23. Merge K Sorted Lists - Explanation

Description

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

Return the sorted linked list that is the result of merging all of the individual linked lists.

Example 1:

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

Output: [1,1,2,3,3,4,5,6]

Example 2:

Input: lists = []

Output: []

Example 3:

Input: lists = [[]]

Output: []

Constraints:

0 <= lists.length <= 1000
0 <= lists[i].length <= 100
-1000 <= lists[i][j] <= 1000

Topics

Recommended Time & Space Complexity

You should aim for a solution as good or better than O(n * k) time and O(1) space, where k is the total number of lists and n is the total number of nodes across all lists.

Hint 1

A brute-force solution would involve storing all n nodes in an array, sorting them, and converting the array back into a linked list, resulting in an O(nlogn) time complexity. Can you think of a better way? Perhaps consider leveraging the idea behind merging two sorted linked lists.

Hint 2

We can merge two sorted linked lists without using any extra space. To handle k sorted linked lists, we can iteratively merge each linked list with a resultant merged list. How can you implement this?

Hint 3

We iterate through the list array with index i, starting at i = 1. We merge the linked lists using mergeTwoLists(lists[i], lists[i - 1]), which returns the head of the merged list. This head is stored in lists[i], and the process continues. Finally, the merged list is obtained at the last index, and we return its head.

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Prerequisites

Before attempting this problem, you should be comfortable with:

Linked Lists - Understanding singly linked list structure, traversal, and node manipulation
Merge Two Sorted Lists - The fundamental operation used repeatedly in all approaches
Min Heap / Priority Queue - Used to efficiently find the smallest element among k lists
Divide and Conquer - Strategy to recursively split and merge lists in pairs
Sorting - Used in the brute force approach to sort all collected values

1. Brute Force

Intuition

The simplest way to merge all linked lists is to ignore the list structure, collect every value, sort them, and then rebuild a single sorted linked list.
This doesn't use any clever merging logic — it is purely based on gathering and sorting.
It's easy to implement but not efficient because sorting dominates the runtime.

Algorithm

Create an empty list nodes.
For each linked list:
- Traverse it and append every node's value to nodes.
Sort the nodes list.
Create a new linked list:
- Use a dummy head.
- For each value in the sorted list, create a new node and attach it.
Return the head of the new merged 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 mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
        nodes = []
        for lst in lists:
            while lst:
                nodes.append(lst.val)
                lst = lst.next
        nodes.sort()

        res = ListNode(0)
        cur = res
        for node in nodes:
            cur.next = ListNode(node)
            cur = cur.next
        return res.next

/**
 * 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; }
 * }
 */

public class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        List<Integer> nodes = new ArrayList<>();
        for (ListNode lst : lists) {
            while (lst != null) {
                nodes.add(lst.val);
                lst = lst.next;
            }
        }
        Collections.sort(nodes);

        ListNode res = new ListNode(0);
        ListNode cur = res;
        for (int node : nodes) {
            cur.next = new ListNode(node);
            cur = cur.next;
        }
        return res.next;
    }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */

class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        vector<int> nodes;
        for (ListNode* lst : lists) {
            while (lst) {
                nodes.push_back(lst->val);
                lst = lst->next;
            }
        }
        sort(nodes.begin(), nodes.end());

        ListNode* res = new ListNode(0);
        ListNode* cur = res;
        for (int node : nodes) {
            cur->next = new ListNode(node);
            cur = cur->next;
        }
        return res->next;
    }
};

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     constructor(val = 0, next = null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */

class Solution {
    /**
     * @param {ListNode[]} lists
     * @return {ListNode}
     */
    mergeKLists(lists) {
        let nodes = [];
        for (let lst of lists) {
            while (lst) {
                nodes.push(lst.val);
                lst = lst.next;
            }
        }
        nodes.sort((a, b) => a - b);

        let res = new ListNode(0);
        let cur = res;
        for (let node of nodes) {
            cur.next = new ListNode(node);
            cur = cur.next;
        }
        return res.next;
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */

public class Solution {
    public ListNode MergeKLists(ListNode[] lists) {
        List<int> nodes = new List<int>();
        foreach (ListNode lst in lists) {
            ListNode curr = lst;
            while (curr != null) {
                nodes.Add(curr.val);
                curr = curr.next;
            }
        }
        nodes.Sort();

