1019. Next Greater Node in Linked List

# 1019. Next Greater Node In Linked List#

## 题目 #

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.

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].

## 代码 #

``````
package leetcode

/**
* type ListNode struct {
*     Val int
*     Next *ListNode
* }
*/
// 解法一 单调栈
res, indexes, nums := make([]int, 0), make([]int, 0), make([]int, 0)
for p != nil {
nums = append(nums, p.Val)
p = p.Next
}
for i := 0; i < len(nums); i++ {
res = append(res, 0)
}
for i := 0; i < len(nums); i++ {
num := nums[i]
for len(indexes) > 0 && nums[indexes[len(indexes)-1]] < num {
index := indexes[len(indexes)-1]
res[index] = num
indexes = indexes[:len(indexes)-1]
}
indexes = append(indexes, i)
}
return res
}

``````

Sep 6, 2020