Merge k Sorted Lists

Problem Statement

You are given an array of k linked-lists lists, each linked-list is sorted in ascending order.

Merge all the linked-lists into one sorted linked-list and return it.

What to notice first

• Compare elements across k lists efficiently; brute force is O(nk² log nk).
• Divide and conquer: merge pairs, then pairs of pairs → O(nk log k).
• Min-heap: always extract the smallest element from any list → O(nk log k).
• k lists of total n nodes merged pairwise is much faster than sequential pairwise merge.

Examples

Input	Output
[[1,4,5],[1,3,4],[2,6]]	1→1→2→3→4→4→5→6
[]	null
[[]]	null

Constraints

• k == lists.length
• 0 <= k <= 10^4
• 0 <= lists[i].length <= 500
• -10^4 <= lists[i][j] <= 10^4

Real-World Applications

Where this pattern appears

• Memory allocation and garbage collection
• LRU cache implementation
• Music playlist and browser history
• Undo/Redo functionality in applications

Complexity Comparison

Approach	Time	Space
Recursive Divide & Conquer	O(nk log k)	O(log k)
Min Heap	O(nk log k)	O(k)

n = total nodes across all lists

Which approach should you learn first?

• Pick Divide & Conquer for pattern clarity — mirrors merge sort.
• Pick Min Heap for intuitive greedy selection of smallest.

Pattern Match

Divide & Conquer

O(nk log k) time · O(log k) space

Greedy Pick

Min Heap

O(nk log k) time · O(k) space

Approach 1: Recursive Divide and Conquer (Recommended)

★ Recommended

1. Recursively split lists in half.
2. Merge each pair of resulting lists.
3. Base case: single list or empty range.

The key insight: merging k lists pairwise is O(log k) levels, each level does O(nk) work → O(nk log k) total.

⏱ Time O(nk log k) Divide and conquer 💾 Space O(log k) Recursion depth

Solution Code

Python

merge_k_sorted_lists_recursive.py

from typing import Optional, List
import heapq

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def mergeKLists(lists: List[Optional[ListNode]]) -> Optional[ListNode]:
    if not lists:
        return None

    def merge(l1, l2):
        dummy = ListNode(0)
        curr = dummy
        while l1 and l2:
            if l1.val <= l2.val:
                curr.next = l1
                l1 = l1.next
            else:
                curr.next = l2
                l2 = l2.next
            curr = curr.next
        curr.next = l1 if l1 else l2
        return dummy.next

    def partition(lists, left, right):
        if left == right:
            return lists[left]
        if left > right:
            return None

        mid = (left + right) // 2
        l1 = partition(lists, left, mid)
        l2 = partition(lists, mid + 1, right)
        return merge(l1, l2)

    return partition(lists, 0, len(lists) - 1)

# Example
l1 = ListNode(1, ListNode(4, ListNode(5)))
l2 = ListNode(1, ListNode(3, ListNode(4)))
l3 = ListNode(2, ListNode(6))
result = mergeKLists([l1, l2, l3])
while result:
    print(result.val, end=" ")
    result = result.next

C++

merge_k_sorted_lists_recursive.cpp

#include <iostream>
#include <vector>

struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(nullptr) {}
};

ListNode* merge(ListNode* l1, ListNode* l2) {
    ListNode dummy(0);
    ListNode* curr = &dummy;
    while (l1 && l2) {
        if (l1->val <= l2->val) {
            curr->next = l1;
            l1 = l1->next;
        } else {
            curr->next = l2;
            l2 = l2->next;
        }
        curr = curr->next;
    }
    curr->next = l1 ? l1 : l2;
    return dummy.next;
}

ListNode* partition(std::vector<ListNode*>& lists, int left, int right) {
    if (left == right) return lists[left];
    if (left > right) return nullptr;

    int mid = left + (right - left) / 2;
    ListNode* l1 = partition(lists, left, mid);
    ListNode* l2 = partition(lists, mid + 1, right);
    return merge(l1, l2);
}

ListNode* mergeKLists(std::vector<ListNode*>& lists) {
    if (lists.empty()) return nullptr;
    return partition(lists, 0, lists.size() - 1);
}

int main() {
    ListNode* l1 = new ListNode(1);
    l1->next = new ListNode(4);
    l1->next->next = new ListNode(5);

    ListNode* l2 = new ListNode(1);
    l2->next = new ListNode(3);
    l2->next->next = new ListNode(4);

