def kthtoLast(l,k): ## If size if known (non recursive solution)
n=l.size()
if(k>n):
print("invalid k")
return -1
nl=LinkedList()
count=1
current=l.head
while(count<=n-k+1 and current !=None):
print(current.getData())
nl.addNode(current.getData())
current=current.getNext()
count=count+1
return nl
def _treeLinkList(self,Node,l):
if(len(l)==0):
ll=LinkedList()
if(Node.getLeftChild()):
ll.addNode(Node.getLeftChild())
lChild=True
else:
print("End of Tree:No more left Child")
lChild=False
return
if(Node.getRightChild()):
ll.addNode(Node.getRightChild())
rChild=True
else:
print("End of Tree:No more right Child")
rChild=False
return
l.append(ll)
if(lChild):
Node=Node.getLeftChild()
self._treeLinkList(Node, l)
if(len(l)>0):
ll=LinkedList()
self.completeTreeLL(l,ll,Node)
if(Node.getLeftChild()):
Node=Node.getLeftChild()
self._treeLinkList(Node, l)
def twoNo(l1,l2):
if(l1==None or l1.size()==0):
return l2
if(l2==None or l2.size()==0):
return l1
n1=no(l1)
print("n1 is",n1)
n2=no(l2)
print("n2 is",n2)
n=n1+n2
n=str(n)
print("n is",n)
l=LinkedList()
for item in n:
l.addNode(item)
l.show()
return l
n=l.size()
if(k>n):
print("invalid k")
return -1
nl=LinkedList()
count=1
current=l.head
while(count<=n-k+1 and current !=None):
print(current.getData())
nl.addNode(current.getData())
current=current.getNext()
count=count+1
return nl
def kthtolastR(l,k): #Using recursion assuming size is unknown
if(l==None):
return 0
i=kthtolastR(l.getNext(), k)+1
if(i>=k):
print(l.getData())
return i
l=LinkedList()
#random.sample(range(30),10)
#[28, 9, 26, 19, 13, 1, 22, 29, 5]
for item in [1,2,3,4,5,6,7] :
l.addNode(item)
l.show()
print("Using Recursion")
nl=kthtolastR(l.head, 2)
from linkList import LinkedList #Exerc 1 lista = LinkedList() lista.append(7) lista.append(8) lista.append(9) lista.append(10) lista.append(11) print(len(lista)) #Exerc 2 lista2 = LinkedList() lista2.append(7) lista2.append(8) lista2.append(9) lista2.append(10) lista2.append(11) print(lista.compara(lista2)) #Exerc3 print(lista) print(lista.reverse()) #Exerc4 print(lista.maiorMedia())
n=n1+n2
n=str(n)
print("n is",n)
l=LinkedList()
for item in n:
l.addNode(item)
l.show()
return l
def no(l):
current=l.head
s=""
stack=[]
while(current!=None):
stack.append(current.getData())
current=current.getNext()
while(len(stack)!=0):
s=s+str(stack.pop())
return int(s)
l1=LinkedList()
l2=LinkedList()
for item in [6,1,7]:
l1.addNode(item)
for item in [2,9,5]:
l2.addNode(item)
twoNo(l1, l2)
Q:Implement a function to check if a list is palindrome?
-Size known or unknown?
-Use of stack allowed or not
-Doubly link list or singly
'''
from linkList import LinkedList
from linkList import Node
def isPalindrome(l):
fast=l.head
slow=l.head
stack=[]
while(fast!=None and fast.getNext()!=None): #Even Elements
stack.append(slow.getData())
slow=slow.getNext()
fast=fast.getNext().getNext()
if(fast!=None): #odd Elements
slow=slow.getNext()
while(slow!=None):
if(slow.getData()!=stack.pop()):
return False
slow=slow.getNext()
return True
l=LinkedList()
for item in "123321" :
l.addNode(item)
print(isPalindrome(l))
from linkList import Node
from linkList import LinkedList
print 'Testing linkList.py'
print 'Initializing linkList1'
linkList1 = LinkedList()
print 'Inserting "Hello" at head'
linkList1.insertHead("Hello")
print 'Size of linkList1:'
print linkList1.size()
print 'Inserting "world!" at head'
linkList1.insertHead("world")
print 'Size of linkList1:'
print linkList1.size()
print 'Search for "Hello"'
try:
linkList1.search("Hello")
print 'Data in list'
except:
print 'Data not in list'
print 'Search for "hello"'
try:
linkList1.search("hello")
print 'Data in list'
