Description
https://leetcode.com/problems/remove-linked-list-elements/
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 <= k <= 50
Explanation
Iterate the linked list and have a previous variable to track the previous node. If the node value equals to the value, link the previous node to the next node.
Python Solution
# 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: ListNode, val: int) -> ListNode:
dummy = ListNode(0)
dummy.next = head
prev = dummy
while head != None:
if head.val == val:
prev.next = head.next
else:
prev = head
head = head.next
return dummy.next
- Time Complexity: O(N).
- Space Complexity: O(1).