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 thenode_j.val
such thatj > i
,node_j.val > node_i.val
, andj
is the smallest possible choice. If such aj
does not exist, the next larger value is0
.Return an array of integers
answer
, whereanswer[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 <= node.val <= 10^9
for each node in the linked list.- 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
【leetcode】1019. Next Greater Node In Linked List
原文:https://www.cnblogs.com/seyjs/p/10636802.html