def test3():
#scenario: End node is the intersection
LLend1 = LinkedList()
LLend2 = LinkedList()
endNode = Node("end")
for num in range(19):
LLend1.push(num)
LLend2.push(num)
print(num)
LLend1.pushNode(endNode)
LLend2.pushNode(endNode)
result = findIntersection(LLend1, LLend2)
#count = 435
# length of both lists = 19
# This is the worst case, and the worst case of checking for the end node, does not stop with the check.
print(result)
def test2():
#worst case: Loop node is at the end, pointing at the first node.
LLisLoop2 = LinkedList()
loopNode2 = Node("LOOP2")
for i in range(20):
if i == 19:
LLisLoop2.pushNode(loopNode2)
else:
LLisLoop2.push(i)
#set the loop
begin = LLisLoop2.begin
loopNode2.next = begin
result = findLoop(LLisLoop2)
#count = 18
print(result)
#chapter 2 problem 7
from ch2_SetUp import Node, LinkedList
LL1 = LinkedList()
LL2 = LinkedList()
LL3 = LinkedList()
for char in "LIST":
LL3.push(char)
for char2 in "0101linked":
LL2.push(char2)
for char3 in "LINK2D":
LL1.push(char3)
nextNode = LL3.begin
LL1.pushNode(nextNode)
LL2.pushNode(nextNode)
#LL1.printList()
#print('-' * 10)
#LL2.printList()
#worst case is no intersection
LL4 = LinkedList()
LL5 = LinkedList()
for char4 in 'notintersecting':
LL4.push(char4)
LL5.push(char4)
#chapter 2 problem 8
from ch2_SetUp import Node, LinkedList
# CAUTION:
# using printList() *while loop* on a linkedlist with a loop, will cause an infinite loop of printing.
LLisLoop = LinkedList()
loopNode = Node("LOOP")
for i in range(20):
if i == 10:
LLisLoop.pushNode(loopNode)
else:
LLisLoop.push(i)
#set the loop
LLisLoop.end.next = loopNode
def findLoop(linkedlist):
node1 = linkedlist.begin
count = 0
if node1 == None:
print("This linked list is empty")
return None
runner = linkedlist.begin.next
if runner:
while node1 and runner:
if node1 == runner:
print("loop node found.")