LeetCode in Kotlin
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
import com_github_leetcode.ListNode
/*
* Example:
* var li = ListNode(5)
* var v = li.`val`
* Definition for singly-linked list.
* class ListNode(var `val`: Int) {
* var next: ListNode? = null
* }
*/
class Solution {
fun mergeNodes(head: ListNode): ListNode? {
var temp = head.next
var slow = head
var sum = 0
var fast = temp
while (temp != null) {
if (temp.`val` == 0) {
temp.`val` = sum
sum = 0
slow.next = fast!!.next
slow = temp
fast = fast.next
} else {
sum += temp.`val`
fast = temp
}
temp = temp.next
}
return head.next
}
}