i'm trying to create a simple single linked list in python.
(i know there is no need to implement list in python, but that's not the point)
here is my code:
class Node:
def __init__(self,data):
self.data = data
self.next= None
class List:
def __init__(self):
self.firstNode = Node(None)
def inserthead(self,newnode):
if not self.firstNode.next:
newnode.next = None
self.firstNode.next = newnode
else:
newnode.next = self.firstNode.next
self.firstNode.next= newnode
def __show(self,start):
if start.next:
print start.data
self.__show(start.next)
def printlist(self):
self.__show(self.firstNode)
def __reverte_recursive(self,node):
temp = None
if not node.next: return node
else:
temp = self.__reverte_recursive(node.next)
node.next.next= node
node.next = None
return temp
def reverte_list1(self):
self.firstNode=self.__reverte_recursive(self.firstNode)
def __reverte_iterative(self,node):
temp = None
previous = None
while node and node.next:
temp = node.next
node.next= previous
previous = node
node = temp
return previous
def reverte_iterative(self):
self.firstNode=self.__reverte_iterative(self.firstNode)
nodeA = Node("A")
nodeB = Node("B")
nodeC = Node("C")
nodeD = Node("D")
nodeE = Node("E")
list1= List()
list1.inserthead(nodeA)
list1.inserthead(nodeB)
class Node:
def __init__(self,data):
self.data = data
self.next= None
class List:
def __init__(self):
self.firstNode = Node(None)
def inserthead(self,newnode):
if not self.firstNode.next:
newnode.next = None
self.firstNode.next = newnode
else:
newnode.next = self.firstNode.next
self.firstNode.next= newnode
def __show(self,start):
if start.next:
print start.data
self.__show(start.next)
def printlist(self):
self.__show(self.firstNode)
def __reverte_recursive(self,node):
temp = None
if not node.next: return node
else:
temp = self.__reverte_recursive(node.next)
node.next.next= node
node.next = None
return temp
def reverte_list1(self):
self.firstNode=self.__reverte_recursive(self.firstNode)
def __reverte_iterative(self,node):
temp = None
previous = None
while node and node.next:
temp = node.next
node.next= previous
previous = node
node = temp
return previous
def reverte_iterative(self):
self.firstNode=self.__reverte_iterative(self.firstNode)
nodeA = Node("A")
nodeB = Node("B")
nodeC = Node("C")
nodeD = Node("D")
nodeE = Node("E")
list1= List()
list1.inserthead(nodeA)
list1.inserthead(nodeB)
list1.inserthead(nodeC)
list1.inserthead(nodeD)
list1.inserthead(nodeE)
print "list"
list1.printlist()
print "list reverse"
list1.reverte_list1()
list1.printlist()
list1.reverte_iterative()
print "list reverse reverse"
list1.printlist()
and there are the result:
None
E
D
C
B
list reverse
A
B
C
D
E
list reverse reverse
E
D
C
B
for some reason i can't print all the list and in the first case doesn't print the "A" Node
but print the first node(but i checkd and the B node points to A)
the first reverse is OK
but the third again don't print the A node even if it's pointed by the B Node.
probably the problem of the printing is in the __show function.
but i suppose i a conceptual bug.
thanks
def __show(self,start):
if start.next:
print start.data
self.__show(start.next)
that only prints the current node if it has a next node, thats why the last node is never printed.
should be:
def __show(self,start):
if start:
print start.data
self.__show(start.next)
you make similar mistakes of checking/assigning a node instead of it's next and vice versa throughout the code (in inserthead() for example - which is causing the None being printed)
Related
why can't print reversed this double linked list by python?
always print 6 or None
please can anyone help me fast to pass this task
///////////////////////////////////////////////////////////////////////////
class Node:
def __init__(self, data=None, next=None, prev=None):
self.data = data
self.next = next
self.previous = prev
sample methods==>
def set_data(self, newData):
self.data = newData
def get_data(self):
return self.data
def set_next(self, newNext):
self.next = newNext
def get_next(self):
return self.next
def hasNext(self):
return self.next is not None
def set_previous(self, newprev):
self.previous = newprev
def get_previous(self):
return self.previous
def hasPrevious(self):
return self.previous is not None
class double===>
class DoubleLinkedList:
def __init__(self):
self.head = None
self.tail = None
def addAtStart(self, item):
newNode = Node(item)
if self.head is None:
self.head = self.tail = newNode
else:
newNode.set_next(self.head)
newNode.set_previous(None)
self.head.set_previous(newNode)
self.head = newNode
def size(self):
current = self.head
count = 0
while current is not None:
count += 1
current = current.get_next()
return count
here is the wrong method ==>
try to fix it without more changes
def printReverse(self):
current = self.head
while current:
temp = current.next
current.next = current.previous
current.previous = temp
current = current.previous
temp = self.head
self.head = self.tail
self.tail = temp
print("Nodes of doubly linked list reversed: ")
while current is not None:
print(current.data),
current = current.get_next()
call methods==>
new = DoubleLinkedList()
new.addAtStart(1)
new.addAtStart(2)
new.addAtStart(3)
new.printReverse()
Your printReverse seems to do something else than what its name suggests. I would think that this function would just iterate the list nodes in reversed order and print the values, but it actually reverses the list, and doesn't print the result because of a bug.
