203. Remove Linked List Elements

Problem

Remove all elements from a linked list of integers that have value val.

Example:

Input:  1->2->6->3->4->5->6, val = 6
Output: 1->2->3->4->5

Thought Process

• the regular case is simple.

e.g. to delete 6 below, we can use a node(delnode) to store 6’s next node

2->6->3

then 2’s next points to delnode’s next node, then 6’s next points to None

• The tricky part is the edge case:
• if the deleted node is None, then return None
• if the deleted node is head, then delnode = head, delnode.next = head.next, then remove head
• it’s possible that after nth deletion, the new head might still be the target node to be removed. Here we need a loop
• also after deleting nodes, it’s possible that the linked list is empty, then return None

White Board

Below is the white board:

Code

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

class Solution:
def removeElements(self, head, val):
if head is None:

while(head is not None and head.val == val):
delNode = None

if head is None:

while(curr.next is not None):
if (curr.next.val == val):
delNode = curr.next
curr.next = delNode.next
delNode = None
else:
curr = curr.next

Complexity

• Time complexity is O(n)

Optimization

There are just way too many edge case evaluations in the above solution. One good solution is to introduce a dummy head whose next points to head

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

class Solution:
def removeElement(self, head, val):

while curr.next is not None:
if curr.next.val == val:
delNode = curr.next
curr.next = delNode.next
delNode.next = None
else:
curr = curr.next