        ListNode res = new ListNode(0);
        ListNode cur = res;
        foreach (int node in nodes) {
            cur.next = new ListNode(node);
            cur = cur.next;
        }
        return res.next;
    }
}

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeKLists(lists []*ListNode) *ListNode {
    nodes := make([]int, 0)

    for _, list := range lists {
        curr := list
        for curr != nil {
            nodes = append(nodes, curr.Val)
            curr = curr.Next
        }
    }

    sort.Ints(nodes)

    dummy := &ListNode{Val: 0}
    curr := dummy

    for _, val := range nodes {
        curr.Next = &ListNode{Val: val}
        curr = curr.Next
    }

    return dummy.Next
}

/**
 * Example:
 * var li = ListNode(5)
 * var v = li.`val`
 * Definition for singly-linked list.
 * class ListNode(var `val`: Int) {
 *     var next: ListNode? = null
 * }
 */
class Solution {
    fun mergeKLists(lists: Array<ListNode?>): ListNode? {
        val nodes = mutableListOf<Int>()

        for (list in lists) {
            var curr = list
            while (curr != null) {
                nodes.add(curr.`val`)
                curr = curr.next
            }
        }

        nodes.sort()

        val dummy = ListNode(0)
        var curr = dummy

        for (value in nodes) {
            curr.next = ListNode(value)
            curr = curr.next!!
        }

        return dummy.next
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init() { self.val = 0; self.next = nil; }
 *     public init(_ val: Int) { self.val = val; self.next = nil; }
 *     public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
 * }
 */
class Solution {
    func mergeKLists(_ lists: [ListNode?]) -> ListNode? {
        var nodes: [Int] = []

        for list in lists {
            var lst = list
            while lst != nil {
                nodes.append(lst!.val)
                lst = lst?.next
            }
        }

        nodes.sort()

        let dummy = ListNode(0)
        var cur = dummy
        for node in nodes {
            cur.next = ListNode(node)
            cur = cur.next!
        }

        return dummy.next
    }
}

Time & Space Complexity

Time complexity: $O(n \log n)$
Space complexity: $O(n)$

2. Iteration

Intuition

We repeatedly pick the smallest head node among all the lists and attach it to our result list.
At every step:

Look at the first node of each non-empty list.
Choose the one with the smallest value.
Move that list's pointer forward.
Append the chosen node to our merged list.

This is similar to merging k sorted arrays by always picking the smallest available element.

Algorithm

Create a dummy node to build the merged list.
Set a pointer cur to the dummy.
Repeat:
- Search through all lists to find the list whose current node has the smallest value.
- If no list has remaining nodes (all are empty), stop.
- Attach the smallest node to cur.next.
- Move cur forward.
- Move the chosen list's pointer to its next node.
Return the merged list starting from dummy.next.

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class Solution:
    def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
        res = ListNode(0)
        cur = res

        while True:
            minNode = -1
            for i in range(len(lists)):
                if not lists[i]:
                    continue
                if minNode == -1 or lists[minNode].val > lists[i].val:
                    minNode = i

            if minNode == -1:
                break
            cur.next = lists[minNode]
            lists[minNode] = lists[minNode].next
            cur = cur.next

        return res.next

/**
 * 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; }
 * }
 */

public class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        ListNode res = new ListNode(0);
        ListNode cur = res;

        while (true) {
            int minNode = -1;
            for (int i = 0; i < lists.length; i++) {
                if (lists[i] == null) {
                    continue;
                }
                if (minNode == -1 || lists[minNode].val > lists[i].val) {
                    minNode = i;
                }
            }

