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.