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
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