            if (minNode == -1) {
                break;
            }
            cur.next = lists[minNode];
            lists[minNode] = lists[minNode].next;
            cur = cur.next;
        }

        return res.next;
    }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */

class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        ListNode* res = new ListNode(0);
        ListNode* cur = res;

        while (true) {
            int minNode = -1;
            for (int i = 0; i < lists.size(); i++) {
                if (!lists[i]) continue;
                if (minNode == -1 || lists[minNode]->val > lists[i]->val) {
                    minNode = i;
                }
            }

            if (minNode == -1) break;
            cur->next = lists[minNode];
            lists[minNode] = lists[minNode]->next;
            cur = cur->next;
        }
        return res->next;
    }
};

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     constructor(val = 0, next = null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */

class Solution {
    /**
     * @param {ListNode[]} lists
     * @return {ListNode}
     */
    mergeKLists(lists) {
        let res = new ListNode(0);
        let cur = res;

        while (true) {
            let minNode = -1;
            for (let i = 0; i < lists.length; i++) {
                if (!lists[i]) continue;
                if (minNode === -1 || lists[minNode].val > lists[i].val) {
                    minNode = i;
                }
            }

            if (minNode === -1) break;
            cur.next = lists[minNode];
            lists[minNode] = lists[minNode].next;
            cur = cur.next;
        }
        return res.next;
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */

public class Solution {
    public ListNode MergeKLists(ListNode[] lists) {
        ListNode res = new ListNode(0);
        ListNode cur = res;

        while (true) {
            int minNode = -1;
            for (int i = 0; i < lists.Length; i++) {
                if (lists[i] == null) continue;
                if (minNode == -1 || lists[minNode].val > lists[i].val) {
                    minNode = i;
                }
            }

            if (minNode == -1) break;
            cur.next = lists[minNode];
            lists[minNode] = lists[minNode].next;
            cur = cur.next;
        }
        return res.next;
    }
}

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeKLists(lists []*ListNode) *ListNode {
    res := &ListNode{Val: 0}
    cur := res

    for {
        minNode := -1
        for i := range lists {
            if lists[i] == nil {
                continue
            }
            if minNode == -1 || lists[minNode].Val > lists[i].Val {
                minNode = i
            }
        }

        if minNode == -1 {
            break
        }

        cur.Next = lists[minNode]
        lists[minNode] = lists[minNode].Next
        cur = cur.Next
    }

    return res.Next
}

/**
 * Example:
 * var li = ListNode(5)
 * var v = li.`val`
 * Definition for singly-linked list.
 * class ListNode(var `val`: Int) {
 *     var next: ListNode? = null
 * }
 */
class Solution {
    fun mergeKLists(lists: Array<ListNode?>): ListNode? {
        val res = ListNode(0)
        var cur = res

        while (true) {
            var minNode = -1
            for (i in lists.indices) {
                if (lists[i] == null) {
                    continue
                }
                if (minNode == -1 || lists[minNode]!!.`val` > lists[i]!!.`val`) {
                    minNode = i
                }
            }

            if (minNode == -1) {
                break
            }

            cur.next = lists[minNode]
            lists[minNode] = lists[minNode]!!.next
            cur = cur.next!!
        }

        return res.next
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init() { self.val = 0; self.next = nil; }
 *     public init(_ val: Int) { self.val = val; self.next = nil; }
 *     public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
 * }
 */
class Solution {
    func mergeKLists(_ lists: [ListNode?]) -> ListNode? {
        var lists = lists
        let dummy = ListNode(0)
        var cur = dummy

        while true {
            var minNodeIndex: Int? = nil
            for i in 0..<lists.count {
                if let node = lists[i] {
                    if minNodeIndex == nil || node.val < lists[minNodeIndex!]?.val ?? Int.max {
                        minNodeIndex = i
                    }
                }
            }

            if minNodeIndex == nil {
                break
            }

            cur.next = lists[minNodeIndex!]
            lists[minNodeIndex!] = lists[minNodeIndex!]?.next
            cur = cur.next!
        }

        return dummy.next
    }
}

Time & Space Complexity

Time complexity: $O(n * k)$
Space complexity: $O(1)$

Where $k$ is the total number of lists and $n$ is the total number of nodes across $k$ lists.

