# Merge Nodes in Between Zeros

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.
* 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) {
int sum=0;
ListNode curr = head, temp = head.next, lastcheck = null;
while(curr.next!=null)
{
sum=0;
while(temp.val!=0)
{
sum+=temp.val;
temp = temp.next;
}
curr.val = sum;
curr.next = temp;
lastcheck = curr;
curr = curr.next;
temp = curr.next;
}
lastcheck.next = null;
return head;
}
}
```