将多个排序的链接列表合并为一个

Question

我正在尝试编写一个程序来合并两个排序的链表。 但它并没有合并它们。 似乎错误在于表示头部指针。 如何表示头部指针？是我的方式是正确的。 我在使用指针时遇到很多问题，请给我一些建议，尤其是头部指针

合并两个排序链接列表

class Node:
def __init__(self,data):
self.data=data
self.next =None
class linkedlist:
def __init__(self):
self.head=None
def push(self,ndata):
nnode = Node(ndata)
nnode.next = self.head
self.head = nnode
def addNodeToList(self,ndata):
nnode = Node(ndata)
if self.head is None:
self.head = nnode
return
last = self.head
while last.next is not None:
last = last.next
last.next = nnode

def printList(self):
temp = self.head
while temp is not None:
print(temp.data,end = ' ')
temp = temp.next
def merge(first,second):
dummy=linkedlist()
temp = dummy.head
temp1=first.head
temp2=second.head
while temp1 and temp2:
if temp1.data<temp2.data:
temp=temp1
temp.next=None
temp1=temp1.next
print('n1..')
elif temp1.data>=temp2.data:
temp=temp2
temp.next=None
temp2=temp2.next
print('n2..')
if temp1 is not None:
temp.next= temp1
print('n3..')
else:
temp.next=temp2
print('n4..')
print('Done')
return  dummy          
if __name__=='__main__':

list1=linkedlist()
list1.addNodeToList(10)
list1.addNodeToList(20)
list1.addNodeToList(30)
list1.addNodeToList(40)
list1.addNodeToList(50)

list2=linkedlist()
# Create linked list 2 : 5->15->18->35->60 
list2.addNodeToList(5)
list2.addNodeToList(15)
list2.addNodeToList(18)
list2.addNodeToList(35)
list2.addNodeToList(60)
list1.printList()

print()
list2.printList()

a=merge(list1,list2)
a.printList()
Expected output is a single merged linkedlist

Answer 1

你已经有了addNodeToList()函数，为什么不使用它呢？

def merge(first,second):
dummy=linkedlist()
temp1=first.head
temp2=second.head
while temp1 and temp2:
if temp1.data<temp2.data:
dummy.addNodeToList(temp1.data)
temp1=temp1.next
else:
dummy.addNodeToList(temp2.data)
temp2=temp2.next
while temp1:
dummy.addNodeToList(temp1.data)
temp1=temp1.next
while temp2:
dummy.addNodeToList(temp2.data)
temp2=temp2.next
return  dummy

Answer 2

您永远不会在代码中添加dummy- 您执行temp = dummy.head，然后重新分配temp，并且dummy列表永远不会更新。这是一种方法：

class Node:
def __init__(self, data):
self.data = data
self.next = None
class LinkedList: # classes in python must be in CamelCase
def __init__(self):
self.head = None
def push(self, ndata):
nnode = Node(ndata)
nnode.next = self.head
self.head = nnode
def addNodeToList(self, ndata):
nnode = Node(ndata)
if self.head is None:
self.head = nnode
return
last = self.head
while last.next is not None:
last = last.next
last.next = nnode
def printList(self):
temp = self.head
while temp is not None:
print(temp.data, end = ' ')
temp = temp.next
def walk_list(self):
temp = self.head
values = []
while temp is not None:
values.append(temp.data)
temp = temp.next
return values
def merge(first, second):
values1 = first.walk_list()
values1.extend(second.walk_list())
dummy = LinkedList()
for v in sorted(values1):
dummy.push(v)
return dummy
if __name__ == '__main__':
list1 = LinkedList()
list1.addNodeToList(10)
list1.addNodeToList(20)
list1.addNodeToList(30)
list1.addNodeToList(40)
list1.addNodeToList(50)
list2 = LinkedList()
# Create linked list 2 : 5->15->18->35->60
list2.addNodeToList(5)
list2.addNodeToList(15)
list2.addNodeToList(18)
list2.addNodeToList(35)
list2.addNodeToList(60)
list1.printList()
print()
list2.printList()
a = merge(list1, list2)
a.printList()