- Once the fast node.next is null, return slow node data
"""
from LinkedList.singlyLL import SinglyLL
def find_middle(head):
""" Find middle element in a linkedlist """
slow_node = head
fast_node = head
"""taking fast_node not none condition to
prevent nontype error in case of even number
of nodes.
"""
while fast_node and fast_node.next is not None:
slow_node = slow_node.next
fast_node = fast_node.next.next
return slow_node.data
if __name__ == '__main__':
sll = SinglyLL()
for i in range(0, 10):
sll.append(i)
print(sll.display())
print(find_middle(sll.head))
prev_node = cur_node
cur_node = cur_node.next
if cur_node not in node_list:
node_list.append(cur_node)
else:
print('Loop Detected')
prev_node.next = None
return
print('No Loop detected')
if __name__ == '__main__':
ll = SinglyLL()
for i in range(0, 10):
ll.append(i)
print(ll.display())
detect_remove_loop(ll.head)
print('Creating a linked list with loop')
ll2 = SinglyLL()
create_ll_with_loop(ll2)
# print('infinite loop')
# ll2.display()
print('running LL with loop')
detect_remove_loop(ll2.head)
cur_node = ll2.head
while cur_node is not None:
print(cur_node.data)
cur_node = cur_node.next
def remove_dup_sorted_list(head):
""" remove duplicate elements from a sorted list """
cur_node = head
while cur_node.next is not None:
prev_node = cur_node
cur_node = cur_node.next
if cur_node.data == prev_node.data:
prev_node.next = cur_node.next
if __name__ == '__main__':
print('Removing Dups from an unsorted list')
sll = SinglyLL()
sll.append(2)
sll.append(3)
sll.append(4)
sll.append(2)
sll.append(5)
sll.append(3)
print(sll.display())
remove_dup(sll.head)
print(sll.display())
print('*' * 10)
print('Removing dups from a sorted list')
prev = l3.next
if l1 is not None:
l1 = l1.next
if l2 is not None:
l2 = l2.next
if carry < 0:
l3.append(carry)
print(l3.display())
if __name__ == '__main__':
l1 = SinglyLL()
l1.append(3)
l1.append(4)
l1.append(5)
print(l1.display())
l2 = SinglyLL()
l2.append(2)
l2.append(5)
l2.append(9)
l2.append(8)
print(l2.display())
l3 = SinglyLL()
add_two_numbers(l1.head, l2.head, l3.head)