The error in your code is that the final loop has a condition that is guaranteed to be false. current is always None when it reaches that loop, so nothing gets printed there. This is easily fixed by initialising current just before the loop with:
current = self.head
That fixes your issue, but it is not nice to have a function that both reverses the list, and prints it. It is better practice to separate these two tasks. The method that reverses the list could be named reverse. Then add another method that allows iteration of the values in the list. This is done by defining __iter__. The caller can then easily print the list with that iterator.
Here is how that looks:
def reverse(self):
current = self.head
while current:
current.previous, current.next = current.next, current.previous
current = current.previous
self.head, self.tail = self.tail, self.head
def __iter__(self):
node = self.head
while node:
yield node.data
node = node.next
def __repr__(self):
return "->".join(map(repr, self))
The main program can then be:
lst = DoubleLinkedList()
lst.addAtStart(1)
lst.addAtStart(2)
lst.addAtStart(3)
print(lst)
lst.reverse()
print(lst)
I am trying to remove the kth element from the END of a linked list. Here is my code
class LinkedList:
def __init__(self, value):
self.value = value
self.next = None
def removeKthNodeFromEnd(head, k):
if k == 0 or head is None:
return
temp = head
while temp is not None and k > 1:
temp = temp.next
k -= 1
if temp is None or k > 1:
return head
trailing = None
leading = head
while temp.next is not None:
trailing = leading
leading = leading.next
temp = temp.next
if trailing is None:
head = head.next
else:
trailing.next = leading.next
head = LinkedList(0)
head.next = LinkedList(1)
head.next.next = LinkedList(2)
head.next.next.next = LinkedList(3)
head.next.next.next.next = LinkedList(4)
head.next.next.next.next.next = LinkedList(5)
head.next.next.next.next.next.next = LinkedList(6)
head.next.next.next.next.next.next.next = LinkedList(7)
head.next.next.next.next.next.next.next.next = LinkedList(8)
head.next.next.next.next.next.next.next.next.next = LinkedList(9)
removeKthNodeFromEnd(head, 10)
while head is not None:
print(head.value)
head = head.next
But this does not work and prints out all the values in the linked list from 0 to 9. Why is this the case? If the node to delete is the head node which I check by checking trailing is None, then I update head = head.next. I am changing head to be the next node. I can return head after this update and the result of head = removeKthNodeFromEnd(head, k) will give me the desired output but why can't I do it the other way without returning anything? For instance in my solution I can get rid of any element in between the first and last node including the last and works just fine. The original head gets updated and node gets removed. But when trying to update the head node head = head.next it does not work.
One way I achieved this is by doing the following...
if trailing is None:
head.value = head.next.value
head.next = head.next.next
But why must I use the values? To me this seems the same as
if trailing is None:
head = head.next
head.next = head.next.next
but does not work
It doesn't work because head in that function is a local name. Assigning to a local variable never does anything to other variables, even not when they happen to have the same name (like the global head).
The "trick" you have by moving a linked list value with head.value = head.next.value will not work when the list only has one node.
One way to do this, is to return the (potentially modified) value of head, and expect the caller to assign this back to their own head variable.
So:
in removeKthNodeFromEnd make sure all return statements are like return head, and that you add a return head at the very end. In short, removeKthNodeFromEnd should always return head.
in the main program, change the call to head = removeKthNodeFromEnd(head, 10)
That will solve your issue.
