876.链表的中间节点

题目

Python

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:def middleNode(self, head: Optional[ListNode]) -> Optional[ListNode]:slow = fast = headwhile fast and fast.next:slow = slow.nextfast = fast.next.nextreturn slow