Python Linked List

linkedlist.py

card1 = {
    'name': 'John Smith',
    'age': 20,
    'gpa': 3.0,
    'pointer': None
}
card1

card2 = {
    'name': 'Amanda Jones',
    'age': 22,
    'gpa': 3.5,
    'pointer': None
}
card2

# Here's the part where we connect them
card1['pointer'] = card2
card1

# Now card1['pointer'] and card2 are pointing to the same thing

id(card1['pointer'])
id(card2)

# Let's add two more cards to this linked list
card3 = {
    'name': 'Carol Lim',
    'age': 23,
    'gpa': 2.5,
    'pointer': None
}
card3

card4 = {
    'name': 'Bernard Lee',
    'age': 25,
    'gpa': 4.0,
    'pointer': None
}
card4

# And let's connect them
# Note! Modifying card2 will also modify card1['pointer']
card2['pointer'] = card3
card3['pointer'] = card4  # Remember, they're pointing to the same thing!

# So now we have a linked list with four items and each item is connected in
# In the order 1,2,3,4.

# Here's some code that will print the entire list

current_card = card1
while(current_card is not None):
    print('Address: %x' % id(current_card))  # Print in hex format :)
    print('Name: %s' % current_card['name'])
    print('Age: %d' % current_card['age'])
    print('GPA: %f' % current_card['gpa'])
    print('Pointer: %x' % id(current_card['pointer']))
    print('=========================================')  # Print a divider
    # Set the current_card to the next item on the linked list
    current_card = current_card['pointer']

# It will stop at card4 because card4's pointer is None. current_card will be
# None and the loop will terminate

# Note however that card4's pointer prints an address. That may seem strange,
# until you remember that almost everything in Python is a pointer, in which
# case you'll realize that the address it's pointing to has a value of None.

'%x' % id(None)