<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]:
        cumsum = [0]
        remaining = []
        p = head
        while p is not None:
            csum = p.val + cumsum[-1]
            remaining.append(True)
            try:
                start = 0
                index = cumsum.index(csum, start)
                while not remaining[index]:
                    start = index + 1
                    index = cumsum.index(csum, start)
            except ValueError:
                index = -1
            if index >= 0:
                for i in range(index, len(remaining)):
                    remaining[i] = False
            cumsum.append(csum)
            p = p.next
        prev = None
        p = head
        for i in range(len(remaining)):
            if not remaining[i]:
                if prev is None:
                    head = p.next
                else:
                    prev.next = p.next
            else:
                prev = p
            p = p.next
        return head
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