题目描述：
LeetCode 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 <= 10^9for each node in the linked list.- The given list has length in the range
[0, 10000].
题目大意：
给定链表，返回链表每个元素之后距离最近且值更大的元素。
解题思路：
栈（Stack）
记结果数组为ans
遍历链表，用栈维护所有不小于当前元素的数字及下标。
将栈顶所有小于当前元素head.val的数字tv及下标ti弹出，令其ans[ti] = head.val
Python代码：
# 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]
"""
stack = []
ans = []
cnt = 0
while head:
ans.append(0)
while stack and stack[-1][0] < head.val:
tv, ti = stack.pop()
ans[ti] = head.val
stack.append((head.val, cnt))
cnt += 1
head = head.next
return ans
本文链接：http://bookshadow.com/weblog/2019/03/31/leetcode-next-greater-node-in-linked-list/
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