Remove linked list elements

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
        dummy = ListNode()
        dummy.next = head
        prev, curr = dummy, head

        while curr:
            if curr.val == val:
                prev.next = curr.next
            else:
                prev = curr
            curr = curr.next

        return dummy.next

Remove Linked List Elements

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Given the head of a linked list and an integer val, remove all the nodes of the linked list that has Node.val == val, and return the new head.

 

Example 1:

Input: head = [1,2,6,3,4,5,6], val = 6
Output: [1,2,3,4,5]

Example 2:

Input: head = [], val = 1
Output: []

Example 3:

Input: head = [7,7,7,7], val = 7
Output: []

 

Constraints:

• The number of nodes in the list is in the range [0, 104].
• 1 <= Node.val <= 50
• 0 <= val <= 50