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 <= node.val <= 10^9
for each node in the linked list.- The given list has length in the range
[0, 10000]
.
给出一个以头节点 head
作为第一个节点的链表。链表中的节点分别编号为:node_1, node_2, node_3, ...
。
每个节点都可能有下一个更大值(next larger value):对于 node_i
,如果其 next_larger(node_i)
是 node_j.val
,那么就有 j > i
且 node_j.val > node_i.val
,而 j
是可能的选项中最小的那个。如果不存在这样的 j
,那么下一个更大值为 0
。
返回整数答案数组 answer
,其中 answer[i] = next_larger(node_{i+1})
。
注意:在下面的示例中,诸如 [2,1,5]
这样的输入(不是输出)是链表的序列化表示,其头节点的值为 2,第二个节点值为 1,第三个节点值为 5 。
示例 1:
输入:[2,1,5] 输出:[5,5,0]
示例 2:
输入:[2,7,4,3,5] 输出:[7,0,5,5,0]
示例 3:
输入:[1,7,5,1,9,2,5,1] 输出:[7,9,9,9,0,5,0,0]
提示:
- 对于链表中的每个节点,
1 <= node.val <= 10^9
- 给定列表的长度在
[0, 10000]
范围内
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * public var val: Int 5 * public var next: ListNode? 6 * public init(_ val: Int) { 7 * self.val = val 8 * self.next = nil 9 * } 10 * } 11 */ 12 class Solution { 13 var ret:[Int] = [Int]() 14 var d:[(Int,Int)] = [(Int,Int)]() 15 var ans:[Int] = [Int]() 16 17 func nextLargerNodes(_ head: ListNode?) -> [Int] { 18 var head = head 19 while(head != nil) 20 { 21 ret.append(head!.val) 22 head = head?.next 23 } 24 for i in stride(from:ret.count - 1,through:0,by:-1) 25 { 26 while(d.count != 0 && d.first!.0 <= ret[i]) 27 { 28 29 d.removeFirst() 30 } 31 if d.count != 0 32 { 33 ans.append(d.first!.0) 34 } 35 else 36 { 37 ans.append(0) 38 } 39 d.insert((ret[i],i),at:0) 40 } 41 return ans.reversed() 42 } 43 }