#44 Hard Time O(n log k) Space O(log k) Animated

Merge k Sorted Lists

Merge k sorted linked lists and return it as one sorted list.

LeetCode ↗

Tutor Mode Build the mental model before you memorize the code.

1 Predict
2 Trace
3 Explain
4 Implement

Pattern Pointer rewiring with saved next state

Before changing a link, save the only pointer that would otherwise be lost.

Approach: Min-Heap / Priority Queue

Click play to start

Step 0 / 0

Press play to start the visualization.

1×

Core Insight

Before changing a link, save the only pointer that would otherwise be lost.

Common Pitfall

Always store next before redirecting current.next.

Description

Merge k sorted linked lists and return it as one sorted list.

Constraints

Time Complexity: O(n log k)
Space Complexity: O(log k)

Solution — Min-Heap / Priority Queue

TimeO(n log k) SpaceO(log k)

Implementation Plan

Translate the invariant into these small, testable moves.

Initialize the pointers.
Save the next node.
Rewire the current node.
Advance both pointers until the list is consumed.

def mergeKLists(lists):
    head = ListNode(None)
    curr = head
    h = []
    
    for i, node in enumerate(lists):
        if node:
            heapq.heappush(h, (node.val, i, node))
            
    while h:
        val, i, node = heapq.heappop(h)
        curr.next = node
        curr = node
        if node.next:
            heapq.heappush(h, (node.next.val, i, node.next))
            
    return head.next

LRU Cache

Reverse Nodes in k-Group

Merge k Sorted Lists

Steps

Description

Constraints

Tags

Solution — Min-Heap / Priority Queue