1019.py

'''
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(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def nextLargerNodes(self, head):
        """
        :type head: ListNode
        :rtype: List[int]
        """
        result = []
        while head:
            result.append(head.val)
            head = head.next
        
        stack = [result[-1]]
        ans = [0]
        for val in range(len(result)-2, -1, -1):
            if result[val] < stack[-1]:
                ans.append(stack[-1])
            else:
                while stack and stack[-1] <= result[val]:
                    stack.pop()
                if stack:
                    ans.append(stack[-1])
                else:
                    ans.append(0)
            stack.append(result[val])
        return ans[::-1]