<p>You are given the <code>head</code> of a linked list, which contains a series of integers <strong>separated</strong> by <code>0</code>&#39;s. The <strong>beginning</strong> and <strong>end</strong> of the linked list will have <code>Node.val == 0</code>.</p>

<p>For <strong>every </strong>two consecutive <code>0</code>&#39;s, <strong>merge</strong> all the nodes lying in between them into a single node whose value is the <strong>sum</strong> of all the merged nodes. The modified list should not contain any <code>0</code>&#39;s.</p>

<p>Return <em>the</em> <code>head</code> <em>of the modified linked list</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/02/02/ex1-1.png" style="width: 600px; height: 41px;" />
<pre>
<strong>Input:</strong> head = [0,3,1,0,4,5,2,0]
<strong>Output:</strong> [4,11]
<strong>Explanation:</strong> 
The above figure represents the given linked list. The modified list contains
- The sum of the nodes marked in green: 3 + 1 = 4.
- The sum of the nodes marked in red: 4 + 5 + 2 = 11.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/02/02/ex2-1.png" style="width: 600px; height: 41px;" />
<pre>
<strong>Input:</strong> head = [0,1,0,3,0,2,2,0]
<strong>Output:</strong> [1,3,4]
<strong>Explanation:</strong> 
The above figure represents the given linked list. The modified list contains
- The sum of the nodes marked in green: 1 = 1.
- The sum of the nodes marked in red: 3 = 3.
- The sum of the nodes marked in yellow: 2 + 2 = 4.
</pre>

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

<ul>
	<li>The number of nodes in the list is in the range <code>[3, 2 * 10<sup>5</sup>]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 1000</code></li>
	<li>There are <strong>no</strong> two consecutive nodes with <code>Node.val == 0</code>.</li>
	<li>The <strong>beginning</strong> and <strong>end</strong> of the linked list have <code>Node.val == 0</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 mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
        cur = head

        while cur.next:
            node = cur.next
            cur = cur.next
            while cur.next.val != 0:
                node.val += cur.next.val
                cur = cur.next

            cur = cur.next
            node.next = cur.next # fine even if null
        return head.next


        # cur = head
        # dummy = ListNode()
        # tail = dummy

        # while cur.next:
        #     node = ListNode()
        #     while cur.next.val != 0:
        #         node.val += cur.next.val
        #         cur = cur.next
        #     tail.next = node
        #     tail = tail.next
        #     cur = cur.next

        # return dummy.next
891