3. Merge Lists One By One

Intuition

Instead of merging all k lists at once, we can merge them one by one.

First merge list 0 and list 1 → get a sorted list.
Then merge that result with list 2.
Then merge that result with list 3.
Repeat until all lists are merged.

Each merge operation is just like the standard "merge two sorted linked lists" problem:

Compare the heads.
Attach the smaller one.
Move that list's pointer forward.
Continue until one list is empty, then attach the rest of the other list.

Algorithm

If the list of lists is empty, return null (or equivalent).
For each list from index 1 to k - 1:
- Merge lists[i - 1] and lists[i] into a single sorted list.
- Store the merged result back into lists[i].
After the loop, lists[k - 1] will contain the fully merged list.
Return lists[k - 1].

Merging two lists (mergeList(l1, l2)):

Create a dummy node and a tail pointer starting at the dummy.
While both l1 and l2 are non-empty:
- Compare l1.val and l2.val.
- Attach the smaller node to tail.next.
- Move that list's pointer forward.
- Move tail forward.
If one list still has remaining nodes, attach it to tail.next.
Return dummy.next as the merged head.

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class Solution:
    def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
        if len(lists) == 0:
            return None

        for i in range(1, len(lists)):
            lists[i] = self.mergeList(lists[i - 1], lists[i])

        return lists[-1]

    def mergeList(self, l1, l2):
        dummy = ListNode()
        tail = dummy

        while l1 and l2:
            if l1.val < l2.val:
                tail.next = l1
                l1 = l1.next
            else:
                tail.next = l2
                l2 = l2.next
            tail = tail.next
        if l1:
            tail.next = l1
        if l2:
            tail.next = l2
        return dummy.next

/**
 * 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 ListNode mergeKLists(ListNode[] lists) {
        if (lists.length == 0) return null;

        for (int i = 1; i < lists.length; i++) {
            lists[i] = merge(lists[i], lists[i - 1]);
        }
        return lists[lists.length - 1];
    }

    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;
        }

        if (l1 != null) {
            curr.next = l1;
        } else {
            curr.next = l2;
        }

        return dummy.next;
    }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */

class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        if (lists.empty()) return nullptr;

        for (int i = 1; i < lists.size(); i++) {
            lists[i] = merge(lists[i], lists[i - 1]);
        }
        return lists.back();
    }

private:
    ListNode* merge(ListNode* l1, ListNode* l2) {
        ListNode* dummy = new ListNode(0);
        ListNode* curr = dummy;

        while (l1 != nullptr && l2 != nullptr) {
            if (l1->val <= l2->val) {
                curr->next = l1;
                l1 = l1->next;
            } else {
                curr->next = l2;
                l2 = l2->next;
            }
            curr = curr->next;
        }

        if (l1 != nullptr) {
            curr->next = l1;
        } else {
            curr->next = l2;
        }

        return dummy->next;
    }
};

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     constructor(val = 0, next = null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */

class Solution {
    /**
     * @param {ListNode[]} lists
     * @return {ListNode}
     */
    mergeKLists(lists) {
        if (lists.length === 0) return null;

        for (let i = 0; i < lists.length; i++) {
            lists[i] = this.mergeList(lists[i], lists[i - 1]);
        }
        return lists[lists.length - 1];
    }

