707 design linked list

Best Solution

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class :
def __init__(self, val):
self.val = val
self.prev = self.next = None

class MyLinkedList:

def __init__(self):
"""
Initialize your data structure here.
"""

self.head = ListNode(None)
self.tail = ListNode(None)
self.head.next = self.tail
self.tail.prev = self.head
self.size = 0

def get(self, index):
"""
Get the value of the index-th node in the linked list. If the index is invalid, return -1.
:type index: int
:rtype: int
"""
if self._inRange(index, 0, self.size-1):
return self._getNodeAt(index).val
return -1


def addAtHead(self, 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.
:type val: int
:rtype: void
"""
self.addAtIndex(0, val)

def addAtTail(self, val):
"""
Append a node of value val to the last element of the linked list.
:type val: int
:rtype: void
"""
self.addAtIndex(self.size, val)

def addAtIndex(self, 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.
:type index: int
:type val: int
:rtype: void
"""
if self._inRange(index, 0, self.size):
curr = self._getNodeAt(index)
new = ListNode(val)
new.next = curr
new.prev = curr.prev
curr.prev.next = new
curr.prev = new
self.size += 1


def deleteAtIndex(self, index):
"""
Delete the index-th node in the linked list, if the index is valid.
:type index: int
:rtype: void
"""
if self._inRange(index, 0, self.size-1):
curr = self._getNodeAt(index)
curr.prev.next = curr.next
curr.next.prev = curr.prev
self.size -= 1

def _inRange(self, index, start, end):
return start <= index <= end

def _getNodeAt(self, index):
if index < self.size // 2:
curr = self.head
for _ in range(-1, index):
curr = curr.next
else:
# go backwards
curr = self.tail
for _ in range(self.size, index, -1):
curr = curr.prev
return curr