<p>Given the <code>head</code> of a linked list, we repeatedly delete consecutive sequences of nodes that sum to <code>0</code> until there are no such sequences.</p>

<p>After doing so, return the head of the final linked list.&nbsp; You may return any such answer.</p>

<p>&nbsp;</p>
<p>(Note that in the examples below, all sequences are serializations of <code>ListNode</code> objects.)</p>

<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> head = [1,2,-3,3,1]
<strong>Output:</strong> [3,1]
<strong>Note:</strong> The answer [1,2,1] would also be accepted.
</pre>

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

<pre>
<strong>Input:</strong> head = [1,2,3,-3,4]
<strong>Output:</strong> [1,2,4]
</pre>

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

<pre>
<strong>Input:</strong> head = [1,2,3,-3,-2]
<strong>Output:</strong> [1]
</pre>

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

<ul>
	<li>The given linked list will contain between <code>1</code> and <code>1000</code> nodes.</li>
	<li>Each node in the linked list has <code>-1000 &lt;= node.val &lt;= 1000</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 removeZeroSumSublists(self, head: Optional[ListNode]) -> Optional[ListNode]:
        """
        A sublist will sum to 0 if
        you can split the sublist in 
        2 parts that when added sum 
        up to 0
        """
        cur = head
        arr = []
        while cur:
            if cur.val != 0:
                arr.append(cur.val)
            cur = cur.next
        
        def delete():
            nonlocal arr
            for i in range(len(arr)):
                cur = arr[i]
                for j in range(i + 1, len(arr)):
                    cur += arr[j]
                    if cur == 0:
                        arr = arr[:i] + arr[j + 1:]
                        return True
            return False
        
        while True:
            if not delete():
                break
        
        dummy = ListNode()
        cur = dummy
        for n in arr:
            cur.next = ListNode(n)
            cur = cur.next

        return dummy.next

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