leetcode1019 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]

 """
 class ListNode:
     def __init__(self, x):
         self.val = x
         self.next = None
 class Solution1(object):
     def nextLargerNodes(self, head):
         if not head.next:
             return head if head.val != 0 else None  #判断头节点是为否为空
         nums = []
         p = head
         while(p):   #将链表转为数组
             nums.append(p.val)
             p = p.next
         stack = []  #创建一个栈
         res = [0] * len(nums)  #保存结果的数组
         #bug 0 没有加[0]
         for i, n in enumerate(nums):   #
             while stack and nums[stack[-1]] < n: #！！！单调递减的栈，栈中存的是索引
                 res[stack.pop()] = n
             stack.append(i)   #将索引压栈
         return res
 """
 runtime error
 单调递减(增)栈,是一个非常普遍的解法
 传送门https://blog.csdn.net/qq_17550379/article/details/86519771
 """

 """
 我们也可以不将链表中的元素存放到一个list里面，
 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，
 所以我们可以在stack中记录(index, val)数据对。
 """
 class Solution2(object):
     def nextLargerNodes(self, head):
         res, stack = list(), list()
         while head:
             while stack and stack[-1][1] < head.val:
                 res[stack.pop()[0]] = head.val
             stack.append([len(res), head.val])
             res.append(0)
             head = head.next
         return res