Leetcode: Remove element from list

Given the head of a linked list, remove all items with val. Afterwards, return the new head. To solve this problem, we need to consider three cases. The element can be in the beginning, in the middle or at the end. For the middle case, we can just define el.next = el.next.next. This also works for the end, except that we need to make sure that the next element is not None. The beginning is a little bit more difficult. Here we can work around this problem by adding a pseudo-node in front of the beginning and then always look at the next one. At the end we return new_head.next. The other corner case is if two nodes that need to be deleted are right next to each other. In this case we cannot move the pointer forward.

from typing import Optional

# Definition for singly-linked list.
class ListNode:
  def __init__(self, val=0, next=None):
    self.val = val
    self.next = next
  
  def __repr__(self) -> str:
    l = str(self.val)
    if self.next is not None:
      l += f", {self.next}"
    return l


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

    return new_head.next

if __name__ == '__main__':
  s = Solution().removeElements
  l = ListNode(1, ListNode(2, ListNode(6, ListNode(3, ListNode(4, ListNode(5, ListNode(6)))))))
  print(s(l, 6))

Runtime: 169 ms, faster than 5.14% of Python3 online submissions for Remove Linked List Elements. Memory Usage: 17.8 MB, less than 81.69% of Python3 online submissions for Remove Linked List Elements.