    /**
     * @param {ListNode} l1
     * @param  {ListNode} l2
     * @return {ListNode}
     */
    mergeList(l1, l2) {
        const dummy = new ListNode();
        let tail = dummy;

        while (l1 && l2) {
            if (l1.val < l2.val) {
                tail.next = l1;
                l1 = l1.next;
            } else {
                tail.next = l2;
                l2 = l2.next;
            }
            tail = tail.next;
        }
        if (l1) {
            tail.next = l1;
        }
        if (l2) {
            tail.next = l2;
        }
        return dummy.next;
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */

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

        for (int i = 1; i < lists.Length; i++) {
            lists[i] = Merge(lists[i], lists[i - 1]);
        }
        return lists[lists.Length - 1];
    }

    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;
        }

        if (l1 != null) {
            curr.next = l1;
        } else {
            curr.next = l2;
        }

        return dummy.next;
    }
}

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeList(l1 *ListNode, l2 *ListNode) *ListNode {
    dummy := &ListNode{}
    tail := dummy

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

    if l1 != nil {
        tail.Next = l1
    }
    if l2 != nil {
        tail.Next = l2
    }

    return dummy.Next
}

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

    for i := 1; i < len(lists); i++ {
        lists[i] = mergeList(lists[i-1], lists[i])
    }

    return lists[len(lists)-1]
}

/**
 * Example:
 * var li = ListNode(5)
 * var v = li.`val`
 * Definition for singly-linked list.
 * class ListNode(var `val`: Int) {
 *     var next: ListNode? = null
 * }
 */
class Solution {
    private fun mergeList(l1: ListNode?, l2: ListNode?): ListNode? {
        val dummy = ListNode(0)
        var tail = dummy
        var first = l1
        var second = l2

        while (first != null && second != null) {
            if (first.`val` < second.`val`) {
                tail.next = first
                first = first.next
            } else {
                tail.next = second
                second = second.next
            }
            tail = tail.next!!
        }

        if (first != null) {
            tail.next = first
        }
        if (second != null) {
            tail.next = second
        }

        return dummy.next
    }

    fun mergeKLists(lists: Array<ListNode?>): ListNode? {
        if (lists.isEmpty()) {
            return null
        }

        for (i in 1 until lists.size) {
            lists[i] = mergeList(lists[i-1], lists[i])
        }

        return lists.last()
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init() { self.val = 0; self.next = nil; }
 *     public init(_ val: Int) { self.val = val; self.next = nil; }
 *     public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
 * }
 */
class Solution {
    func mergeKLists(_ lists: [ListNode?]) -> ListNode? {
        if lists.isEmpty {
            return nil
        }

        var lists = lists
        for i in 1..<lists.count {
            lists[i] = mergeList(lists[i - 1], lists[i])
        }

        return lists.last!
    }

    private func mergeList(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
        let dummy = ListNode(0)
        var tail = dummy
        var l1 = l1, l2 = l2

        while l1 != nil && l2 != nil {
            if l1!.val < l2!.val {
                tail.next = l1
                l1 = l1?.next
            } else {
                tail.next = l2
                l2 = l2?.next
            }
            tail = tail.next!
        }

        if l1 != nil {
            tail.next = l1
        }
        if l2 != nil {
            tail.next = l2
        }

        return dummy.next
    }
}

Time & Space Complexity

Time complexity: $O(n * k)$
Space complexity: $O(1)$

Where $k$ is the total number of lists and $n$ is the total number of nodes across $k$ lists.

4. Heap

Intuition

We want to always pick the smallest current node among all k lists as efficiently as possible.

Instead of scanning all heads every time (which is slow), we can use a min-heap (priority queue):

Push the head of each non-empty list into the heap.
The heap always gives us the node with the smallest value on top.
We pop that node, attach it to our result list, and then push its next node (if it exists) into the heap.
Repeat until the heap is empty.

This way, at every step we choose the globally smallest node in O(log k) time, where k is the number of lists.