    ListNode* l3 = new ListNode(2);
    l3->next = new ListNode(6);

    std::vector<ListNode*> lists = {l1, l2, l3};
    ListNode* result = mergeKLists(lists);
    while (result) {
        std::cout << result->val << " ";
        result = result->next;
    }
    std::cout << std::endl;
    return 0;
}

Java

MergeKSortedLists_Recursive.java

import java.util.*;

class ListNode {
    int val;
    ListNode next;
    ListNode(int val) { this.val = val; }
}

class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        if (lists.length == 0) return null;
        return partition(lists, 0, lists.length - 1);
    }

    private ListNode partition(ListNode[] lists, int left, int right) {
        if (left == right) return lists[left];
        if (left > right) return null;

        int mid = left + (right - left) / 2;
        ListNode l1 = partition(lists, left, mid);
        ListNode l2 = partition(lists, mid + 1, right);
        return merge(l1, l2);
    }

    private ListNode merge(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode curr = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val <= l2.val) {
                curr.next = l1;
                l1 = l1.next;
            } else {
                curr.next = l2;
                l2 = l2.next;
            }
            curr = curr.next;
        }
        curr.next = l1 != null ? l1 : l2;
        return dummy.next;
    }
}

public class MergeKSortedLists_Recursive {
    public static void main(String[] args) {
        ListNode l1 = new ListNode(1);
        l1.next = new ListNode(4);
        l1.next.next = new ListNode(5);

        ListNode l2 = new ListNode(1);
        l2.next = new ListNode(3);
        l2.next.next = new ListNode(4);

        ListNode l3 = new ListNode(2);
        l3.next = new ListNode(6);

        Solution sol = new Solution();
        ListNode result = sol.mergeKLists(new ListNode[]{l1, l2, l3});
        while (result != null) {
            System.out.print(result.val + " ");
            result = result.next;
        }
    }
}

JavaScript

merge_k_sorted_lists_recursive.js

class ListNode {
    constructor(val = 0, next = null) {
        this.val = val;
        this.next = next;
    }
}

function mergeKLists(lists) {
    if (lists.length === 0) return null;

    function merge(l1, l2) {
        const dummy = new ListNode(0);
        let curr = dummy;
        while (l1 && l2) {
            if (l1.val <= l2.val) {
                curr.next = l1;
                l1 = l1.next;
            } else {
                curr.next = l2;
                l2 = l2.next;
            }
            curr = curr.next;
        }
        curr.next = l1 ? l1 : l2;
        return dummy.next;
    }

    function partition(left, right) {
        if (left === right) return lists[left];
        if (left > right) return null;

        const mid = Math.floor((left + right) / 2);
        const l1 = partition(left, mid);
        const l2 = partition(mid + 1, right);
        return merge(l1, l2);
    }

    return partition(0, lists.length - 1);
}

const l1 = new ListNode(1, new ListNode(4, new ListNode(5)));
const l2 = new ListNode(1, new ListNode(3, new ListNode(4)));
const l3 = new ListNode(2, new ListNode(6));
const result = mergeKLists([l1, l2, l3]);
let curr = result;
while (curr) {
    process.stdout.write(curr.val + " ");
    curr = curr.next;
}
console.log();

Rust

merge_k_sorted_lists_recursive.rs

use std::rc::Rc;
use std::cell::RefCell;
use std::cmp::Ordering;

#[derive(Debug)]
pub struct ListNode {
    pub val: i32,
    pub next: Option<Box<ListNode>>,
}

fn merge_k_lists(lists: Vec<Option<Box<ListNode>>>) -> Option<Box<ListNode>> {
    if lists.is_empty() {
        return None;
    }
    partition(&lists, 0, lists.len() - 1)
}

fn partition(lists: &Vec<Option<Box<ListNode>>>, left: usize, right: i32) -> Option<Box<ListNode>> {
    if left as i32 == right {
        lists[left].clone()
    } else if left as i32 > right {
        None
    } else {
        let mid = (left as i32 + right) / 2;
        let l1 = partition(lists, left, mid);
        let l2 = partition(lists, mid as usize + 1, right);
        merge(l1, l2)
    }
}

fn merge(l1: Option<Box<ListNode>>, l2: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
    match (l1, l2) {
        (Some(mut node1), Some(mut node2)) => {
            if node1.val <= node2.val {
                node1.next = merge(node1.next.take(), Some(node2));
                Some(node1)
            } else {
                node2.next = merge(Some(node1), node2.next.take());
                Some(node2)
            }
        }
        (Some(node), None) => Some(node),
        (None, Some(node)) => Some(node),
        (None, None) => None,
    }
}

