how to values of a linked are in another linked list python code example
Example 1: linked list in python
class Node:
def __init__(self, data = None, next_node = None):
self.data = data
self.nextNode = next_node
def get_data(self):
return self.data
def set_data(self, data):
self.data = data
def get_nextNode(self):
return self.nextNode
def set_nextNode(self, nextNode):
self.nextNode = nextNode
class LinkedList:
def __init__(self, head = None):
self.head = head
def add_Node(self, data):
# if empty
if self.head == None:
self.head = Node(data)
# not empty
else:
curr_Node = self.head
# if node added is at the start
if data < curr_Node.get_data():
self.head = Node(data, curr_Node)
# not at start
else:
while data > curr_Node.get_data() and curr_Node.get_nextNode() != None:
prev_Node = curr_Node
curr_Node = curr_Node.get_nextNode()
# if node added is at the middle
if data < curr_Node.get_data():
prev_Node.set_nextNode(Node(data, curr_Node))
# if node added is at the last
elif data > curr_Node.get_data() and curr_Node.get_nextNode() == None:
curr_Node.set_nextNode(Node(data))
def search(self, data):
curr_Node = self.head
while curr_Node != None:
if data == curr_Node.get_data():
return True
else:
curr_Node = curr_Node.get_nextNode()
return False
def delete_Node(self, data):
if self.search(data):
# if data is found
curr_Node = self.head
#if node to be deleted is the first node
if curr_Node.get_data() == data:
self.head = curr_Node.get_nextNode()
else:
while curr_Node.get_data() != data:
prev_Node = curr_Node
curr_Node = curr_Node.get_nextNode()
#node to be deleted is middle
if curr_Node.get_nextNode() != None:
prev_Node.set_nextNode(curr_Node.get_nextNode())
# node to be deleted is at the end
elif curr_Node.get_nextNode() == None:
prev_Node.set_nextNode(None)
else:
return "Not found."
def return_as_lst(self):
lst = []
curr_Node = self.head
while curr_Node != None:
lst.append(curr_Node.get_data())
curr_Node = curr_Node.get_nextNode()
return lst
def size(self):
curr_Node = self.head
count = 0
while curr_Node:
count += 1
curr_Node = curr_Node.get_nextNode()
return count
## TEST CASES #
test1 = LinkedList()
test2 = LinkedList()
test1.add_Node(20)
test1.add_Node(15)
test1.add_Node(13)
test1.add_Node(14)
test1.delete_Node(17)
print(test1.return_as_lst())
print(test2.size())
Example 2: linked list python example
class Node:
def __init__(self, data=None):
self.data = data
self.next = None
class SLinkedList:
def __init__(self):
self.head = None
def Atbegining(self, data_in):
NewNode = Node(data_in)
NewNode.next = self.head
self.head = NewNode
# Function to remove node
def RemoveNode(self, Removekey):
HeadVal = self.head
if (HeadVal is not None):
if (HeadVal.data == Removekey):
self.head = HeadVal.next
HeadVal = None
return
while (HeadVal is not None):
if HeadVal.data == Removekey:
break
prev = HeadVal
HeadVal = HeadVal.next
if (HeadVal == None):
return
prev.next = HeadVal.next
HeadVal = None
def LListprint(self):
printval = self.head
while (printval):
print(printval.data),
printval = printval.next
llist = SLinkedList()
llist.Atbegining("Mon")
llist.Atbegining("Tue")
llist.Atbegining("Wed")
llist.Atbegining("Thu")
llist.RemoveNode("Tue")
llist.LListprint()