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

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.

Author: Gowsika23

2181. Merge Nodes in Between Zeros

@@ -0,0 +1,36 @@
+/**
+ * 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; }
+ * }
+ */
+public class Solution {
+
+    public ListNode mergeNodes(ListNode head) {
+        // Initialize a sentinel/dummy node with the first non-zero value.
+        ListNode current = head.next;
+        ListNode nextSum = current;
+
+        while (nextSum != null) {
+            int sum = 0;
+            // Find the sum of all nodes until you encounter a 0.
+            while (nextSum.val != 0) {
+                sum += nextSum.val;
+                nextSum = nextSum.next;
+            }
+
+            // Assign the sum to the current node's value.
+            current.val = sum;
+            // Move nextSum to the first non-zero value of the next block.
+            nextSum = nextSum.next;
+            // Move current also to this node.
+            current.next = nextSum;
+            current = current.next;
+        }
+        return head.next;
+    }
+}