LeetCode #1019 — MEDIUM

Next Greater Node In Linked List

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given the head of a linked list with n nodes.

For each node in the list, find the value of the next greater node. That is, for each node, find the value of the first node that is next to it and has a strictly larger value than it.

Return an integer array answer where answer[i] is the value of the next greater node of the i^th node (1-indexed). If the i^th node does not have a next greater node, set answer[i] = 0.

Example 1:

Input: head = [2,1,5]
Output: [5,5,0]

Example 2:

Input: head = [2,7,4,3,5]
Output: [7,0,5,5,0]

Constraints:

The number of nodes in the list is n.
1 <= n <= 10⁴
1 <= Node.val <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given the head of a linked list with n nodes. For each node in the list, find the value of the next greater node. That is, for each node, find the value of the first node that is next to it and has a strictly larger value than it. Return an integer array answer where answer[i] is the value of the next greater node of the ith node (1-indexed). If the ith node does not have a next greater node, set answer[i] = 0.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Linked List · Stack

Example 1

[2,1,5]

Example 2

[2,7,4,3,5]

Step 02

Core Insight

What unlocks the optimal approach

We can use a stack that stores nodes in monotone decreasing order of value. When we see a node_j with a larger value, every node_i in the stack has next_larger(node_i) = node_j .

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1019: Next Greater Node In Linked List
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public int[] nextLargerNodes(ListNode head) {
        List<Integer> nums = new ArrayList<>();
        for (; head != null; head = head.next) {
            nums.add(head.val);
        }
        Deque<Integer> stk = new ArrayDeque<>();
        int n = nums.size();
        int[] ans = new int[n];
        for (int i = n - 1; i >= 0; --i) {
            while (!stk.isEmpty() && stk.peek() <= nums.get(i)) {
                stk.pop();
            }
            if (!stk.isEmpty()) {
                ans[i] = stk.peek();
            }
            stk.push(nums.get(i));
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1019: Next Greater Node In Linked List
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func nextLargerNodes(head *ListNode) []int {
	nums := []int{}
	for ; head != nil; head = head.Next {
		nums = append(nums, head.Val)
	}
	stk := []int{}
	n := len(nums)
	ans := make([]int, n)
	for i := n - 1; i >= 0; i-- {
		for len(stk) > 0 && stk[len(stk)-1] <= nums[i] {
			stk = stk[:len(stk)-1]
		}
		if len(stk) > 0 {
			ans[i] = stk[len(stk)-1]
		}
		stk = append(stk, nums[i])
	}
	return ans
}

# Accepted solution for LeetCode #1019: Next Greater Node In Linked List
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def nextLargerNodes(self, head: Optional[ListNode]) -> List[int]:
        nums = []
        while head:
            nums.append(head.val)
            head = head.next
        stk = []
        n = len(nums)
        ans = [0] * n
        for i in range(n - 1, -1, -1):
            while stk and stk[-1] <= nums[i]:
                stk.pop()
            if stk:
                ans[i] = stk[-1]
            stk.append(nums[i])
        return ans

// Accepted solution for LeetCode #1019: Next Greater Node In Linked List
// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
use std::collections::VecDeque;
impl Solution {
    pub fn next_larger_nodes(head: Option<Box<ListNode>>) -> Vec<i32> {
        let mut nums = Vec::new();
        let mut current = &head;
        while let Some(node) = current {
            nums.push(node.val);
            current = &node.next;
        }

        let mut stk = VecDeque::new();
        let n = nums.len();
        let mut ans = vec![0; n];
        for i in (0..n).rev() {
            while !stk.is_empty() && stk.back().copied().unwrap() <= nums[i] {
                stk.pop_back();
            }
            if let Some(&top) = stk.back() {
                ans[i] = top;
            }
            stk.push_back(nums[i]);
        }
        ans
    }
}

// Accepted solution for LeetCode #1019: Next Greater Node In Linked List
/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function nextLargerNodes(head: ListNode | null): number[] {
    const nums: number[] = [];
    while (head) {
        nums.push(head.val);
        head = head.next;
    }
    const stk: number[] = [];
    const n = nums.length;
    const ans: number[] = Array(n).fill(0);
    for (let i = n - 1; ~i; --i) {
        while (stk.length && stk.at(-1)! <= nums[i]) {
            stk.pop();
        }
        ans[i] = stk.length ? stk.at(-1)! : 0;
        stk.push(nums[i]);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

COPY TO ARRAY

O(n) time

O(n) space

Copy all n nodes into an array (O(n) time and space), then use array indexing for random access. Operations like reversal or middle-finding become trivial with indices, but the O(n) extra space defeats the purpose of using a linked list.

IN-PLACE POINTERS

O(n) time

O(1) space

Most linked list operations traverse the list once (O(n)) and re-wire pointers in-place (O(1) extra space). The brute force often copies nodes to an array to enable random access, costing O(n) space. In-place pointer manipulation eliminates that.

Shortcut: Traverse once + re-wire pointers → O(n) time, O(1) space. Dummy head nodes simplify edge cases.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Losing head/tail while rewiring

Wrong move: Pointer updates overwrite references before they are saved.

Usually fails on: List becomes disconnected mid-operation.

Fix: Store next pointers first and use a dummy head for safer joins.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.