# Unfortunately, we can't iterate backwards through the list b/c it's singly linked.....
def is_palindrome(start):
contents = []
current = start
while current != None:
contents.append(current.getData())
current = current.getNext()
if len(contents) == 0 or len(contents) == 1:
return True
midpoint = int(math.floor(len(contents) / 2))
for i in range(midpoint):
if contents[i] != contents[len(contents) - 1 - i]:
return False
return True
root = Node(1)
root.addNext(Node(2))
root.addNext(Node(1))
root.addNext(Node(4))
root.addNext(Node(2))
root.addNext(Node(1))
print(is_palindrome(root))
pointer1 = pointer1.getNext()
pointer2 = pointer2.getNext()
exponent += 1
# At this point, either pointer1 = None or pointer2 = None or both equal None
# If both, we can just return the sum
if pointer1 == None and pointer2 == None:
return list_sum
elif pointer1 == None:
while pointer2 != None:
power = 10**exponent
list_sum += pointer2.getData() * power
pointer2 = pointer2.getNext()
exponent += 1
else:
while pointer1 != None:
power = 10**exponent
list_sum += pointer1.getData() * power
pointer1 = pointer1.getNext()
exponent += 1
return list_sum
root1 = Node(14)
root1.addNext(Node(1))
root2 = Node(15)
print(sum_lists(root1, root2))
# Detect the *first* node in a cycle in a circular linkedlist. # BFS is an obvious approach, but we can try something else def detect_cycle(start): # Idea: 1) We can maintain a list of nodes that we've seen and return the first node seen twice # 2) We can have two pointers run through the list, one twice the rate of the other. # If, at any point, the two pointers point to the same node, that's the one pointer1 = start pointer2 = start check = 0 while pointer1 != pointer2 or check == 0: # They both start at the same node check = 1 pointer1 = pointer1.getNext() pointer2 = pointer2.getNext().getNext() # Moves twice as fast return pointer1.getData() root = Node(1) root.addNext(Node(2)) root.addNext(Node(3)) begin = Node(4) # Where the cycle starts root.addNext(begin) root.addNext(Node(5)) root.addNext(Node(6)) root.addNext(begin) print(detect_cycle(root))
while current != None:
val = current.getData()
if val in seen:
# Remove this node from the list
next_node = current.getNext()
previous.setNext(next_node)
else:
seen.append(val)
previous = current
current = current.getNext()
return start_node # We can't return "current".....
root = Node(14)
root.addNext(Node(1))
root.addNext(Node(13))
root.addNext(Node(14))
root.addNext(Node(2))
root.addNext(Node(3))
root.addNext(Node(5))
root.addNext(Node(11))
root.addNext(Node(25))
print(root.toStr())
mod = remove_dups1(root)
print(mod.toStr())
