Find the first shared node between two linked lists

linked-list-shared.py

from operator import itemgetter

class ListNode:
    def __init__(self, value):
        self.value = value

    # Append a node to this list node by setting the next property
    def append(self, node):
        self.next = node


# Build a list from an iterable
def build_list(elements):
    head = None
    current = None
    for v in elements:
        n = ListNode(v)
        
        if head == None:
            head = n
            current = head
        else:
            current.append(n)
            current = n

    # Return both the head and the tail of the list
    return (head, current)

# Get the length of the list by traversing it
def list_length(n):
    count = 1;
    try:
        while n.next != None:
            n = n.next
            count += 1
    except AttributeError:
        return count


(first_list, first_tail) = build_list(range(1, 6))
(second_list, second_tail) = build_list(range(5, 7))
(shared_list, _) = build_list(range(100, 160))

first_tail.append(shared_list)
second_tail.append(shared_list)

# Get the length of each list an an array
lists_with_length = map(lambda l: (l, list_length(l)), [first_list, second_list])

# Sort the lists by length in descending order
lists_with_length.sort(key = itemgetter(1), reverse=True)

longer = lists_with_length[0][0]
shorter = lists_with_length[1][0]

diff = lists_with_length[0][1] - lists_with_length[1][1]

# Traverse the longer list the number of nodes by which it is longer
# than the shorter list.
while diff > 0:
    longer = longer.next
    diff -= 1


# Traverse both lists comparing the values.
try:
    while 1:
        if longer.value == shorter.value:
            print("Shared node with value", shorter.value)
            break
    
        longer = longer.next
        shorter = shorter.next
except AttributeError:
    pass