1019. 链表中的下一个更大节点

思路：用栈记录当前较小元素的下标。 时间超越100%的python提交用户，空间超越62%的Python提交用户

 1 # Definition for singly-linked list.
 2 # class ListNode(object):
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.next = None
 6 
 7 class Solution(object):
 8     def nextLargerNodes(self, head):
 9         if not head:
10             return []
11         nums=[]
12         while head:
13             nums.append(head.val)
14             head=head.next
15         stacks,res=[],[0]*len(nums)
16         for i in range(len(nums)):
17             while stacks and nums[i]>nums[stacks[-1]]:
18                 res[stacks.pop()]=nums[i]
19             stacks.append(i)
20         return res