,9,4,5,6,8,999,666]
 s1 = [1,3,5,666,4,5]
 for i in s1:
     listnode1.add(i)
 listnode1.print_node(listnode1.head)

 s = Solution()
 head = s.middleNode(listnode1.head)
 listnode1.print_node(head)

1 3 5 666 4 5
666 4 5