# 707-design-linked-list Try it on leetcode ## Description

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:

 

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
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

 

Constraints:

## Solution(Python) ```Python class ListNode: def __init__(self, val): self.val = val self.prev = None self.next = None class MyLinkedList: def __init__(self): """ Initialize your data structure here. """ self.head = None self.tail = None self.length = 0 def get(self, index: int) -> int: """ Get the value of the index-th node in the linked list. If the index is invalid, return -1. """ if index < 0 or index >= self.length: return -1 cur = self.head while index != 0: cur = cur.next index -= 1 return cur.val def addAtHead(self, val: int) -> None: """ 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. """ new_node = ListNode(val) new_node.next = self.head if self.head: self.head.prev = new_node self.head = new_node self.length += 1 if self.length == 1: self.tail = new_node ### trace and debug # self.print_linked_list() def addAtTail(self, val: int) -> None: """ Append a node of value val to the last element of the linked list. """ new_node = ListNode(val) new_node.prev = self.tail if self.tail: self.tail.next = new_node self.tail = new_node self.length += 1 if self.length == 1: self.head = new_node ### trace and debug # self.print_linked_list() def addAtIndex(self, index: int, val: int) -> None: """ 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. """ if index < 0 or index > self.length: return elif index == 0: self.addAtHead(val) elif index == self.length: self.addAtTail(val) else: cur = self.head while index - 1 != 0: cur = cur.next index -= 1 new_node = ListNode(val) new_node.next = cur.next cur.next.prev = new_node cur.next = new_node new_node.prev = cur self.length += 1 ### trace and debug # self.print_linked_list() def deleteAtIndex(self, index: int) -> None: """ Delete the index-th node in the linked list, if the index is valid. """ if index < 0 or index >= self.length: return elif index == 0: if self.head.next: self.head.next.prev = None self.head = self.head.next self.length -= 1 if self.length == 0: self.tail = None elif index == self.length - 1: if self.tail.prev: self.tail.prev.next = None self.tail = self.tail.prev self.length -= 1 if self.length == 0: self.head = None else: cur = self.head while index - 1 != 0: cur = cur.next index -= 1 cur.next = cur.next.next cur.next.prev = cur self.length -= 1 ### trace and debug # self.print_linked_list() def print_linked_list(self): cur = self.head while cur: print(f" {cur.val} -> ", end="") cur = cur.next print("\n") return # Your MyLinkedList object will be instantiated and called as such: # obj = MyLinkedList() # param_1 = obj.get(index) # obj.addAtHead(val) # obj.addAtTail(val) # obj.addAtIndex(index,val) # obj.deleteAtIndex(index) ```