Algorithm

Create a min-heap (priority queue).
For each non-empty linked list:
- Push its head node into the heap (keyed by node value).
Create a dummy node to start the result list, and a pointer cur pointing to it.
While the heap is not empty:
- Pop the node with the smallest value from the heap.
- Attach this node to cur.next, and move cur forward.
- If this node has a next node, push that next node into the heap.
When the heap is empty, all nodes have been merged in sorted order.
Return dummy.next as the head of the merged list.

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class NodeWrapper:
    def __init__(self, node):
        self.node = node

    def __lt__(self, other):
        return self.node.val < other.node.val

class Solution:
    def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
        if len(lists) == 0:
            return None

        res = ListNode(0)
        cur = res
        minHeap = []

        for lst in lists:
            if lst is not None:
                heapq.heappush(minHeap, NodeWrapper(lst))

        while minHeap:
            node_wrapper = heapq.heappop(minHeap)
            cur.next = node_wrapper.node
            cur = cur.next

            if node_wrapper.node.next:
                heapq.heappush(minHeap, NodeWrapper(node_wrapper.node.next))

        return res.next

/**
 * 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 ListNode mergeKLists(ListNode[] lists) {
        if (lists.length == 0) return null;

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

        ListNode res = new ListNode(0);
        ListNode cur = res;
        while (!minHeap.isEmpty()) {
            ListNode node = minHeap.poll();
            cur.next = node;
            cur = cur.next;

            node = node.next;
            if (node != null) {
                minHeap.offer(node);
            }
        }
        return res.next;
    }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */

class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        if (lists.empty()) return nullptr;

        auto cmp = [](ListNode* a, ListNode* b) { return a->val > b->val; };
        priority_queue<ListNode*, vector<ListNode*>, decltype(cmp)> minHeap(cmp);

        for (ListNode* list : lists) {
            if (list != nullptr) {
                minHeap.push(list);
            }
        }

        ListNode* res = new ListNode(0);
        ListNode* cur = res;
        while (!minHeap.empty()) {
            ListNode* node = minHeap.top();
            minHeap.pop();
            cur->next = node;
            cur = cur->next;

            node = node->next;
            if (node != nullptr) {
                minHeap.push(node);
            }
        }
        return res->next;
    }
};

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     constructor(val = 0, next = null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */

class Solution {
    /**
     * @param {ListNode[]} lists
     * @return {ListNode}
     */
    mergeKLists(lists) {
        if (lists.length === 0) return null;
        const minHeap = new MinPriorityQueue((x) => x.val);
        for (let list of lists) {
            if (list != null) minHeap.enqueue(list);
        }

        let res = new ListNode(0);
        let cur = res;
        while (minHeap.size() > 0) {
            let node = minHeap.dequeue();
            cur.next = node;
            cur = cur.next;

            node = node.next;
            if (node != null) {
                minHeap.enqueue(node);
            }
        }
        return res.next;
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */

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

        var minHeap = new PriorityQueue<ListNode, int>();
        foreach (var list in lists) {
            if (list != null) {
                minHeap.Enqueue(list, list.val);
            }
        }

        var res = new ListNode(0);
        var cur = res;
        while (minHeap.Count > 0) {
            var node = minHeap.Dequeue();
            cur.next = node;
            cur = cur.next;

            node = node.next;
            if (node != null) {
                minHeap.Enqueue(node, node.val);
            }
        }
        return res.next;
    }
}

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeKLists(lists []*ListNode) *ListNode {
    if len(lists) == 0 {
        return nil
    }

    minHeap := priorityqueue.NewWith(func(a, b interface{}) int {
        return a.(*ListNode).Val - b.(*ListNode).Val
    })

    for _, list := range lists {
        if list != nil {
            minHeap.Enqueue(list)
        }
    }

    res := &ListNode{Val: 0}
    cur := res

    for !minHeap.Empty() {
        node, _ := minHeap.Dequeue()
        cur.Next = node.(*ListNode)
        cur = cur.Next

