Formatted question description: https://leetcode.ca/all/707.html

Medium

## Description

Design your implementation of the linked list. You can choose to use the singly linked list or the 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.

• get(index): Get the value of the index-th node in the linked list. If the index is invalid, return -1.
• addAtHead(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.
• addAtTail(val): Append a node of value val to the last element of the linked list.
• addAtIndex(index, val): Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted.
• deleteAtIndex(index): Delete the index-th node in the linked list, if the index is valid.

Example:

Input:
[[],[1],[3],[1,2],[1],[1],[1]]
Output:
[null,null,null,null,2,null,3]

Explanation:


Constraints:

• 0 <= index,val <= 1000
• At most 2000 calls will be made to get, addAtHead, addAtTail, addAtIndex and deleteAtIndex.

## Solution

This solution uses the doubly linked list.

Create a class Node that has data fields int val that is the node’s value, Node prev that is the node’s previous node, and Node next that is the node’s next node.

In class MyLinkedList, there are three data fields, which are Node head that is the first node, Node tail that is the last node, and int size that is the number of nodes in the linked list.

For the constructor, initialize head and tail to null, and size to 0.

For get(index), if index >= size, then return -1. Otherwise, find the index-th node starting from head, and return the node’s value.

For addAtHead(val), create a new node with value val. If size == 0, set both head and tail to be the new node. Otherwise, update the nodes around head and assign the new node to head. Increase size by 1.

For addAtTail(val), create a new node with value val. If size == 0, set both head and tail to be the new node. Otherwise, update the nodes around tail and assign the new node to tail. Increase size by 1.

For addAtIndex(index, val), first check index. If index == 0, then call addAtHead(val). Else, if index == size, then call addAtTail(val). Else, create a new node with value val, find the index-th node starting from head, and insert the new node at position index, with the nodes around the position updated.

For deleteAtIndex(index, val), first check index. If index == 0, then delete the node at head with the nodes around head updated. Else, if index == size - 1, then delete the node at tail with the nodes around tail updated. Else, find the index-th node starting from head, and delete the node, with the nodes around the position updated.