leetcode1019 Next Greater Node In Linked List

 1 """
 2 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.
 3
 4 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.
 5
 6 Return an array of integers answer, where answer[i] = next_larger(node_{i+1}).
 7
 8 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.
 9
10
11
12 Example 1:
13
14 Input: [2,1,5]
15 Output: [5,5,0]
16
17 Example 2:
18
19 Input: [2,7,4,3,5]
20 Output: [7,0,5,5,0]
21
22 Example 3:
23
24 Input: [1,7,5,1,9,2,5,1]
25 Output: [7,9,9,9,0,5,0,0]
26
27 """
28 class ListNode:
29     def __init__(self, x):
30         self.val = x
31         self.next = None
32 class Solution1(object):
33     def nextLargerNodes(self, head):
34         if not head.next:
35             return head if head.val != 0 else None  #判断头节点是为否为空
36         nums = []
37         p = head
38         while(p):   #将链表转为数组
39             nums.append(p.val)
40             p = p.next
41         stack = []  #创建一个栈
42         res = [0] * len(nums)  #保存结果的数组
43         #bug 0 没有加[0]
44         for i, n in enumerate(nums):   #
45             while stack and nums[stack[-1]] < n: #！！！单调递减的栈，栈中存的是索引
46                 res[stack.pop()] = n
47             stack.append(i)   #将索引压栈
48         return res
49 """
50 runtime error
51 单调递减(增)栈,是一个非常普遍的解法
52 传送门https://blog.csdn.net/qq_17550379/article/details/86519771
53 """
54
55 """
56 我们也可以不将链表中的元素存放到一个list里面，
57 而是直接去处理链表，不过对于链表我们无法快速索引具体位置的值，
58 所以我们可以在stack中记录(index, val)数据对。
59 """
60 class Solution2(object):
61     def nextLargerNodes(self, head):
62         res, stack = list(), list()
63         while head:
64             while stack and stack[-1][1] < head.val:
65                 res[stack.pop()[0]] = head.val
66             stack.append([len(res), head.val])
67             res.append(0)
68             head = head.next
69         return res

原文地址：https://www.cnblogs.com/yawenw/p/12250489.html