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Merge k Sorted Lists

Hard Heaps Merge K Sorted Original on LeetCode

Merge k Sorted Lists is a hard heaps problem solved with the merge k sorted pattern. The best approach, optimal (min-heap of heads), runs in O(N log k) time and O(k) space. Below are 2 approaches in Java, from merge one list at a time up.

Problem

Given an array of k sorted linked lists, merge them into one sorted linked list and return its head.

Examples

Example 1

Input
lists = [[1, 4, 5], [1, 3, 4], [2, 6]]
Output
[1, 1, 2, 3, 4, 4, 5, 6]

Example 2

Input
lists = [[], []]
Output
[]

Constraints

  • 0 <= k <= 10^4; total nodes N up to 10^4.
  • Each list is sorted.

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

ApproachTimeSpace
Merge one list at a timeO(k · N)O(1)
Optimal (min-heap of heads)O(N log k)O(k)

1Merge one list at a time

TimeO(k · N)The growing result is re-walked for every list.
SpaceO(1)

Merge list 0 with list 1, then the result with list 2, and so on.

  1. result = null; for each list: result = mergeTwo(result, list).
Java
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        ListNode result = null;
        for (ListNode l : lists) result = merge(result, l);
        return result;
    }

    private ListNode merge(ListNode a, ListNode b) {
        ListNode dummy = new ListNode(), t = dummy;
        while (a != null && b != null) {
            if (a.val <= b.val) { t.next = a; a = a.next; } else { t.next = b; b = b.next; }
            t = t.next;
        }
        t.next = a != null ? a : b;
        return dummy.next;
    }
}

2Optimal (min-heap of heads)

TimeO(N log k)
SpaceO(k)

Put the head of each non-empty list in a min-heap. Repeatedly poll the smallest, append it, and push its next node.

  1. Offer all non-null heads.
  2. While the heap is not empty: n = poll(); tail.next = n; if n.next != null offer(n.next).
Java
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        PriorityQueue<ListNode> heap = new PriorityQueue<>((a, b) -> Integer.compare(a.val, b.val));
        for (ListNode l : lists) if (l != null) heap.offer(l);
        ListNode dummy = new ListNode(), tail = dummy;
        while (!heap.isEmpty()) {
            ListNode n = heap.poll();
            tail.next = n;
            tail = n;
            if (n.next != null) heap.offer(n.next);
        }
        return dummy.next;
    }
}

Edge cases to test

  • Empty lists inside the array
  • k = 0

Hints

Hint 1

The next node of the answer is always the smallest current head. A min-heap finds it in O(log k).

FAQ

What is the best time complexity for Merge k Sorted Lists?

Optimal (min-heap of heads) runs in O(N log k) time and O(k) extra space.

Which pattern does Merge k Sorted Lists use?

It is a heaps problem that uses the merge k sorted pattern. Other problems with the same pattern: Merge k Sorted Arrays.

Is there a brute force solution for Merge k Sorted Lists?

Yes. Merge one list at a time takes O(k · N) time and O(1) space. Merge list 0 with list 1, then the result with list 2, and so on.

Which edge cases should I test for Merge k Sorted Lists?

Empty lists inside the array; k = 0.