[LeetCode] Next Greater Node In Linked List

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].

解题思路

• 思路1：对于每个节点，遍历其后的节点，找到第一个大于它的节点。
• 思路2：先把链表转为数组，然后借助栈来实现。当栈为空或者当前数组元素不大于栈顶指针所指向的元素时，把当前元素的下表入栈，否则依次把栈顶指针出站，并把相应位置的值设为数组的当前元素。

实现代码

实现1：

// Runtime: 1115 ms, faster than 5.12% of Java online submissions for Next Greater Node In Linked List.
// Memory Usage: 42.8 MB, less than 100.00% of Java online submissions for Next Greater Node In Linked List.
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public int[] nextLargerNodes(ListNode head) {
        List result = new ArrayList<>();
        while (head != null) {
            ListNode cur = head;
            while (head != null && head.val <= cur.val) {
                head = head.next;
            }

            if (head != null) {
                result.add(head.val);
            } else {
                result.add(0);
            }
            head = cur.next;
        }
        
        int[] nums = new int[result.size()];
        for (int i = 0; i < result.size(); i++) {
            nums[i] = result.get(i);
        }
        return nums;
    }
}

实现2：

// Runtime: 39 ms, faster than 64.33% of Java online submissions for Next Greater Node In Linked List.
// Memory Usage: 42.6 MB, less than 100.00% of Java online submissions for Next Greater Node In Linked List.
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public int[] nextLargerNodes(ListNode head) {
        List nums = new ArrayList<>();
        for (ListNode node = head; node != null; node = node.next) {
            nums.add(node.val);
        }
        
        int[] res = new int[nums.size()];
        Stack stack = new Stack<>();
        for (int i = 0; i < nums.size(); i++) {
            while (!stack.empty() && nums.get(i) > nums.get(stack.peek())) {
                res[stack.pop()] = nums.get(i);
            }
            stack.push(i);
        }
        return res;
    }
}