2181. Merge Nodes in Between Zeros

[DNPotapov]

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def mergeNodes(self, head):
        """
        :type head: Optional[ListNode]
        :rtype: Optional[ListNode]
        """
        cur = head
        res = ListNode(None)
        lst = res
        count = 0
        while cur:
            if cur.val != 0:
                count += cur.val
            else:
                lst.next = ListNode(count)
                lst = lst.next
                count = 0
            cur = cur.next
        return res.next.next

[DNPotapov]

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:

Input: head = [0,3,1,0,4,5,2,0] 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:

Input: head = [0,1,0,3,0,2,2,0] 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.