Here are some ideas to make your code more elegant:
Create a separate class for maintaining the head of the linked list
Improve the constructor, so that it is easier to initialise a linked list
Add an iterator so it is easier to print a linked list
class Node: # renamed
def __init__(self, value, nxt=None):
self.value = value
self.next = nxt # is now an argument
class LinkedList: # added
def __init__(self, *values):
self.head = None
if values: # populate the list with values
for value in reversed(values):
self.push_front(value)
def push_front(self, value):
self.head = Node(value, self.head)
def __iter__(self):
node = self.head
while node:
yield node.value
node = node.next
def removeKthNodeFromEnd(self, k):
if k <= 0:
return
temp = self.head
for _ in range(k):
if not temp: # k is out of range
return
temp = temp.next
if temp:
trailing = self.head
while temp.next:
trailing = trailing.next
temp = temp.next
trailing.next = trailing.next.next
else:
self.head = self.head.next
lst = LinkedList(*range(10))
print(*lst)
lst.removeKthNodeFromEnd(0)
print(*lst)
And if you maintain a size attribute, you can do:
class LinkedList: # added
def __init__(self, *values):
self.head = None
self.size = 0
if values:
for value in reversed(values):
self.push_front(value)
def push_front(self, value):
self.head = Node(value, self.head)
self.size += 1
def __iter__(self):
node = self.head
while node:
yield node.value
node = node.next
def removeAt(self, k):
if 1 < k <= self.size:
trailing = self.head
for _ in range(k - 2):
trailing = trailing.next
trailing.next = trailing.next.next
elif 1 == k <= self.size:
self.head = self.head.next
def removeKthNodeFromEnd(self, k):
self.removeAt(self.size + 1 - k)
How can I create a simple function that merges two linked lists in a way that allows me to do the following with something like 'merge(self, other)', also I do not need my merged list to necessarily be sorted - I would like the merge function to simply just add and I've included driver code to give an idea
ls = [2,3,4,5]
ls2 = [42, 17]
ls.merge(ls2) # should change ls to [2,3,4,5,42,17]
ls2.head.data = 24 # should change ls2 to [24,17] and ls to [2,3,4,5,24,17]
class Node:
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
def __init__(self):
self.head = None
def merge_sorted(self, llist):
p = self.head
q = llist.head
s = None
if not p:
return q
if not q:
return p
if p and q:
if p.data <= q.data:
s = p
p = s.next
else:
s = q
q = s.next
new_head = s
while p and q:
if p.data <= q.data:
s.next = p
s = p
p = s.next
else:
s.next = q
s = q
q = s.next
if not p:
s.next = q
if not q:
s.next = p
return new_head
llist_1 = LinkedList()
llist_2 = LinkedList()
llist_1.append(1)
llist_1.append(5)
llist_1.append(7)
llist_1.append(9)
llist_1.append(10)
llist_2.append(2)
llist_2.append(3)
llist_2.append(4)
llist_2.append(6)
llist_2.append(8)
llist_1.merge_sorted(llist_2)
llist_1.print_list()
I guess append is a better name for simple merge.
class Node:
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
def __init__(self):
self.head = None
self.tail = None # If we don't store tail, `append` would be O(n)
def append(self, other): # `other` is also a `LinkedList`
if self.head:
self.tail.next = other.head
self.tail = other.tail
else:
self.head = other.head
self.tail = other.tail
To allow for an efficient append and merge implementation you would need to add the tail attribute to your linked list implementation
I would vote against a print_list method, because printing should not be managed by such a class. Instead provide a method that will give a string representation of the list, and let the main program decide whether to print that.
Here is how that could work:
class Node:
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
def __init__(self):
self.head = None
self.tail = None # <-- add this to have an efficient append/merge method
def append(self, data):
node = Node(data)
if not self.head: # When this list is empty
self.head = node
else:
self.tail.next = node
self.tail = node
def merge(self, llist):
if not self.head: # When this list is empty
self.head = llist.head
else:
self.tail.next = llist.head
self.tail == llist.tail
def __iter__(self): # To facilitate any need to iterate through the list
node = self.head
while node:
yield node.data
node = node.next
def __str__(self): # Don't make a print method; instead provide a string
return "->".join(map(str, self)) # This calls self.__iter__()
llist_1 = LinkedList()
llist_2 = LinkedList()
llist_1.append(1)
llist_1.append(5)
llist_1.append(7)
llist_1.append(9)
llist_1.append(10)
llist_2.append(2)
llist_2.append(3)
llist_2.append(4)
llist_2.append(6)
llist_2.append(8)
llist_1.merge(llist_2)
print(llist_1) # Only call `print` here
I learned about linked-lists today. I learned how to insert and remove nodes from them. In the following code, it teaches me how to insert nodes with three distinct functions: at beginning (the head node), in between and at the end. However, they teach me how to remove the node in and single function. I don't find the code in the remove function to be very clear. Can anyone help make an easier to understand code under the remove function? Is there a way to make it into three functions just like the inserting ones? I'm open to any suggestion or explanation. Thanks in advance.
