def merge_two_sorted_lists(L1, L2):
# head of both lists
curr_1 = L1.head
curr_2 = L2.head
# empty linked list
new_ll = LinkedList()
# traverse while still nodes and compare which the least to new_ll
while curr_1 and curr_2:
if curr_1.data <= curr_2.data:
new_ll.addNode(Node(curr_1.data))
curr_1 = curr_1.next
else:
new_ll.addNode(Node(curr_2.data))
curr_2 = curr_2.next
# appends remaining nodes of L1 or L2
if not curr_1:
while curr_2:
new_ll.addNode(Node(curr_2.data))
curr_2 = curr_2.next
elif not curr_2:
while curr_1:
new_ll.addNode(Node(curr_1.data))
curr_1 = curr_1.next
# return the new list
return new_ll
def add_two_numbers(L1: LinkedList, L2: LinkedList):
final_ll = LinkedList()
L1_curr = L1.head
L2_curr = L2.head
carry = 0
while L1_curr or L2_curr or carry:
val = carry + (L1_curr.data if L1_curr else 0) + (L2_curr.data
if L2_curr else 0)
L1_curr = L1_curr.next if L1_curr else None
L2_curr = L2_curr.next if L2_curr else None
final_ll.addNode(Node(val % 10))
carry = val // 10
return final_ll
# naive: append everything and sort in place
# time: O(n+m)
# space: O(1)
l1 = LinkedList()
l2 = LinkedList()
node1 = Node(1)
node2 = Node(2)
node3 = Node(3)
node4 = Node(4)
node5 = Node(5)
node6 = Node(6)
node7 = Node(7)
l1.addNode(node1)
l2.addNode(node2)
l1.addNode(node3)
l2.addNode(node4)
l1.addNode(node5)
l2.addNode(node6)
l1.addNode(node7)
print("\n")
l1.print_list()
print("\n")
l2.print_list()
print("\n")
x = merge_two_sorted_lists(l1, l2)
# set temp next to current
# and previous next should be temp
else:
temp.set_next(current)
previous.set_next(temp)
return L1
if __name__ == "__main__":
# two pointers previous and current with a special condition for empty linkedlist
# time: O(n)
# space: O(1)
l1 = LinkedList()
node1 = Node(1)
node2 = Node(2)
node4 = Node(4)
node5 = Node(5)
l1.addNode(node1)
l1.addNode(node2)
l1.addNode(node4)
l1.addNode(node5)
print("\n")
x = orderedInsert(l1, 3)
x.print_list()
return L1
if __name__ == "__main__":
# naive: traverse two times, first to record length of list and to subtract k from it to see which element to stop at
# the first is k up ahead, by the time it stops the second will be at kth last
# time: O(n)
# space: O(1)
l1 = LinkedList()
node1 = Node(1)
node2 = Node(2)
node3 = Node(3)
node4 = Node(4)
node5 = Node(5)
node6 = Node(6)
node7 = Node(7)
l1.addNode(node1)
l1.addNode(node2)
l1.addNode(node3)
l1.addNode(node4)
l1.addNode(node5)
l1.addNode(node6)
l1.addNode(node7)
x = remove_kth_last(l1, 2)
print("\n")
x.print_list()
if __name__ == "__main__":
# even list before odd list
# first is index 0
# time: O(n)
# space: O(1)
l1 = LinkedList()
node0 = Node(0)
node1 = Node(1)
node2 = Node(2)
node3 = Node(3)
node4 = Node(4)
node5 = Node(5)
node6 = Node(6)
node7 = Node(7)
l1.addNode(node0)
l1.addNode(node1)
l1.addNode(node2)
l1.addNode(node3)
l1.addNode(node4)
l1.addNode(node5)
l1.addNode(node6)
l1.addNode(node7)
print("\n")
x = even_odd_merge(l1)
x.print_list()
# naive:
# grade school algorithm
# time: O(max(n,m)) - n,m length of two lists
# space: O(1)
l1 = LinkedList()
l2 = LinkedList()
node1 = Node(3)
node2 = Node(1)
node3 = Node(4)
node4 = Node(7)
node5 = Node(0)
node6 = Node(9)
l1.addNode(node1)
l1.addNode(node2)
l1.addNode(node3)
l2.addNode(node4)
l2.addNode(node5)
l2.addNode(node6)
print("\n")
l1.print_list()
print("\n")
l2.print_list()
print("\n")
x = add_two_numbers(l1, l2)
x.print_list()
if __name__ == "__main__":
# naive: maintain 3 lists and create new nodes for each.
# instead attach existing nodes to chain
# time: O(n)
# space: O(1)
ll = LinkedList()
ll.print_list()
node1 = Node(3)
node2 = Node(2)
node3 = Node(2)
node4 = Node(11)
node5 = Node(7)
node6 = Node(5)
node7 = Node(11)
ll.addNode(node1)
ll.addNode(node2)
ll.addNode(node3)
ll.addNode(node4)
ll.addNode(node5)
ll.addNode(node6)
ll.addNode(node7)
ll.print_list()
print("\n")
x=list_pivoting(ll,7)
x.print_list()
last = current
current = nextNode
# set the head to the very last node
L1.head = last
return L1
if __name__ == "__main__":
# uses three pointers one for current, last, and nextnode
# time: O(n)
# space: O(1)
linkedlst = LinkedList()
linkedlst.addNode(Node(1))
linkedlst.addNode(Node(2))
linkedlst.addNode(Node(3))
linkedlst.addNode(Node(4))
x = reverseList(linkedlst)
# last=None,curr=1->2->3->4
# next=2->3->4, curr=1,last=1,curr=2->3->4
# next=3->4,curr=2->1,last=2->1,curr=3->4
# next=4,curr=3->2->1,last=3->2->1,curr=4
# next=None,curr=4->3->2->1,last=4->3->2-1,curr=None
x.print_list()
# time: O(n)
# space: O(1)
l1 = LinkedList()
l2 = LinkedList()
node0 = Node(0)
node1 = Node(1)
node2 = Node(2)
node3 = Node(3)
node4 = Node(4)
node5 = Node(5)
node6 = Node(6)
node7 = Node(7)
l1.addNode(node0)
l1.addNode(node1)
l1.addNode(node2)
l1.addNode(node3)
l1.addNode(node4)
l1.addNode(node5)
print("\n")
l1.print_list()
print(l1.length())
l2.addNode(node6)
l2.addNode(node7)
l2.addNode(node3)
l2.addNode(node4)
l2.addNode(node5)