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One Pass

One pointer​

Array to Linked List

Easy1 solutionanalysis1 playground

Array to Linked List​

Solutions:

Interactive Visualization

Analysis

def array_to_linked_list(arr):
head = ListNode(arr[0])
curr = head
for i in range(1, len(arr)):
curr.next = ListNode(arr[i])
curr = curr.next
return head
Easy2 solutions2 playgrounds

1290. Convert Binary Number in a Linked List to Integer​

Solutions:

Interactive Visualization

def getDecimalValue(self, head: Optional[ListNode]) -> int:
number = 0
curr = head
while curr:
number = number * 2 + curr.val
curr = curr.next
return number
Easy2 solutions1 playground

3263. Convert Doubly Linked List to Array I​

Solutions:

Interactive Visualization

def toArray(self, root: "Optional[Node]") -> List[int]:
array = []
curr = root
while curr:
array.append(curr.val)
curr = curr.next
return array
Medium2 solutions1 playground

3294. Convert Doubly Linked List to Array II​

Solutions:

Interactive Visualization

def toArray(self, node: "Optional[Node]") -> List[int]:
array = collections.deque()
curr = node
while curr:
array.appendleft(curr.val)
curr = curr.prev
curr = node.next if node else None
while curr:
array.append(curr.val)
curr = curr.next
return arrays

237. Delete Node in a Linked List

Medium1 solution1 playground

237. Delete Node in a Linked List​

Solutions:

Interactive Visualization

def deleteNode(self, node) -> None:
node.val = node.next.val
node.next = node.next.next

Search In Linked List

Basic1 solution1 playground

Search In Linked List​

Solutions:

Interactive Visualization

def searchLinkedList(self, head, x):
while head:
if head.data == x:
return True
head = head.next
return False

Is Linked List Length Even?

Basic1 solution1 playground

Is Linked List Length Even?​

Solutions:

Interactive Visualization

def isLengthEven(self, head):
length = 0
while head:
head = head.next
length += 1
return length % 2 == 0

Frequency in a Linked List

Easy1 solution1 playground

Frequency in a Linked List​

Solutions:

Interactive Visualization

def count(self, head, key):
count = 0
while head:
count += head.data == key
head = head.next
return count

Modular Node

Basic1 solution1 playground

Modular Node​

Solutions:

Interactive Visualization

def modularNode(self, head, k):
mod_node = None
curr = head
index = 1
while curr:
if index % k == 0:
mod_node = curr
curr = curr.next
index += 1
return mod_node.data if mod_node else -1

Remove Nodes​

203. Remove Linked List Elements

Easy2 solutions2 playgrounds

203. Remove Linked List Elements​

Solutions:

Interactive Visualization

def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
sentinel = ListNode(None, next=head)
current_node = sentinel
while current_node and current_node.next:
if current_node.next.val == val:
current_node.next = current_node.next.next
else:
current_node = current_node.next
return sentinel.next

83. Remove Duplicates from Sorted List

83. Remove Duplicates from Sorted List​

:::

Solutions:

Interactive Visualization

def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
current_node = head
while current_node and current_node.next:
if current_node.val == current_node.next.val:
current_node.next = current_node.next.next
else:
current_node = current_node.next
return head
Easy1 solution1 playground

1474. Delete N Nodes After M Nodes of a Linked List​

Solutions:

Interactive Visualization

def deleteNodes(self, head: Optional[ListNode], m: int, n: int) -> Optional[ListNode]:
sentinel = ListNode(None)
curr = sentinel
while curr:
for _ in range(m):
if curr:
curr = curr.next
for _ in range(n):
if curr:
curr.next = curr.next.next
return sentinel.next

Modify Nodes​

Medium1 solution1 playground

2046. Sort Linked List Already Sorted Using Absolute Values​

Solutions:

Interactive Visualization

def sortLinkedList(self, head: Optional[ListNode]) -> Optional[ListNode]:
sentinel = ListNode(None, next=head)
current_node = sentinel.next
while current_node and current_node.next:
if current_node.next.val < 0:
next_node = current_node.next
current_node.next = next_node.next
next_node.next = sentinel.next
sentinel.next = next_node
else:
current_node = current_node.next
return sentinel.next

Counter​

3063. Linked List Frequency

Easy2 solutions2 playgrounds

3063. Linked List Frequency​

Solutions:

Interactive Visualization

def frequenciesOfElements(self, head: Optional[ListNode]) -> Optional[ListNode]:
counter = Counter()
curr = head
while curr:
counter[curr.val] += 1
curr = curr.next
sentinel = ListNode(None)
curr = sentinel
for freq in counter.values():
curr.next = ListNode(freq)
curr = curr.next
return sentinel.nexts