Here's my code:
class Node:
def __init__(self, data=None):
self.data = data
self.nextnode = None
class LinkedList:
def __init__(self):
self.headnode = None
def printlist(self):
node = self.headnode
while node is not None:
print (node.data)
node = node.nextnode
def atbegining(self,new_node):
new_node.nextnode = self.headnode
self.headnode = new_node
# Function to add newnode
def AtEnd(self, newnode):
if self.headnode is None:
self.headnode = newnode
return
node = self.headnode
while(node.nextnode):
node = node.nextnode
node.nextnode=newnode
# Function to add node
def Inbetween(self,preNode,newNode):
if preNode is None:
print("The mentioned node is absent")
return
newNode.nextnode = preNode.nextnode
preNode.nextnode = newNode
# Function to remove node
def RemoveNode(self, RemoveVal):
node = self.headnode
if (node is not None):
if (node.data == RemoveVal):
self.headnode = node.nextnode
node = None
return
while (node is not None):
if node.data == RemoveVal:
break
prevnode = node
node = node.nextnode
if (node == None):
return
prevnode.nextnode = node.nextnode
node = None
list1 = LinkedList()
list1.headnode = Node("Mon")
n2 = Node("Tue")
n3 = Node("Wed")
# Link first Node to second node
list1.headnode.nextnode = n2
# Link second Node to third node
n2.nextnode = n3
n4 = Node("Sun")
n5 = Node("Tur")
n6 = Node("Newdate")
list1.atbegining(n4)
list1.AtEnd(n5)
list1.Inbetween(list1.headnode,n6)
list1.RemoveNode("Newdate")
list1.printlist()
RemoveNode is complicated by the fact that there are two structurally distinct kinds of LinkedLists: one whose head is None, and one whose head is not None. You can fix this by making sure every LinkedList contains at least one node. This is typically referred to as a dummy node, and you can use this node to store metadata (such as the length of the list).
The Node class itself does not change.
class Node:
def __init__(self, data=None):
self.data = data
self.nextnode = None
The LinkedList, however, simplifies by creating a dummy node. This provides
the guarantee that every node that stores real data is point to by another node.
class LinkedList:
def __init__(self):
self.headnode = Node(0)
def insert(self, preNode, newNode):
newNode.nextnode = preNode.nextnode
preNode.nextnode = newNode
self.headnode.data += 1
def append(self, newNode):
curr = self.headnode
while curr.nextNode is not None:
curr = curr.nextNode
self.insert(curr, newNode)
def prepend(self, newNode):
self.insert(self.headnode, newNode)
def _find_before(self, val):
pre = self.headnode
while pre.nextnode is not None:
curr = pre.nextnode
if curr.data == val:
return pre
pre = curr
def remove(self, RemoveVal):
pre = self._find_before(RemoveVal)
if pre is None:
return
pre.nextnode = pre.nextnode.nextnode
self.headnode.data -= 1
This simplifies all three insertions. The general case can always apply, since there is always a node that comes before the node you insert. append and prepend are simple wrappers that find the appropriate node to pass to insert.
Likewise, remove simply finds the node before the given value, and if the search succeeds, handles updating the prior node's nextnode attribute.
insert and remove also update the size of the list stored in the dummy node.
A find method becomes a simple wrapper around _find_before; if you find a node before the value you are looking for, just return the node that follows it.
I think that an alternative design will make the code much clearer. Consider for example the following:
class Node:
def __init__(self, data=None):
self.data = data
self.nextnode = None
def printlist(self):
print(self.data)
if self.nextnode is not None:
self.nextnode.printlist()
def push(self, node):
node.nextnode = self
return node
def insertafter(self, node):
node.nextnode = self.nextnode
self.nextnode = node
return self
def append(self, node):
lastnode = self
while lastnode.nextnode is not None:
lastnode = lastnode.nextnode
lastnode.nextnode = node
return self
def remove(self, value):
prev = None
walk = self
while walk is not None:
if walk.data == value:
if prev is None:
return walk.nextnode
else:
prev.nextnode = walk.nextnode
return self
else:
prev = walk
walk = walk.nextnode
return self
list1 = Node("Mon")
n2 = Node("Tue")
n3 = Node("Wed")
# Link first Node to second node
list1 = list1.insertafter(n2)
# Link second Node to third node
n2 = n2.insertafter(n3)
n4 = Node("Sun")
n5 = Node("Tur")
n6 = Node("Newdate")
list1 = list1.push(n4)
list1 = list1.append(n5)
list1 = list1.insertafter(n6)
list1 = list1.remove("Newdate")
list1.printlist()
The main idea is that a Node is the linked list. As long as you have the head of the list kept in a variable, you can have access to the entire list, without the need for a separate data structure.
I am trying to write a program to merge two sorted linked lists.
But its not merging them.
it seems like the error is in representing head pointer.
how to represent head pointer? is my way correct.
I am facing a lot of problems using pointers pls give me some suggestions especially head pointer
merge two sorted linkedlist
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
You already have the addNodeToList() function so why not use it?
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
You are never adding to dummy in your code - you do temp = dummy.head and then you reassign temp and the dummy list is never updated. Here is an approach:
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()