第一次写的,如果fast.next是none,那么fast.next.next会报错
def middleNode(self, head): # write your code here i = 0 if head.val != None: slow = head fast = head while fast.next.val != None: slow = slow.next fast = fast.next while fast.next.val != None: fast = fast.next return slow.val else: return None第二次写的。需要先判断fast.next.next不为None，再给slow往下走一步。判断链表为空，应该判断head是否为none ,不应该判断head.next。
""" Definition of ListNode class ListNode(object): def __init__(self, val, next=None): self.val = val self.next = next """ class Solution: """ @param head: the head of linked list. @return: a middle node of the linked list """ def middleNode(self, head): # write your code here if not head.next: return None else: slow = head fast = head while fast.next: slow = slow.next fast = fast.next while fast.next: fast = fast.next break return(slow)第三次
""" Definition of ListNode class ListNode(object): def __init__(self, val, next=None): self.val = val self.next = next """ class Solution: """ @param head: the head of linked list. @return: a middle node of the linked list """ def middleNode(self, head): # write your code here if not head: return None else: slow = head fast = head while fast.next: fast = fast.next if fast.next: slow = slow.next fast = fast.next return(slow)通过 查看别人的写法，先判断fast.next，用and连接判断fast.next.next。一条就搞定。
class Solution: """ @param head: the head of linked list. @return: a middle node of the linked list """ def middleNode(self, head): # write your code here if not head: return None else: slow,fast = head,head while fast.next and fast.next.next: slow = slow.next fast= fast.next.next return slow