#Complexity: O(n + m) where m + n are the lenghts of the 2 lists, space O(1), since no additional memory
from ListNode import ListNode
def merge_two_sorted_lists(l1: [ListNode], l2: [ListNode]) ->[ListNode]:
# Creates a placeholder for the result
head = ListNode()
tail = head
# Start building tail with the smaller value and move pointer of the list where it got used
while l1 and l2:
if l1.data <= l2.data:
tail.next = l1
l1 = l1.next
else:
tail.next = l2
l2 = l2.next
# move tail to the actual node that was smaller
tail = tail.next
# Append the remaining nodes
tail.next = l1 or l2
return head.next
a = ListNode(2, ListNode(5, ListNode(7)))
b = ListNode(3, ListNode(11))
a.print_list()
b.print_list()
print("Merge and Sorted:")
c = merge_two_sorted_lists(a,b)
c.print_list()
# coding=utf-8
"""Delete Node in a Linked List."""
from __future__ import print_function
from ListNode import ListNode
def _solve(node):
node.val = node.next.val
node.next = node.next.next
if __name__ == '__main__':
node1 = ListNode(1)
node2 = ListNode(2)
node3 = ListNode(3)
node4 = ListNode(4)
node1.next = node2
node2.next = node3
node3.next = node4
_solve(node3)
node1.print_list()
