Good afternoon! Here's our prompt for today.

Now that we've implemented a Linked List, let's start operating on it! Assume you have a Linked List implementation with this definition:

1class LinkedList {
2	constructor() {
3		this.head = null;
4		this.tail = null;
5	}
6
7	prepend(newVal) {
8		const currentHead = this.head;
9		const newNode = new Node(newVal);
10		newNode.next = currentHead;
11		this.head = newNode;
12
13		if (!this.tail) {
14			this.tail = newNode;
15		}
16	}
17
18	append(newVal) {
19		const newNode = new Node(newVal);
20		if (!this.head) {
21			this.head = newNode;
22			this.tail = newNode;
23		} else {
24			this.tail.next = newNode;
25			this.tail = newNode;
26		}
27	}
28}

Can you write a method getUnion to find the union of two linked lists? A union of two sets includes everything in both sets.

So given 1 -> 2 -> 3 and 3 -> 4 -> 5, we'd get 1 -> 2 -> 3 -> 4 -> 5.

Constraints

• Length of both the linkedlist <= 1000
• The nodes will always contain integer values between -1000000000 and 1000000000
• Let m, n be the lengths of the two lists
• Expected time complexity : O(n*m)
• Expected space complexity : O(n+m)

Try to solve this here or in Interactive Mode.

How do I practice this challenge?

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def get_union(l1, l2):

    # fill in this method

    return l1

​

​

# Node definition

class Node:

    def __init__(self, val):

        self.val = val

        self.next = None

​

​

def create_nodes(head, nodes):

    for val in nodes:

        new_node = Node(val)

        head.next = new_node

        head = new_node

​

​

list1 = Node(3)

nodes1 = [4, 5, 6, 7, 8, 9, 10]

create_nodes(list1, nodes1)

​

list2 = Node(1)

nodes2 = [2, 3, 4, 5, 6, 7, 8]

create_nodes(list2, nodes2)

​

​

def list_to_str(head):

Here's our guided, illustrated walk-through.

How do I use this guide?