        if cur.Next != nil {
            minHeap.Enqueue(cur.Next)
        }
    }

    return res.Next
}

/**
 * Example:
 * var li = ListNode(5)
 * var v = li.`val`
 * Definition for singly-linked list.
 * class ListNode(var `val`: Int) {
 *     var next: ListNode? = null
 * }
 */
class Solution {
    fun mergeKLists(lists: Array<ListNode?>): ListNode? {
        if (lists.isEmpty()) return null

        val minHeap = PriorityQueue<ListNode>(compareBy { it.`val` })

        for (list in lists) {
            list?.let { minHeap.offer(it) }
        }

        val res = ListNode(0)
        var cur = res

        while (minHeap.isNotEmpty()) {
            val node = minHeap.poll()
            cur.next = node
            cur = cur.next!!

            node.next?.let { minHeap.offer(it) }
        }

        return res.next
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init() { self.val = 0; self.next = nil; }
 *     public init(_ val: Int) { self.val = val; self.next = nil; }
 *     public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
 * }
 */
struct NodeWrapper: Comparable {
    let node: ListNode

    init(_ node: ListNode) {
        self.node = node
    }

    static func < (lhs: NodeWrapper, rhs: NodeWrapper) -> Bool {
        return lhs.node.val < rhs.node.val
    }

    static func == (lhs: NodeWrapper, rhs: NodeWrapper) -> Bool {
        return lhs.node.val == rhs.node.val
    }
}

class Solution {
    func mergeKLists(_ lists: [ListNode?]) -> ListNode? {
        var heap = Heap<NodeWrapper>()

        for list in lists {
            if let node = list {
                heap.insert(NodeWrapper(node))
            }
        }

        let dummy = ListNode(0)
        var tail = dummy

        while let wrapper = heap.popMin() {
            tail.next = wrapper.node
            tail = tail.next!

            if let next = wrapper.node.next {
                heap.insert(NodeWrapper(next))
            }
        }

        return dummy.next
    }
}

Time & Space Complexity

Time complexity: $O(n \log k)$
Space complexity: $O(k)$

Where $k$ is the total number of lists and $n$ is the total number of nodes across $k$ lists.

5. Divide And Conquer (Recursion)

Intuition

Instead of merging all k lists at once or one by one in order, we can use a divide and conquer strategy, similar to how merge sort works.

Idea:

Split the list of linked lists into two halves.
Recursively merge the left half into one sorted list.
Recursively merge the right half into one sorted list.
Finally, merge these two sorted lists into a single sorted list.

By always merging pairs of lists, we reduce the total work compared to merging k lists sequentially.
Each merge of two lists is linear in their total length, and we do about log k levels of merging.

This makes the approach both clean and efficient.

Algorithm

Base Cases
- If the input list lists is empty, return null.
- Use a recursive function divide(lists, l, r):
  - If l > r, return null.
  - If l == r, return lists[l] (only one list to return).
Divide Step
- Compute mid = (l + r) // 2.
- Recursively compute:
  - left = divide(lists, l, mid)
  - right = divide(lists, mid + 1, r)
Conquer Step (merge two lists)
- Merge left and right using the standard merge two sorted linked lists routine:
  - Create a dummy node and use a curr pointer.
  - While both lists are non-empty:
    - Attach the smaller node to curr.next.
    - Move that list's pointer forward.
    - Move curr forward.
  - Attach any remaining nodes from either list.
- Return the merged list.
Final Answer
- Call divide(lists, 0, len(lists) - 1) and return the resulting list.

This approach efficiently merges k sorted linked lists in a structured, recursive way.

