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: List

Skip the first zero, replace every zero node with the sum of values of its previous nodes.

Time complexity: O(n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  ListNode* mergeNodes(ListNode* head) {
    ListNode dummy;    
    head = head->next;
    for (ListNode* prev = &dummy; head; head = head->next) {
      int sum = 0;
      while (head->val != 0) {
        sum += head->val;
        head = head->next;
      }
      prev->next = head;
      head->val = sum;
      prev = head;      
    }
    return dummy.next;
  }
};

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