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.
Input: head = [1,2,6,3,4,5,6], val = 6 Output: [1,2,3,4,5]
Input: head = , val = 1 Output: 
Input: head = [7,7,7,7], val = 7 Output: 
- The number of nodes in the list is in the range
1 <= Node.val <= 50
0 <= k <= 50
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.
# 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).