Solving Common Linked List Problems on LeetCode: Deletion, Reversal, and Custom Implementation

For removing nodes with a given value, using a sentinel node avoids handling the head as a special case. A pointer starts at the sentinel and examines each successor, unlinking any node whose data matches the target.

class Solution:
    def removeElements(self, head: Optional[ListNode], target: int) -> Optional[ListNode]:
        sentinel = ListNode(0, head)
        prev = sentinel
        while prev.next is not None:
            if prev.next.val == target:
                prev.next = prev.next.next
            else:
                prev = prev.next
        return sentinel.next

Reversing a singly linked list is most intuitive with two pointers. A previous pointer (None initially) and a current pointer (initially the head) move forward together, redirecting each node's next reference. Its critical to capture current.next before the link is overwritten.

class Solution:
    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        previous = None
        current = head
        while current:
            nxt = current.next
            current.next = previous
            previous = current
            current = nxt
        return previous

The same idea translates into a recursive helper that carries the accumulating reversed prefix. The base case returns the new head when the current node is exhausted.

class Solution:
    def reverseListRec(self, head: Optional[ListNode]) -> Optional[ListNode]:
        def helper(node, acc):
            if not node:
                return acc
            nxt = node.next
            node.next = acc
            return helper(nxt, node)
        return helper(head, None)

Implementing a custom linked list with index-based operations bneefits from a dummy head and a tracked size. When traversing to an index, the current reference must stop at the node before the point of insertion or deletion. Using a for loop over the index value makes the logic explicit and avoids off-by-one errors.

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

class MyLinkedList:
    def __init__(self):
        self._sentinel = ListNode()
        self._size = 0

    def get(self, idx: int) -> int:
        if idx < 0 or idx >= self._size:
            return -1
        current = self._sentinel.next
        for _ in range(idx):
            current = current.next
        return current.val

    def addAtHead(self, val: int) -> None:
        node = ListNode(val, self._sentinel.next)
        self._sentinel.next = node
        self._size += 1

    def addAtTail(self, val: int) -> None:
        cur = self._sentinel
        while cur.next:
            cur = cur.next
        cur.next = ListNode(val)
        self._size += 1

    def addAtIndex(self, idx: int, val: int) -> None:
        if idx < 0 or idx > self._size:
            return
        prev = self._sentinel
        for _ in range(idx):
            prev = prev.next
        node = ListNode(val, prev.next)
        prev.next = node
        self._size += 1

    def deleteAtIndex(self, idx: int) -> None:
        if idx < 0 or idx >= self._size:
            return
        prev = self._sentinel
        for _ in range(idx):
            prev = prev.next
        if prev.next:
            prev.next = prev.next.next
        self._size -= 1

Tags: Linked List LeetCode Data Structures algorithms python

Posted on Sun, 09 Aug 2026 16:49:44 +0000 by mrgym