head = None
place = ln
if place.nextNode == None:
return place
nextplace = place.nextNode
while nextplace != None:
nextplace = place.nextNode
if place.value < x:
if front == None:
front = place
head = front
else:
front.nextNode = place
front = front.nextNode
else:
place.nextNode = back
back = place
place = nextplace
front.nextNode = back
return head
a = LN(3)
b = LN(5,a)
c = LN(9,b)
d = LN(2,c)
e = LN(6,d)
printList(e)
print()
printList(partition(e,6))
nextNode = nextnext
nextnext = nextNode.nextNode
nextNode.nextNode = currentNode
return nextNode
a = LN(1)
b = LN(2)
c = LN(3)
a.nextNode = b
b.nextNode = c
printList(a)
printList(reverse(a))
#I looked at the solution on InterviewCake after doing this, and their
#implementation is simpler to read. Oh well, I'll try harder next time.
'''
solution from InterviewCake, for my own reference
def reverse(head_of_list):
current = head_of_list
previous = None
next = None
# until we have 'fallen off' the end of the list
while current:
stack1.append(0)
#creating answer linked list
carry = 0
answer = None
while len(stack1) > 0:
value = stack1.pop()+stack2.pop()+carry
carry = value//10
temp = LN(value%10,answer)
answer = temp
if carry != 0:
temp = LN(carry,answer)
answer = temp
return answer
a = LN(6)
b = LN(1,a)
c = LN(7,b) #716
d = LN(2)
e = LN(9,d)
f = LN(5,e) #592
g = sumList(c,f)
printList(g) #1308
#Cracking the Coding Interview 4.3
#Given a binary tree, design an algorithm which creates a linked list
#of all the nodes at each depth
#I'm going to store each depth in an index of a list, with a header linked list
from TN import TN, printTree
from LN import LN, printList
from codingInterview42 import minimalTree
def listDepth(TN,DL,index):
if TN != None:
DL[index].nextNode = LN(TN.value,DL[index].nextNode)
listDepth(TN.left,DL,index+1)
listDepth(TN.right,DL,index+1)
l = [1,2,3,4,5,6,7,8,9]
root = minimalTree(l)
print(printTree(root,""))
DL = [LN(),LN(),LN(),LN()]
listDepth(root,DL,0)
for a in DL:
printList(a.nextNode)
print()
#I assume I have the linked list as well as the length
def palindrome(listNode, l: int):
ln,stack = listNode,[]
half = ceil(l/2)
#loading half the nodes into the stack
for i in range(half):
stack.append(ln.value)
ln = ln.nextNode
#dequeueing if it's an odd number of nodes
if half%2 == 1:
stack.pop()
#checking to see if the second half mirrors the first
while ln != None:
if ln.value != stack.pop():
return False
ln = ln.nextNode
return True
a = LN(3)
b = LN(2,a)
c = LN(1,b)
d = LN(2,c)
e = LN(3,d)
f = LN(4,e)
printList(f)
print()
print(palindrome(f,6))
