Python 为什么我的程序最终没有打印?

Python 为什么我的程序最终没有打印?,python,python-3.x,list,linked-list,Python,Python 3.x,List,Linked List,我自己在学习算法和数据结构,并想编写一个程序,以以下方式合并两个链表:给定两个整数a和b,它将从第一个链表中删除从a到b的节点,并插入第二个链表 我从中获取了一个链表的实现,并编写了一个函数 def merge(list1: LinkedList, a: int, b: int, list2: LinkedList) -> LinkedList: current = list1.head previous = None insertion = list2.head

我自己在学习算法和数据结构,并想编写一个程序,以以下方式合并两个链表:给定两个整数a和b,它将从第一个链表中删除从a到b的节点,并插入第二个链表

我从中获取了一个链表的实现,并编写了一个函数

def merge(list1: LinkedList, a: int, b: int, list2: LinkedList) -> LinkedList:

    current = list1.head
    previous = None
    insertion = list2.head
    counter = 0

    while insertion.next_node:
        insertion = insertion.next_node

    while current:
        if counter == a:
            previous.next_node = list2.head
        elif counter == b:
            insertion.next_node = current.next_node
        previous = current
        counter += 1
        current = current.next_node

    return list1.printList()
这是生成程序的完整代码:

class Node(object):
 
    def __init__(self, data=None, next_node=None):
        self.data = data
        self.next_node = next_node
 
 
class LinkedList(object):
    def __init__(self, head=None):
        self.head = head
 
    def size(self):
     current = self.head
     count = 0
     while current:
        count += 1
        current = current.next_node
     return count
       
    def printList(self): 
        temp = self.head 
        while (temp): 
            print (temp.data, " -> ", end = '') 
            temp = temp.next_node
        print("")
 
    def insert_at_head(self, data):
      new_node = Node(data)
      new_node.next_node = self.head
      self.head = new_node
 
    def get_next_node (self,node):
      return node.next_node.data

list1 = LinkedList(Node(0))
s = list1.head
for i in range(1,6):
    s.next_node = Node(i)
    s = s.next_node
list1.printList()

list2 = LinkedList(Node(99))
w = list2.head
for j in range(98,96,-1):
    w.next_node = Node(j)
    w = w.next_node
list2.printList()

def merge(list1: LinkedList, a: int, b: int, list2: LinkedList) -> LinkedList:

    current = list1.head
    previous = None
    insertion = list2.head
    counter = 0

    while insertion.next_node:
        insertion = insertion.next_node

    while current:
        if counter == a:
            previous.next_node = list2.head
        elif counter == b:
            insertion.next_node = current.next_node
        previous = current
        counter += 1
        current = current.next_node

    return list1.printList()

print(merge(list1, 1, 3, list2))
它的工作原理与预期相同,只是在输出的末尾打印了一个额外的None。为什么不打印?我遗漏了什么?

您的最后一行打印您的
合并呼叫的返回。
merge
的返回是
list.printList()
。但是,
list.printList()
没有定义返回,因此它默认为
None
。请注意,
list.printList()
调用时仍会打印;它只是没有定义的返回

听起来您只是想调用
merge
,而不需要打印。所以最后一行可能是:

merge(list1, 1, 3, list2)