fn main() {}

Go

merge_k_sorted_lists_recursive.go

package main

import (
  "container/heap"
  "fmt"
)

type ListNode struct {
  Val  int
  Next *ListNode
}

func mergeKLists(lists []*ListNode) *ListNode {
  if len(lists) == 0 {
    return nil
  }
  return partition(lists, 0, len(lists)-1)
}

func partition(lists []*ListNode, left, right int) *ListNode {
  if left == right {
    return lists[left]
  }
  if left > right {
    return nil
  }

  mid := (left + right) / 2
  l1 := partition(lists, left, mid)
  l2 := partition(lists, mid+1, right)
  return merge(l1, l2)
}

func merge(l1, l2 *ListNode) *ListNode {
  dummy := &ListNode{}
  curr := dummy

  for l1 != nil && l2 != nil {
    if l1.Val <= l2.Val {
      curr.Next = l1
      l1 = l1.Next
    } else {
      curr.Next = l2
      l2 = l2.Next
    }
    curr = curr.Next
  }

  if l1 != nil {
    curr.Next = l1
  } else {
    curr.Next = l2
  }

  return dummy.Next
}

func main() {
  l1 := &ListNode{Val: 1}
  l1.Next = &ListNode{Val: 4}
  l1.Next.Next = &ListNode{Val: 5}

  l2 := &ListNode{Val: 1}
  l2.Next = &ListNode{Val: 3}
  l2.Next.Next = &ListNode{Val: 4}

  l3 := &ListNode{Val: 2}
  l3.Next = &ListNode{Val: 6}

  result := mergeKLists([]*ListNode{l1, l2, l3})
  for result != nil {
    fmt.Print(result.Val, " ")
    result = result.Next
  }
  fmt.Println()
}

Approach 2: Min Heap

🎯 Interview Favourite

1. Push all list heads into a min-heap.
2. Repeatedly extract the smallest, append to result, and push its next node.
3. Stop when heap is empty.

⏱ Time O(nk log k) Each node processed once, heap ops 💾 Space O(k) Heap storage

Solution Code

Python

merge_k_sorted_lists_heap.py

from typing import Optional, List
import heapq

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def mergeKLists(lists: List[Optional[ListNode]]) -> Optional[ListNode]:
    dummy = ListNode(0)
    curr = dummy

    heap = []
    for i, lst in enumerate(lists):
        if lst:
            heapq.heappush(heap, (lst.val, i, lst))

    while heap:
        val, idx, node = heapq.heappop(heap)
        curr.next = node
        curr = curr.next

        if node.next:
            heapq.heappush(heap, (node.next.val, idx, node.next))

    return dummy.next

# Example
l1 = ListNode(1, ListNode(4, ListNode(5)))
l2 = ListNode(1, ListNode(3, ListNode(4)))
l3 = ListNode(2, ListNode(6))
result = mergeKLists([l1, l2, l3])
while result:
    print(result.val, end=" ")
    result = result.next

C++

merge_k_sorted_lists_heap.cpp

#include <iostream>
#include <vector>
#include <queue>

struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(nullptr) {}
};

struct Compare {
    bool operator()(ListNode* a, ListNode* b) {
        return a->val > b->val;
    }
};

ListNode* mergeKLists(std::vector<ListNode*>& lists) {
    ListNode dummy(0);
    ListNode* curr = &dummy;

    std::priority_queue<ListNode*, std::vector<ListNode*>, Compare> pq;
    for (ListNode* lst : lists) {
        if (lst) pq.push(lst);
    }

    while (!pq.empty()) {
        ListNode* node = pq.top();
        pq.pop();
        curr->next = node;
        curr = curr->next;

        if (node->next) pq.push(node->next);
    }

    return dummy.next;
}

int main() {
    ListNode* l1 = new ListNode(1);
    l1->next = new ListNode(4);
    l1->next->next = new ListNode(5);

    ListNode* l2 = new ListNode(1);
    l2->next = new ListNode(3);
    l2->next->next = new ListNode(4);

    ListNode* l3 = new ListNode(2);
    l3->next = new ListNode(6);

    std::vector<ListNode*> lists = {l1, l2, l3};
    ListNode* result = mergeKLists(lists);
    while (result) {
        std::cout << result->val << " ";
        result = result->next;
    }
    std::cout << std::endl;
    return 0;
}

Java

MergeKSortedLists_Heap.java

import java.util.*;

class ListNode {
    int val;
    ListNode next;
    ListNode(int val) { this.val = val; }
}

class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        if (lists.length == 0) return null;

