203. Remove Linked List Elements
Easy
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
# 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]:
# we can use sentinel head , this will solve removing head issue
sentinel_head = ListNode(0)
sentinel_head.next = head
prev = sentinel_head
node = head
while node is not None:
if node.val == val:
prev.next = node.next
else:
prev = node
node = node.next
return sentinel_head.next