2181. Merge Nodes in Between Zeros
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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.
Hints
Hint 1
How can you use two pointers to modify the original list into the new list?
Hint 2
Have a pointer traverse the entire linked list, while another pointer looks at a node that is currently being modified.
Hint 3
Keep on summing the values of the nodes between the traversal pointer and the modifying pointer until the former comes across a ‘0’. In that case, the modifying pointer is incremented to modify the next node.
Hint 4
Do not forget to have the next pointer of the final node of the modified list point to null.