题目如下：

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. 1 <= node.val <= 10^9 for each node in the linked list.
2. The given list has length in the range [0, 10000].

解题思路：本题是找出离自己最近的大于自己的数，和以前做过的 【leetcode】84. Largest Rectangle in Histogram 非常相似，区别在于【leetcode】84. Largest Rectangle in Histogram 是找出最近的比自己小的数，但是原理是一样的。我的解法就是把链表转成list，然后参照【leetcode】84. Largest Rectangle in Histogram的解法。

代码如下：

# 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]

        """

        val = []

        while head != None:

            val.append(head.val)

            head = head.next

        res = [0] * len(val)

        for i in range(len(val)-2,-1,-1):

            next = i+1

            while res[next] != 0 and val[i] >= val[next] :

                next = res[next]

            if val[i] >= val[next]:

                res[i] = 0

            else:

                res[i] = next

        #print res

        for i in range(len(res)):

            if res[i] == 0:

                continue

            res[i] = val[res[i]]

        return res