#### QUESTION:

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.
```

#### EXPLANATION:

虽然是medium的题目, 但是其实没有多难, 就是每次遇到0将前面的累加的值创建一个node链接上, 这样就可以了.

#### SOLUTION:

```
class Solution {
func mergeNodes(_ head: ListNode?) -> ListNode? {
var result:ListNode = ListNode(-1)
var tmpResult:ListNode = result
var tmp:ListNode? = head
var tmpVal:Int = 0;
while tmp != nil {
if (tmp?.val == 0) {
tmpResult.next = ListNode(tmpVal)
tmpResult = tmpResult.next!
tmpVal = 0
} else {
tmpVal += tmp!.val
}
tmp = tmp?.next
}
return result.next?.next
}
}
```