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python - 如何合并两个链表

转载 作者:塔克拉玛干 更新时间:2023-11-03 06:38:45 26 4
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我正在尝试将两个链接列表合并在一起,这个问题是我们破解编码面试的问题。问题2.4

在分区中,我创建了两个链表llink_A和llink_B。从主链接中获取值并将它们分开在 llink_A 和 llink_B 中。我想如果我遍历 llink_A 到最后然后指向 llink_B 应该工作。但是当我运行该程序时它不起作用。

当我运行程序时,我得到[3, 5, 8, 5, 10, 2, 1]链接列表长度:7[3, 5, 5, 2, 1] --> llistA[8, 10]--> llistB

class Node:
#Singly link list
def __init__(self,data = None):
self.data = data
self.next = None

class linklist:
#linkList class
def __init__(self):
self.head = None
self.size = 0

def push(self,data):
node = Node(data)# create a new node

if self.head == None: #check to see if the head node is
#empty
self.head = node # If its empty add it to the new node
self.size = 1
return
#if the head of the linklist is filled

current = self.head
while current.next is not None:#Check the current postion is
#empty
#Move to the next line if nothing is there
current = current.next


current.next = node #point self.head to a new node
self.size+=1

def lenght(self):
#note the count doesn't start at zero
cur = self.head
counter = 0
while cur is not None:
counter+=1
cur = cur.next
print('Linklist len: '+str(counter))
return counter

def printList(self):
curr = self.head
elem = []

while(curr != None):
elem.append(curr.data)
curr = curr.next
print(elem)
#1->2->3
def remove_node(self,data):
#1->2->3
curr = self.head
if curr is not None and curr.data == data:
self.head = curr.next
curr = None
#look for the node that has the data we are looking for
while curr is not None:
if curr.data == data:
break
prev = curr
curr = curr.next

#if data isn't found just reutrn
if(curr == None):
return

#allow us to unlink the nodes
prev.next = curr.next
curr = None

def partition(self,num):
#llist_A and llist_b to be continers
list_A = linklist()
list_B = linklist()
curr = self.head
# Idea: Make two seperate llink and smash them together
while curr is not None:
if curr.data <= num:
list_A.push(curr.data)
elif curr.data > num:
list_B.push(curr.data)
curr = curr.next
list_A.printList()
list_B.printList()

#Go through all of A the point the end of A to the begining
#of B
self.head = list_A.head
head_A = list_A.head
while head_A:
if head_A.next is not None:
head_A = head_A.next

head_A.next = list_B.head
head_A = head_A.next

















llist = linklist()
llist.push(3)
llist.push(5)
llist.push(8)
llist.push(5)
llist.push(10)
llist.push(2)
llist.push(1)
llist.printList()
llist.lenght()
llist.partition(5)
llist.printList()

最佳答案

你忘了在这个循环中将 curr 更新为 curr.next:

    while curr is not None:
if curr.data <= num:
llist_A.push(curr.data)
else:
llist_B.push(curr.data)

导致无限循环的原因,只需将其更新为:

    while curr is not None:
if curr.data <= num:
llist_A.push(curr.data)
else:
llist_B.push(curr.data)
curr = curr.next

关于python - 如何合并两个链表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54219758/

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