1. Topic
Source of the topic: https://leetcode-cn.com/problems/next-greater-node-in-linked-list/
Given a linked list with the head node headas the first node. The nodes in the linked list are respectively numbered: node_1, node_2, node_3, ....
Each node may have the next larger value ( next larger value ): for node_i, if it next_larger(node_i)is node_j.val, then there is j > iand node_j.val > node_i.val, but jis the smallest possible option. If no such exists j, then the next greater value is 0.
Returns an array of integer answers answerwhere answer[i] = next_larger(node_{i+1}).
Note: In the example below, [2,1,5]an input such as this (not an output) is a serialized representation of a linked list with the value 2 for the head node, 1 for the second node, and 5 for the third node.
- 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]
- hint:
For each node in the linked list, 1 <= node.val <= 10^9
The length of the given list is in the range [0, 10000]
Two, solution
Solve using a monotonic stack:
For specific ideas, see the author's diagram and code comments
The diagram is as follows:
the code
func nextLargerNodes(head *ListNode) []int {
//首先遍历链表，将链表的值入数组arr
var arr []int
for head != nil{
arr = append(arr, head.Val)
head = head.Next
}
//定义最后的返回数组result和栈stack,长度为链表也就是arr的长度
//注意：go语言中int类型默认值为0，[]int，即默认为[0]
result, stack := make([]int, len(arr)), make([]int, len(arr))
//遍历数组arr
for index := 0; index < len(arr); index++{
//当栈为空的时候将数组下标入栈
//之所以入下标而不直接入值是为了在result中找到并修改对应index下标的位置
if len(stack) == 0{
stack = append(stack, index)
}else{
//如果栈中已有元素，那么开始循环比较
//若此时的index对应的值大于stack栈顶元素坐标对应的值，就将栈顶元素出栈
for len(stack) > 0 && arr[index] > arr[stack[len(stack)-1]]{
//将栈顶元素坐标赋给pop，用于出栈
pop := len(stack) - 1
//并改变result中对应栈顶元素坐标位置的值为index对应值（也就是当前最大值）
result[stack[pop]] = arr[index]
//弹出栈顶元素
stack = stack[:pop]
}
//循环完毕，直至栈中没有比index对应值更大的数，就将该最大值坐标index入栈
stack = append(stack, index)
}
}
//返回result
return result
}