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

# @lc code=start
using LeetCode

function next_larger_nodes(head::ListNode{Int})
    head = reverse_list(head)
    stk = Int[]
    res = Int[]
    while !isnothing(head)
        while !isempty(stk) && stk[end] <= val(head)
            pop!(stk)
        end
        pushfirst!(res, isempty(stk) ? 0 : stk[end])
        push!(stk, val(head))
        head = next(head)
    end
    return res
end
# @lc code=end

next_larger_nodes (generic function with 1 method)

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