題目如下：

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].

解題思路：本題是找出離自己最近的大於自己的數，和以前做過的 【leetcode】84. Largest Rectangle in Histogram 非常相似，區別在於【leetcode】84. Largest Rectangle in Histogram 是找出最近的比自己小的數，但是原理是一樣的。我的解法就是把鏈表轉成list，然後參照【leetcode】84. Largest Rectangle in Histogram的解法。

代碼如下：

# 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

【leetcode】1019. Next Greater Node In Linked List