2181. Merge Nodes in Between Zeros
Medium
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 a singly-linked list.
* class ListNode {
* public $val = 0;
* public $next = null;
* function __construct($val = 0, $next = null) {
* $this->val = $val;
* $this->next = $next;
* }
* }
*/
class Solution {
/**
* @param ListNode $head
* @return ListNode
*/
function mergeNodes($head) {
$dummy = new ListNode(0);
$current = $dummy;
$sum = 0;
// Skip the first zero node
$head = $head->next;
while ($head !== null) {
if ($head->val == 0) {
$current->next = new ListNode($sum);
$current = $current->next;
$sum = 0;
} else {
$sum += $head->val;
}
$head = $head->next;
}
return $dummy->next;
}
}
