Reputation: 163
So I have this code
class Node():
def __init__(self, data):
self.data = data
self.next = None
class LinkedList():
def __init__(self):
self.head = None
self.length = 0
def prepend(self, data):
new_node = Node(data)
new_node.next = self.head
self.head = new_node
self.length += 1
def print_list(self):
if self.head == None:
print('Empty')
else:
currentNode = self.head
while currentNode != None:
print(currentNode.data, end = ' ')
currentNode = currentNode.next
def reverse(self):
first = self.head
second = first.next
while second:
temp = second.next
second.next = first
first = second
second = temp
I prepend and then have a list of [8,7,11,6,5,10]. When I try to use the reverse function and try to print out I get infinite loop of 8 and 7. I'm not sure why. I thought the first already points to the self.head so that when it gets changed, the self.head gets changed as well. But its not the case. Could someone please help me with this?
Upvotes: 0
Views: 172
Reputation: 3722
Your reverse
method is focussing on only the first two elements, and setting up cyclic references between them.
Here's the modified reverse()
method, that works. (Note that the reverse()
method contains an enclosed helper function called rev_()
, which recurses through the linked list):
def reverse (self): #------------MODIFIED--------------
def rev_ (head):
rest = head.next
if rest is None:
return head
else:
reversed_rest = rev_(rest)
rest.next = head
head.next = None
return reversed_rest
if self.head is not None:
self.head = rev_(self.head)
Testing it out:
my_list = LinkedList()
my_list.print_list()
print()
my_list.reverse()
my_list.print_list()
print()
my_list.prepend(21)
my_list.print_list()
print()
my_list.reverse()
my_list.print_list()
print()
my_list.prepend(22)
my_list.prepend(23)
my_list.print_list()
print()
my_list.reverse()
my_list.print_list()
Output:
Empty
Empty
21
21
23 22 21
21 22 23
Upvotes: 2