问题
I am running below code to remove duplicates from Linked List. But my code only prints linked List before removing duplicates. Once removeDup method is called, it does not print anything. Below is my code. Please tell me what am I missing.
class Node:
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
def __init__(self):
self.head = None
def insert(self, data):
node = Node(data)
node.next=self.head
self.head = node
def printl(self):
current = self.head
while current:
print current.data
current= current.next
def removeDups(self):
current = self.head
while current is not None:
second = current.next
while second:
if second.data == current.data:
current.next = second.next.next
second = second.next
current = current.next
# l.printl()
l= LinkedList()
l.insert(15)
l.insert(14)
l.insert(16)
l.insert(15)
l.insert(15)
l.insert(14)
l.insert(18)
l.insert(159)
l.insert(12)
l.insert(10)
l.insert(15)
l.insert(14)
l.printl()
print "==============="
l.removeDups()
l.printl()
回答1:
Your logic for removing the duplicated items you find is not right. It causes you to cut out all the items between the first occurrence of a value and a point past its last occurrence. For your example list, that results in a single item, 14 being printed after the deduplication runs (it cuts from just after the first value to the end, though it makes some smaller cuts along the way too).
Here's a fixed version of your removeDups method.
def removeDups(self):
current = second = self.head
while current is not None:
while second.next is not None: # check second.next here rather than second
if second.next.data == current.data: # check second.next.data, not second.data
second.next = second.next.next # cut second.next out of the list
else:
second = second.next # put this line in an else, to avoid skipping items
current = second = current.next
The main change is that second points to the node before the second node we're actually interested in checking. We do all our work on second.next. We need to keep the reference to second so we can easily cut second.next out of the list. Doing it this way requires that we don't advance second if we've cut out a node, so the second = second.next line needs to be in an else clause.
Since current and second always start with the same value after each update to current, I changed the logic to assign both of them in a single statement. It would work fine the original way, I just think this way looks nicer.
回答2:
I think it is confusing to use the "second" variable.
def removeDups(self):
current = self.head
while current: #First loop
while current.next and current.data == current.next.data: #Second loop
current.next = current.next.next #Deletion
current = current.next
You start at the head of the list and for each node in your list until you hit the None at the end (while current) you enter another loop. That loops checks to make sure there is a next node (while current.next) and if that next node has the same data as the current node (current.data == current.next.data). Each time this second loop is true, it means we have a duplicate. The next line (current.next = current.next.next) is what does the actual deletion. It also conveniently updates current.next to the next node in the list that we want to compare so that the second loop can immediately check again to see if we have another duplicate. Once that second loop has found and deleted all the duplicates for that particular node, we will drop down to the next line (current = current.next), update our current node to the next one and start checking that node for duplicates.
回答3:
We can use a list or dictionary to check whether the item inserted is already there or not
class Node:
def __init__(self,data):
self.data=data
self.next=None
class LinkedList:
def __init__(self):
self.head=None
def append(self,data):
new_Node=Node(data)
if self.head is None:
self.head=new_Node
return
last_node=self.head
while last_node.next:
last_node=last_node.next
last_node.next=new_Node
def printing(self):
current_node=self.head
while current_node:
print(current_node.data)
current_node=current_node.next
def remove_dup(self):
curr=self.head
glist=[] #list to store the values
while curr:
if curr.data in glist: #checking the value already exists in list
prev.next=curr.next
else:
glist.append(curr.data)
prev=curr
curr=curr.next
llist=LinkedList()
llist.append(1)
llist.append(6)
llist.append(1)
llist.append(4)
llist.append(2)
llist.append(2)
llist.append(4)
llist.remove_dup()
llist.printing()
回答4:
This is how removeDuplicates function should be written:
class node:
def __init__(self):
self.data = None
self.next = None
class Linked_List:
def __init__(self):
self.head = None
def get_head(self):
return self.head
def insert(self, data):
if self.head == None:
self.head = node()
self.head.data = data
else:
new_node = node()
new_node.data = data
new_node.next = None
temp = self.head
while(temp.next):
temp=temp.next
temp.next = new_node
def printlist(self):
temp = self.head
while temp!=None:
print(temp.data, end=" ")
temp= temp.next
def removeDuplicates(head):
current = head
element_list = []
prev = None
while (current is not None):
if current.data not in element_list:
element_list.append(current.data)
prev = current
current = current.next
else:
prev.next = current.next
current = current.next
if __name__ == '__main__':
values_list = [[5,2,2,4], [2,2,2,2,2]]
t = len(values_list)
for index in range(t):
list1 = Linked_List()
for i in values_list[index]:
list1.insert(i)
print('Input:')
list1.printlist()
print()
removeDuplicates(list1.head)
print('Output')
list1.printlist()
print('')
It removes duplicate nodes from sorted/unsorted singly linked list
回答5:
you can use an additional data structure to hold the unique values(e.g. a list)
class Node:
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
def __init__(self):
self.head = None
def insert_node(self, data):
new_node = Node(data)
if self.head is None:
self.head = new_node
else:
new_node.next = self.head
self.head = new_node
def remove_dups(self):
uniques = []
prev = None
curr = self.head
while curr is not None:
if curr.data in uniques:
prev.next = curr.next
else:
uniques.append(curr.data)
prev = curr
curr = curr.next
def print_list(self):
output = ""
tmp = self.head
while tmp is not None:
output += str(tmp.data) + " "
tmp = tmp.next
print(output)
`
来源：https://stackoverflow.com/questions/42728271/remove-duplicates-from-linked-list-python