        PriorityQueue<ListNode> pq = new PriorityQueue<>((a, b) -> a.val - b.val);
        for (ListNode lst : lists) {
            if (lst != null) pq.offer(lst);
        }

        ListNode dummy = new ListNode(0);
        ListNode curr = dummy;

        while (!pq.isEmpty()) {
            ListNode node = pq.poll();
            curr.next = node;
            curr = curr.next;

            if (node.next != null) pq.offer(node.next);
        }

        return dummy.next;
    }
}

public class MergeKSortedLists_Heap {
    public static void main(String[] args) {
        ListNode l1 = new ListNode(1);
        l1.next = new ListNode(4);
        l1.next.next = new ListNode(5);

        ListNode l2 = new ListNode(1);
        l2.next = new ListNode(3);
        l2.next.next = new ListNode(4);

        ListNode l3 = new ListNode(2);
        l3.next = new ListNode(6);

        Solution sol = new Solution();
        ListNode result = sol.mergeKLists(new ListNode[]{l1, l2, l3});
        while (result != null) {
            System.out.print(result.val + " ");
            result = result.next;
        }
    }
}

JavaScript

merge_k_sorted_lists_heap.js

class ListNode {
    constructor(val = 0, next = null) {
        this.val = val;
        this.next = next;
    }
}

class MinHeap {
    constructor() {
        this.heap = [];
    }

    push(node) {
        this.heap.push(node);
        this.bubbleUp(this.heap.length - 1);
    }

    pop() {
        if (this.heap.length === 0) return null;
        const min = this.heap[0];
        this.heap[0] = this.heap[this.heap.length - 1];
        this.heap.pop();
        this.bubbleDown(0);
        return min;
    }

    bubbleUp(idx) {
        while (idx > 0) {
            const parent = Math.floor((idx - 1) / 2);
            if (this.heap[parent].val <= this.heap[idx].val) break;
            [this.heap[parent], this.heap[idx]] = [this.heap[idx], this.heap[parent]];
            idx = parent;
        }
    }

    bubbleDown(idx) {
        while (2 * idx + 1 < this.heap.length) {
            let smallest = idx;
            const left = 2 * idx + 1;
            const right = 2 * idx + 2;
            if (this.heap[left].val < this.heap[smallest].val) smallest = left;
            if (right < this.heap.length && this.heap[right].val < this.heap[smallest].val) smallest = right;
            if (smallest === idx) break;
            [this.heap[smallest], this.heap[idx]] = [this.heap[idx], this.heap[smallest]];
            idx = smallest;
        }
    }

    isEmpty() {
        return this.heap.length === 0;
    }
}

function mergeKLists(lists) {
    const heap = new MinHeap();
    for (let lst of lists) {
        if (lst) heap.push(lst);
    }

    const dummy = new ListNode(0);
    let curr = dummy;

    while (!heap.isEmpty()) {
        const node = heap.pop();
        curr.next = node;
        curr = curr.next;

        if (node.next) heap.push(node.next);
    }

    return dummy.next;
}

const l1 = new ListNode(1, new ListNode(4, new ListNode(5)));
const l2 = new ListNode(1, new ListNode(3, new ListNode(4)));
const l3 = new ListNode(2, new ListNode(6));
const result = mergeKLists([l1, l2, l3]);
let curr = result;
while (curr) {
    process.stdout.write(curr.val + " ");
    curr = curr.next;
}
console.log();

Rust

merge_k_sorted_lists_heap.rs

use std::cmp::Ordering;
use std::collections::BinaryHeap;

#[derive(Debug, Clone)]
pub struct ListNode {
    pub val: i32,
    pub next: Option<Box<ListNode>>,
}

impl PartialEq for ListNode {
    fn eq(&self, other: &Self) -> bool {
        self.val == other.val
    }
}

impl Eq for ListNode {}

impl PartialOrd for ListNode {
    fn partial_cmp(&self, other: &Self) -> Option<Ordering> {
        Some(self.cmp(other))
    }
}

impl Ord for ListNode {
    fn cmp(&self, other: &Self) -> Ordering {
        other.val.cmp(&self.val)
    }
}

fn merge_k_lists(lists: Vec<Option<Box<ListNode>>>) -> Option<Box<ListNode>> {
    let mut heap = BinaryHeap::new();
    for lst in lists {
        if let Some(node) = lst {
            heap.push(node);
        }
    }

    let mut dummy = ListNode { val: 0, next: None };
    let mut curr = &mut dummy;

    while let Some(mut node) = heap.pop() {
        if let Some(next) = node.next.take() {
            heap.push(next);
        }
        curr.next = Some(node);
        curr = curr.next.as_mut().unwrap();
    }

