1019 Next Greater Node In Linked List.py

#!/usr/bin/python3
"""
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].
"""

# Definition for singly-linked list.
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None


from typing import List


class Solution:
    def nextLargerNodes(self, head: ListNode) -> List[int]:
        """
        If input is an array, use stack from right to left. Maintain a decreasing stack

        How to make it one-pass? Maintain a stack from left to right for the element
        waiting for the next larger node
        """
        ret = []
        stk = []  # [[index, value]]
        i = 0
        cur = head
        while cur:
            while stk and stk[-1][1] < cur.val:
                idx, _ = stk.pop()
                ret[idx] = cur.val

            stk.append([i, cur.val])
            ret.append(0)
            cur = cur.next
            i += 1

        return ret