LeetCode 1019 - Next Greater Node In Linked List

https://leetcode.com/problems/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. 1 <= node.val <= 10^9 for each node in the linked list.
2. The given list has length in the range [0, 10000]

X. https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/265508/JavaC%2B%2BPython-Next-Greater-Element

Transform the linked list to an arraylist,
then it's a normal "next larger element" problem,
solved by stack.

    public int[] nextLargerNodes(ListNode head) {
        ArrayList<Integer> A = new ArrayList<>();
        for (ListNode node = head; node != null; node = node.next)
            A.add(node.val);
        int[] res = new int[A.size()];
        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < A.size(); ++i) {
            while (!stack.isEmpty() && A.get(stack.peek()) < A.get(i))
                res[stack.pop()] = A.get(i);
            stack.push(i);
        }
        return res;
    }

X. https://leetcode.com/problems/next-greater-node-in-linked-list/discuss/265687/Java-simple-one-pass-solution-O(n)-time-complexity

int[] res;

public int[] nextLargerNodes(ListNode head) {
    calNode(head, 0, new Stack<>());
    return res;
}

public void calNode(ListNode node, int index, Stack<Integer> stack) {
    if(node == null) {
        res = new int[index];
        return;
    }
    calNode(node.next, index + 1, stack);
    while(!stack.empty() && stack.peek() <= node.val)
        stack.pop();
    res[index] = stack.empty() ? 0 : stack.peek();
    stack.push(node.val);
}