LeetCode Q 1019 - 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_i, next_larger(node_i) is the node_j.val such that j > i, node_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]

Note:

1 <= node.val <= 10^9 for each node in the linked list.
The given list has length in the range [0, 10000].

Solution

Code

public int[] nextLargerNodes(ListNode head) {
  
  int len = 0;
  
  ListNode curr = head;
  while (curr != null) {
    len++; curr = curr.next;
  }

  Stack<int[]> stack = new Stack<>();

  int[] res = new int[len];
  int index = 0;

  while (head != null) {
    while (!stack.isEmpty() && stack.peek()[0] < head.val) {
      res[stack.pop()[1]] = head.val;
    }

    stack.push(new int[]{head.val, index++});
    head = head.next;
  }

  return res;
}

Reprint policy