Merge Nodes in Between Zeros

#leetcode #dsa #theabbie

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.

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]:
        if not head:
            return head
        head = head.next
        curr = head
        total = 0
        while curr:
            total += curr.val
            if curr.val == 0:
                head = ListNode(val = total)
                head.next = self.mergeNodes(curr)
                return head
            curr = curr.next