LeetCode 1019. Next Greater Node In Linked List

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:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public int[] nextLargerNodes(ListNode head) {
        if(head == null){
            return new int[0];
        }
        
        if(head.next == null){
            return new int[]{0};
        }
        
        int len = getLength(head);
        int [] res = new int[len];
        head = reverse(head);
        Stack<Integer> stk = new Stack<Integer>();
        
        for(int i = 0; i<len; i++){
            while(!stk.isEmpty() && head.val>=stk.peek()){
                stk.pop();
            }
            
            res[len-i-1] = stk.isEmpty() ? 0 : stk.peek();
            stk.push(head.val);
            head = head.next;
        }
        
        return res;
    }
    
    private ListNode reverse(ListNode head){
        if(head == null || head.next == null){
            return head;
        }
        
        ListNode tail = head;
        ListNode cur = head;
        ListNode pre;
        ListNode temp;
        while(tail.next != null){
            pre = cur;
            cur = tail.next;
            temp = cur.next;
            cur.next = pre;
            tail.next = temp;
        }
        
        return cur;
    }
    
    private int getLength(ListNode head){
        int res = 0;
        while(head != null){
            head = head.next;
            res++;
        }
        
        return res;
    }
}