LeetCode 1019. Next Greater Node In Linked List

Posted 2021-12-18 dylan-java-nyc

原题链接在这里：https://leetcode.com/problems/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.valsuch 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. 1 <= node.val <= 10^9 for each node in the linked list.
2. The given list has length in the range [0, 10000].

题解：

类似Next Greater Element I, Next Greater Element II.

利用stack记录从后向前渐渐变小的栈.

reverse list后就可以从后向前iterate list. 遇到一个点，在stack中一直pop到比自己大的点就是next greater element. 如果stack空了也没找到，就是没有.

记下结果后再把当前点push到stack中.

Time Complexity: O(n).

Space: O(n).

AC Java:

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode 
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x)  val = x; 
 7  * 
 8  */
 9 class Solution 
10     public int[] nextLargerNodes(ListNode head) 
11         if(head == null)
12             return new int[0];
13         
14         
15         if(head.next == null)
16             return new int[]0;
17         
18         
19         int len = getLength(head);
20         int [] res = new int[len];
21         head = reverse(head);
22         Stack<Integer> stk = new Stack<Integer>();
23         
24         for(int i = 0; i<len; i++)
25             while(!stk.isEmpty() && head.val>=stk.peek())
26                 stk.pop();
27             
28             
29             res[len-i-1] = stk.isEmpty() ? 0 : stk.peek();
30             stk.push(head.val);
31             head = head.next;
32         
33         
34         return res;
35     
36     
37     private ListNode reverse(ListNode head)
38         if(head == null || head.next == null)
39             return head;
40         
41         
42         ListNode tail = head;
43         ListNode cur = head;
44         ListNode pre;
45         ListNode temp;
46         while(tail.next != null)
47             pre = cur;
48             cur = tail.next;
49             temp = cur.next;
50             cur.next = pre;
51             tail.next = temp;
52         
53         
54         return cur;
55     
56     
57     private int getLength(ListNode head)
58         int res = 0;
59         while(head != null)
60             head = head.next;
61             res++;
62         
63         
64         return res;
65     
66