2181. Merge Nodes in Between Zeros

Problem Statement

2181. Merge Nodes in Between Zeros

Medium

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.

Java

Source file

/**
 * Definition for singly-linked list.
 * public class ListNode {
 * int val;
 * ListNode next;
 * ListNode() {}
 * ListNode(int val) { this.val = val; }
 * ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeNodes(ListNode head) {
        ListNode curr = head.next, modder = head;
        int runningSum = 0;
        while (curr != null) {
            if (curr.val == 0) {
                modder.next = new ListNode(runningSum, null);
                modder = modder.next;
                runningSum = 0;
            } else
                runningSum += curr.val;
            curr = curr.next;
        }
        return head.next;
    }
}