<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>

# 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], target: int) -> Optional[ListNode]:
        dummy = ListNode(0)
        dummy.next = head
        temp = dummy

        while temp and temp.next:
            if temp.next.val == target:
                temp.next = temp.next.next
            else:
                temp = temp.next
        
        return dummy.next
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