[TS] Implement a singly linked list in TypeScript
In a singly linked list each node in the list stores the contents of the node and a reference (or pointer in some languages) to the next node in the list. It is one of the simplest way to store a collection of items.
In this lesson we cover how to create a linked list data structure and how to use its strengths to implement an O(1) FIFO queue.
/**
* Linked list node
*/
export interface LinkedListNode<T> {
value: T
next?: LinkedListNode<T>
} /**
* Linked list for items of type T
*/
export class LinkedList<T> {
public head?: LinkedListNode<T> = undefined;
public tail?: LinkedListNode<T> = undefined; /**
* Adds an item in O(1)
**/
add(value: T) {
const node = {
value,
next: undefined
}
if (!this.head) {
this.head = node;
}
if (this.tail) {
this.tail.next = node;
}
this.tail = node;
} /**
* FIFO removal in O(1)
*/
dequeue(): T | undefined {
if (this.head) {
const value = this.head.value;
this.head = this.head.next;
if (!this.head) {
this.tail = undefined;
}
return value;
}
} /**
* Returns an iterator over the values
*/
*values() {
let current = this.head;
while (current) {
yield current.value;
current = current.next;
}
}
}
import { LinkedList } from './linkedList'; test('basic', () => {
const list = new LinkedList<number>();
list.add(1);
list.add(10);
list.add(5);
expect(Array.from(list.values())).toMatchObject([1, 10, 5]);
expect(list.dequeue()).toBe(1);
expect(Array.from(list.values())).toMatchObject([10, 5]);
expect(list.dequeue()).toBe(10);
expect(list.dequeue()).toBe(5);
expect(list.dequeue()).toBe(undefined);
expect(Array.from(list.values())).toMatchObject([]);
list.add(5);
expect(Array.from(list.values())).toMatchObject([5]);
});
We can also see the beautiy of Generator with Array.from partten.
Array.from(list.values())
It saves lots of code to keep maintaing the indexing of the array.
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