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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.

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:

```/**
* 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 {
int sum=0;
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;
}
}```