Question

LeetCode Link | 707. Design Linked List | Medium

Design your implementation of the linked list. You can choose to use a singly or doubly linked list.
A node in a singly linked list should have two attributes: val and next. val is the value of the current node, and next is a pointer/reference to the next node.
If you want to use the doubly linked list, you will need one more attribute prev to indicate the previous node in the linked list. Assume all nodes in the linked list are 0-indexed.

Implement the MyLinkedList class:

  • MyLinkedList() Initializes the MyLinkedList object.
  • int get(int index) Get the value of the indexth node in the linked list. If the index is invalid, return -1.
  • void addAtHead(int val) Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list.
  • void addAtTail(int val) Append a node of value val as the last element of the linked list.
  • void addAtIndex(int index, int val) Add a node of value val before the indexth node in the linked list. If index equals the length of the linked list, the node will be appended to the end of the linked list. If index is greater than the length, the node will not be inserted.
  • void deleteAtIndex(int index) Delete the indexth node in the linked list, if the index is valid.

Constrains

  • 0 <= index, val <= 1000
  • Please do not use the built-in LinkedList library.
  • At most 2000 calls will be made to get, addAtHead, addAtTail, addAtIndex and deleteAtIndex.

Example

Example 1

Input:
[“MyLinkedList”, “addAtHead”, “addAtTail”, “addAtIndex”, “get”, “deleteAtIndex”, “get”]
[[], [1], [3], [1, 2], [1], [1], [1]]

Output:
[null, null, null, null, 2, null, 3]

Explanation:

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MyLinkedList myLinkedList = new MyLinkedList();
myLinkedList.addAtHead(1);
myLinkedList.addAtTail(3);
myLinkedList.addAtIndex(1, 2);    // linked list becomes 1->2->3
myLinkedList.get(1);              // return 2
myLinkedList.deleteAtIndex(1);    // now the linked list is 1->3
myLinkedList.get(1);              // return 3

Solution Approach

Method: Manual / Simulate

Constrains

  • None

This method does not have any limitations.

Concept Explanation

This problem involves designing a linked list with five interfaces:

  1. Get the value of the node at the nth index in the linked list.
  2. Insert a node at the beginning of the linked list.
  3. Insert a node at the end of the linked list.
  4. Insert a node before the nth node in the linked list.
  5. Delete the node at the nth index in the linked list.

These five interfaces cover the common operations of linked lists, making it a very good problem for practicing linked list operations.

Code

  • Time complexity: Operations involving the index are O(index), others are O(1)
  • Space complexity: O(n)

TypeScript

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// class ListNode {
//     val: number
//     next: ListNode | null
//     constructor(val?: number, next?: ListNode | null) {
//         this.val = val === undefined ? 0 : val;
//         this.next = next === undefined ? null : next;
//     }
// }

class MyLinkedList {
	private size: number;
	private head: ListNode | null;
	private tail: ListNode | null;

	constructor() {
		this.size = 0;
		this.head = null;
		this.tail = null;
	}

	get(index: number): number {
		if (index < 0 || index >= this.size) return -1;
		let node = this.getNode(index);
		return node.val;
	}

	addAtHead(val: number): void {
		let node = new ListNode(val, this.head);
		this.head = node;
		if (!this.tail) {
			this.tail = node;
		}
		this.size++;
	}

	addAtTail(val: number): void {
		const tailNode = new ListNode(val, null);
		if (this.tail === null) {
			this.head = tailNode;
		} else {
			this.tail.next = tailNode;
		}
		this.tail = tailNode;
		this.size++;
	}

	addAtIndex(index: number, val: number): void {
		if (index > this.size) return;
		if (index === this.size) {
			this.addAtTail(val);
			return;
		}
		if (index === 0) {
			this.addAtHead(val);
			return;
		}
		let curNode = this.getNode(index - 1);
		let node = new ListNode(val, curNode.next);
		curNode.next = node;
		this.size++;
	}

	deleteAtIndex(index: number): void {
		if (index < 0 || index >= this.size) return;
		if (index === 0) {
			this.head = this.head!.next;
			if (index === this.size - 1) {
				this.tail = null
			}
			this.size--;
			return;
		}
		const node = this.getNode(index - 1);
		node.next = node.next.next;
		if (index === this.size - 1) {
			this.tail = node;
		}
		this.size--;
	}

	private getNode(index: number): ListNode {
		if (index === this.size - 1) return this.tail;
		let curNode: ListNode = new ListNode(0, this.head);
		for (let i = 0; i <= index; i++) {
			curNode = curNode.next!;
		}
		return curNode;
	}
}

Conclusion

There’s nothing really new to learn, just simple but effort-required questions. Just remembering the structure and operations of the linked list should be good.