Leetcode 1030. Next Greater Node In Linked List

最新推荐文章于 2024-08-09 18:00:00 发布

Simple_ R

最新推荐文章于 2024-08-09 18:00:00 发布

阅读量295

点赞数

分类专栏： LeedCode365 算法和数据结构

本文链接：https://blog.csdn.net/M_sdn/article/details/88927732

版权

LeedCode365 同时被 2 个专栏收录

68 篇文章

订阅专栏

算法和数据结构

44 篇文章

订阅专栏

本文介绍了一种算法，该算法遍历链表并找出每个节点的下一个更大的节点值。对于给定的链表，算法返回一个整数数组，数组中的每个元素对应原链表中节点的下一个更大值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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].

public int[] nextLargerNodes(ListNode head) {
        ArrayList<Integer> nodes = new ArrayList<>();
        while (head.next != null) {
            nodes.add(head.val);
            head = head.next;
        }

        nodes.add(head.val);
        int[] ans = new int[nodes.size()];
        b: for (int i = 0; i < nodes.size(); i++) {
            for (int j = i+1; j < nodes.size(); j++) {
                if (nodes.get(j) > nodes.get(i)) {
                    ans[i] = nodes.get(j);
                    continue b;
                }
            }
        }
        return ans;
    }