我正在尝试编写代码以从排序的链接列表“ head”中删除重复项。如果列表以重复结尾,则下面的我的代码始终返回最后一个重复。例如[1,2,2,3,3]
将返回[1,2,3,3]
。我不知道为什么。有人有主意吗?
class Solution(object):
def deleteDuplicates(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if not head:
return head
l1=newhead=ListNode(head.val)
head=head.next
while head:
if head.val!=l1.val:
l1.next=head
l1=l1.next
head=head.next
return newhead
答案 0 :(得分:1)
您应该跟踪每个新值的前导节点,并继续获取下一个节点,直到获得具有不同值的节点为止,此时,您将该节点分配为前导节点的下一个节点:
class Solution(object):
def deleteDuplicates(self, head):
node = head
while node:
lead = node
while node.next and node.next.val == lead.val:
node = node.next
node = lead.next = node.next
return head
答案 1 :(得分:0)
问题解决方案
程序/源代码
这是Python程序的源代码,用于从链接列表中删除重复项。
class Node:
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
def __init__(self):
self.head = None
self.last_node = None
def append(self, data):
if self.last_node is None:
self.head = Node(data)
self.last_node = self.head
else:
self.last_node.next = Node(data)
self.last_node = self.last_node.next
def get_prev_node(self, ref_node):
current = self.head
while (current and current.next != ref_node):
current = current.next
return current
def remove(self, node):
prev_node = self.get_prev_node(node)
if prev_node is None:
self.head = self.head.next
else:
prev_node.next = node.next
def display(self):
current = self.head
while current:
print(current.data, end = ' ')
current = current.next
def remove_duplicates(llist):
current1 = llist.head
while current1:
data = current1.data
current2 = current1.next
while current2:
if current2.data == data:
llist.remove(current2)
current2 = current2.next
current1 = current1.next
a_llist = LinkedList()
data_list = input('Please enter the elements in the linked list: ').split()
for data in data_list:
a_llist.append(int(data))
remove_duplicates(a_llist)
print('The list with duplicates removed: ')
a_llist.display()
程序说明