題目

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.

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].

題目大意

給出一個鏈表，要求找出每個結點後面比該結點值大的第一個結點，如果找不到這個結點，則輸出 0 。

解題思路

這一題和第 739 題、第 496 題、第 503 題類似。也有 2 種解題方法。先把鏈表中的數字存到數組中，整道題的思路就和第 739 題完全一致了。普通做法就是 2 層循環。優化的做法就是用單調棧，維護一個單調遞減的棧即可。

�棧即可。

參考代碼

package leetcode

import (
	"github.com/halfrost/LeetCode-Go/structures"
)

// ListNode define
type ListNode = structures.ListNode

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
// 解法一 單調棧
func nextLargerNodes(head *ListNode) []int {
	res, indexes, nums := make([]int, 0), make([]int, 0), make([]int, 0)
	p := head
	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
}