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Abhishek Chaudhary

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# Merge Nodes in Between Zeros

You are given the `head` of a linked list, which contains a series of integers separated by `0`'s. The beginning and end of the linked list will have `Node.val == 0`.

For every two consecutive `0`'s, merge all the nodes lying in between them into a single node whose value is the sum of all the merged nodes. The modified list should not contain any `0`'s.

Return the `head` of the modified linked list.

Example 1:

Output: [4,11]
Explanation:
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.

Example 2:

Output: [1,3,4]
Explanation:
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.

Constraints:

• The number of nodes in the list is in the range `[3, 2 * 105]`.
• `0 <= Node.val <= 1000`
• There are no two consecutive nodes with `Node.val == 0`.
• The beginning and end of the linked list have `Node.val == 0`.

SOLUTION:

``````# 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]: