LeetCode - LinkedList - Next Greater Node In Linked List

We are given a linked list with head as the first node.  Let's number the nodes in the list: node_1, node_2, node_3, ... etc.

Each node may have a next larger value: for node_inext_larger(node_i) is the node_j.val such that j > inode_j.val > node_i.val, and j is the smallest possible choice.  If such a j does not exist, the next larger value is 0.

Return an array of integers answer, where answer[i] = next_larger(node_{i+1}).

Note that in the example inputs (not outputs) below, arrays such as [2,1,5] represent the serialization of a linked list with a head node value of 2, second node value of 1, and third node value of 5.

 

Example 1:

Input: [2,1,5]
Output: [5,5,0]

Example 2:

Input: [2,7,4,3,5]
Output: [7,0,5,5,0]

Example 3:

Input: [1,7,5,1,9,2,5,1]
Output: [7,9,9,9,0,5,0,0]
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public int[] nextLargerNodes(ListNode head) {
        List<Integer> resultList = new ArrayList<Integer>();
        ListNode temp = head;
        while(temp != null){
            ListNode t = temp.next;
            int val = temp.val;
            boolean foundGreaterVal = false;
            while(t != null){
                if(t.val > val){
                  resultList.add(t.val);
                  foundGreaterVal = true;  
                  break;  
                }
                t = t.next;
            }
            if(foundGreaterVal == false) resultList.add(0);
            temp = temp.next;
        }
        //System.out.println(resultList);
        int[] result = resultList.stream().mapToInt(i -> i).toArray();
        //System.out.println(Arrays.toString(result));
        return result;
    }
}

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