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 * 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.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var mergeNodes = function(head) {
    var newHead = new ListNode();
    var resNow = newHead;
    var now = head;
    while (now) {
        if (now.val === 0 && now.next) {
            resNow.next = new ListNode();
            resNow = resNow.next;
        } else if (now.val !== 0) {
            resNow.val += now.val;
        }
        now = now.next;
    }
    return newHead.next;
};

Explain:

nope.

Complexity:

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