The O(n) Club: Merge k Sorted Lists—Because Your Timeline Isn’t Going to Sort Itself
⚡ TL;DR
Got k sorted linked lists and the urge to merge? Forget merging one-at-a-time—unless you like being mocked by your CPU. Use a heap (PriorityQueue) for O(n log k) speed, or go full merge sort with divide & conquer. Your code, your coffee, your call:
// Heap-based merge in Java PriorityQueue<ListNode> heap = new PriorityQueue<>((a, b) -> a.val - b.val); for (ListNode node : lists) if (node != null) heap.offer(node); ListNode dummy = new ListNode(0), tail = dummy; while (!heap.isEmpty()) { ListNode node = heap.poll(); tail.next = node; tail = node; if (node.next != null) heap.offer(node.next); } return dummy.next;
🧠 How Devs Usually Mess This Up
Here’s what most of us (yes, even you, reader) try first: We merge the first two lists, then toss in the third, then the fourth, and so on. What could possibly go wrong? Well, by the end your runtime is gasping for air at O(kn) and you’re wondering why even bubble sort is laughing at you. Or you rush to PriorityQueue and forget: “Hey Java, do you like ListNode?” Java squints and yells: ‘I can’t compare these things!’ Heap tantrum ensues. Moral: sequential merging is slow, and heaps need a Comparator—otherwise, it’s compilation carnage.
🔍 What’s Actually Going On
This isn’t just a coding question—it’s a metaphor for life in the modern software age: everyone streams their own feed, and you (the ambitious CPU of social aggregation) must blend them into a single stream of coherent, well-ordered brilliance. The trick? Always pull the smallest current element (you know, like picking the best donut from each box) until they’re all gone.
- Min-Heap Magic: Toss the head of each list into a PriorityQueue. Every time you pull the smallest, push its successor. Because at most k elements sit in the heap, operations stay fast. (O(n log k); that n is total nodes.)
- Divide & Conquer (a.k.a. Merge Sort for grownups): Pair up lists, merge each pair, halve the problem, and repeat until only one list reigns. Great if you think recursively (or want to teach your interviewer you’re a merge-sort snob).
Both get the job done way quicker than the vanilla O(kn) shuffle.
🛠️ PseudoCode
- Start with an empty
PriorityQueue<ListNode>
(a.k.a. your little heap of happiness). - Push the head of every non-empty list into your heap, with a Comparator that compares
val
. - Set up a
dummy
node (because Java likes a boring starting anchor). - While the heap isn’t empty:
- Pop the smallest node off the heap.
- Attach it at the end of your output list.
- If that node has a
next
, toss it into the heap for future consideration.
- Once the heap is empty, return
dummy.next
—otherwise, you’ll hand back a lonely anchor.
💻 The Code
// Definition for singly-linked list.
class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}
import java.util.PriorityQueue;
public class Solution {
public ListNode mergeKLists(ListNode[] lists) {
PriorityQueue<ListNode> heap = new PriorityQueue<>((a, b) -> a.val - b.val);
for (ListNode node : lists) {
if (node != null) heap.offer(node);
}
ListNode dummy = new ListNode(0);
ListNode tail = dummy;
while (!heap.isEmpty()) {
ListNode minNode = heap.poll();
tail.next = minNode;
tail = minNode;
if (minNode.next != null) heap.offer(minNode.next);
}
return dummy.next;
}
}
⚠️ Pitfalls, Traps & Runtime Smacks
- Comparator Faceplant: Forget to add a Comparator, and Java has a meltdown. Just do it!
- Null Lists: Some lists in
lists[]
will benull
. If you heapify those, expectNullPointerException
and existential questions. - Heap Size Worries: Maximum heap size is k, not n. If k is huge, buy a new laptop. Otherwise, you’re fine.
- Space & Time: Heap and D&C both rock O(n log k) time, O(k) space (for heap); sequential merge is so O(kn) it’s basically retro.
🧠 Interviewer Brain-Tattoo
“If you merge k lists sequentially, may your next interview question be an n-queens backtracking brute-force marathon. Use a heap—keep your timelines, and sanity, orderly.”