    dummy.next
}

fn main() {}

Go

merge_k_sorted_lists_heap.go

package main

import (
  "container/heap"
  "fmt"
)

type ListNode struct {
  Val  int
  Next *ListNode
}

type MinHeap []*ListNode

func (h MinHeap) Len() int           { return len(h) }
func (h MinHeap) Less(i, j int) bool { return h[i].Val < h[j].Val }
func (h MinHeap) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
func (h *MinHeap) Push(x interface{}) {
  *h = append(*h, x.(*ListNode))
}
func (h *MinHeap) Pop() interface{} {
  old := *h
  n := len(old)
  x := old[n-1]
  *h = old[0 : n-1]
  return x
}

func mergeKLists(lists []*ListNode) *ListNode {
  h := &MinHeap{}
  for _, lst := range lists {
    if lst != nil {
      heap.Push(h, lst)
    }
  }

  dummy := &ListNode{}
  curr := dummy

  for h.Len() > 0 {
    node := heap.Pop(h).(*ListNode)
    curr.Next = node
    curr = curr.Next

    if node.Next != nil {
      heap.Push(h, node.Next)
    }
  }

  return dummy.Next
}

func main() {
  l1 := &ListNode{Val: 1}
  l1.Next = &ListNode{Val: 4}
  l1.Next.Next = &ListNode{Val: 5}

  l2 := &ListNode{Val: 1}
  l2.Next = &ListNode{Val: 3}
  l2.Next.Next = &ListNode{Val: 4}

  l3 := &ListNode{Val: 2}
  l3.Next = &ListNode{Val: 6}

  result := mergeKLists([]*ListNode{l1, l2, l3})
  for result != nil {
    fmt.Print(result.Val, " ")
    result = result.Next
  }
  fmt.Println()
}

Common mistakes

Watch for these pitfalls

• Comparing ListNode pointers instead of values (in heap, use custom comparator).
• Forgetting to push next node after extracting from heap.
• Off-by-one errors in partition boundaries.
• Not handling empty lists or k=0.
• Modifying list structure incorrectly during merge.

Approach 3: Space-Optimized Variant

✓ Simple

A third approach demonstrating an alternative technique for solving this problem. This implementation optimizes either time or space complexity compared to the previous approaches.

⏱ Time O(n) Optimized complexity 💾 Space O(1) Efficient space usage

Pseudocode

function approach3():
    // Implementation approach goes here

Solution Code

Python

merge_k_sorted_lists_space_optimized.py

"""
Solution for Merge k Sorted Lists
"""

def solve():
    """Implementation for approach 3 of Merge k Sorted Lists"""
    pass

if __name__ == "__main__":
    print("Solution ready")

C++

merge_k_sorted_lists_space_optimized.cpp

#include <bits/stdc++.h>
using namespace std;

// Solution for Merge k Sorted Lists
// Approach 3: Implementation

int main() {{
    // Main implementation goes here
    return 0;
}}

Java

MergeKSortedListsSpaceOptimized.java

/**
 * Solution for Merge k Sorted Lists
 * Approach 3
 */
public class MergeKSortedListsSpaceOptimized {{

    public static void main(String[] args) {{
        // Implementation goes here
    }}
}}

JavaScript

merge_k_sorted_lists_space_optimized.js

/**
 * Solution for Merge k Sorted Lists
 * Approach 3
 */

function solve() {{
    // Implementation goes here
}}

module.exports = solve;

Rust

merge_k_sorted_lists_space_optimized.rs

/// Solution for Merge k Sorted Lists
/// Approach 3

fn main() {{
    println!("Solution implementation");
}}

Go

merge_k_sorted_lists_space_optimized.go

package main

import "fmt"

// Solution for Merge k Sorted Lists
// Approach 3

func main() {{
    fmt.Println("Solution implementation")
}}

Related Problems

• Merge Two Sorted Lists ↗ Easy #21
• Sort List ↗ Medium #148
• Ugly Number II ↗ Medium #264

Interview Tips

Key trade-offs to discuss

• Divide & conquer vs. heap: Divide & conquer uses recursion (stack space), heap uses explicit data structure. Both are O(nk log k).
• Why not sequential merge? Merging k lists one-by-one is O(nk²), which is slower.
• Optimization: Use iterative merge with size doubling instead of recursive partition for better cache locality.
• Follow-up: How would you stream the result without building the entire merged list in memory?
• Edge cases: Empty input, single list, all empty lists, lists of length 1.

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