2181. Merge Nodes in Between Zeros

Problem

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 * 10^5].

• 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 (Java)

/**
 * 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 temp = head.next;
        ListNode slow = head;
        int sum = 0;
        ListNode fast = temp;
        while (temp != null) {
            if (temp.val == 0) {
                temp.val = sum;
                sum = 0;
                slow.next = fast.next;
                slow = temp;
                fast = fast.next;
            } else {
                sum += temp.val;
                fast = temp;
            }
            temp = temp.next;
        }
        return head.next;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).