Solution Explanation: Merge Nodes in Between Zeros

This problem involves manipulating a linked list where nodes with value 0 act as separators between groups of nodes. The goal is to merge the nodes between each pair of 0s into a single node representing the sum of their values.

Approach: Iterative Traversal and Summation

The most efficient approach is an iterative traversal of the linked list. We maintain a running sum and create a new linked list to store the results.

1. Initialization: Create a dummy node dummy to simplify handling the head of the new list. We'll also have a tail pointer to track the end of the new list and a sum variable to accumulate values between zeros.

2. Iteration: Iterate through the input linked list starting from the node after the initial 0.

3. Summation: If the current node's value is not 0, add it to the sum.

4. Zero Encountered: If a node with value 0 is encountered:

   • Create a new node in the result list with the accumulated sum.
   • Reset the sum to 0.
   • Advance the tail pointer to the newly created node.

5. Final Node: After the loop, there might be a remaining sum. Handle this by creating a final node if necessary.

6. Return: Return the next node of the dummy node, which is the head of the modified linked list.

Time and Space Complexity

• Time Complexity: O(N), where N is the number of nodes in the input linked list. We traverse the list once.
• Space Complexity: O(M), where M is the number of zero separators in the input list. This is because we create a new linked list with at most M nodes (one for each sum). In the worst case, M could be close to N/2, but it's still linear with respect to the input size.

Code Implementation (Python)

# 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]:
        dummy = ListNode(0)  # Dummy node
        tail = dummy
        current_sum = 0
        curr = head.next  # Start from the node after the initial 0
 
        while curr:
            if curr.val != 0:
                current_sum += curr.val
            else:
                new_node = ListNode(current_sum)
                tail.next = new_node
                tail = new_node
                current_sum = 0
            curr = curr.next
 
        return dummy.next  # Return the head of the new list
 

The code in other languages (Java, C++, Go, TypeScript, Rust, C) follows a very similar structure, adapting the syntax and data structures specific to each language. The core logic of iterative traversal, summation, and new node creation remains consistent across all implementations.