## 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 = 6Output:[1,2,3,4,5]

**Example 2:**

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

**Example 3:**

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

**Constraints:**

- The number of nodes in the list is in the range
`[0, 10`

.^{4}] `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).