<p>Given the <code>head</code> of a linked list and an integer <code>val</code>, remove all the nodes of the linked list that has <code>Node.val == val</code>, and return <em>the new head</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/06/removelinked-list.jpg" style="width: 500px; height: 142px;" />
<pre>
<strong>Input:</strong> head = [1,2,6,3,4,5,6], val = 6
<strong>Output:</strong> [1,2,3,4,5]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> head = [], val = 1
<strong>Output:</strong> []
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> head = [7,7,7,7], val = 7
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the list is in the range <code>[0, 10<sup>4</sup>]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 50</code></li>
	<li><code>0 &lt;= val &lt;= 50</code></li>
</ul>


class Solution:
    def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:

        dummy = ListNode(-1)
        dummy.next = head
        current = dummy

   
        while current and current.next:
            if current.next.val == val:
              
                current.next = current.next.next
            else:
              
                current = current.next
        
       
        return dummy.next

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