<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]:
        if head is None:
            return None
        temp=head
        k=[]

        while temp:
            k.append(temp.val)
            temp=temp.next
        for i in range(len(k)):
            a=0
            for j in range(i,len(k)):
                a+=k[j]
                if a==0:
                    for l in range(i,j+1):
                        k[l]=0
                    break
        q=[]
        for i in k:
            if i!=0:
                q.append(i)
        
       
        if len(q)==0:
            return None
        l=ListNode(q[0])
        r=l
    
        for i in range(1,len(q)):
            u=ListNode(q[i])
            l.next=u
            l=l.next
        return r
74