LeetCode-in-Java
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;
/*
* 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) {
ListNode temp = head.next;
ListNode slow = head;
int sum = 0;
ListNode 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;
}
}