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class Solution:
    def mergeKLists(self, lists):
        if not lists or len(lists) == 0:
            return None
        return self.divide(lists, 0, len(lists) - 1)

    def divide(self, lists, l, r):
        if l > r:
            return None
        if l == r:
            return lists[l]

        mid = l + (r - l) // 2
        left = self.divide(lists, l, mid)
        right = self.divide(lists, mid + 1, r)

        return self.conquer(left, right)

    def conquer(self, 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

        if l1:
            curr.next = l1
        else:
            curr.next = l2

        return dummy.next

/**
 * 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 ListNode mergeKLists(ListNode[] lists) {
        if (lists == null || lists.length == 0) {
            return null;
        }
        return divide(lists, 0, lists.length - 1);
    }

    private ListNode divide(ListNode[] lists, int l, int r) {
        if (l > r) {
            return null;
        }
        if (l == r) {
            return lists[l];
        }

        int mid = l + (r - l) / 2;
        ListNode left = divide(lists, l, mid);
        ListNode right = divide(lists, mid + 1, r);

        return conquer(left, right);
    }

    private ListNode conquer(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;
        }

        if (l1 != null) {
            curr.next = l1;
        } else {
            curr.next = l2;
        }

        return dummy.next;
    }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */

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

private:
    ListNode* divide(vector<ListNode*>& lists, int l, int r) {
        if (l > r) {
            return nullptr;
        }
        if (l == r) {
            return lists[l];
        }

        int mid = l + (r - l) / 2;
        ListNode* left = divide(lists, l, mid);
        ListNode* right = divide(lists, mid + 1, r);

        return conquer(left, right);
    }

    ListNode* conquer(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;
        }

        if (l1) {
            curr->next = l1;
        } else {
            curr->next = l2;
        }

        return dummy.next;
    }
};

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     constructor(val = 0, next = null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */

class Solution {
    /**
     * @param {ListNode[]} lists
     * @return {ListNode}
     */
    mergeKLists(lists) {
        if (!lists || lists.length === 0) {
            return null;
        }
        return this.divide(lists, 0, lists.length - 1);
    }

    /**
     * @param {ListNode[]} lists
     * @param {number} l
     * @param {number} r
     * @return {ListNode}
     */
    divide(lists, l, r) {
        if (l > r) {
            return null;
        }
        if (l === r) {
            return lists[l];
        }

        const mid = Math.floor(l + (r - l) / 2);
        const left = this.divide(lists, l, mid);
        const right = this.divide(lists, mid + 1, r);

        return this.conquer(left, right);
    }

    /**
     * @param {ListNode} l1
     * @param  {ListNode} l2
     * @return {ListNode}
     */
    conquer(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;
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */

public class Solution {
    public ListNode MergeKLists(ListNode[] lists) {
        if (lists == null || lists.Length == 0) {
            return null;
        }
        return Divide(lists, 0, lists.Length - 1);
    }

    private ListNode Divide(ListNode[] lists, int l, int r) {
        if (l > r) {
            return null;
        }
        if (l == r) {
            return lists[l];
        }

        int mid = l + (r - l) / 2;
        ListNode left = Divide(lists, l, mid);
        ListNode right = Divide(lists, mid + 1, r);

        return Conquer(left, right);
    }

    private ListNode Conquer(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;
        }

        if (l1 != null) {
            curr.next = l1;
        } else {
            curr.next = l2;
        }

        return dummy.next;
    }
}

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeKLists(lists []*ListNode) *ListNode {
    if len(lists) == 0 {
        return nil
    }
    return divide(lists, 0, len(lists)-1)
}

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

    mid := left + (right-left)/2
    l1 := divide(lists, left, mid)
    l2 := divide(lists, mid+1, right)

    return conquer(l1, l2)
}

func conquer(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
}

/**
 * Example:
 * var li = ListNode(5)
 * var v = li.`val`
 * Definition for singly-linked list.
 * class ListNode(var `val`: Int) {
 *     var next: ListNode? = null
 * }
 */
class Solution {
    fun mergeKLists(lists: Array<ListNode?>): ListNode? {
        if (lists.isEmpty()) return null
        return divide(lists, 0, lists.size - 1)
    }

    private fun divide(lists: Array<ListNode?>, left: Int, right: Int): ListNode? {
        if (left > right) return null
        if (left == right) return lists[left]

        val mid = left + (right - left) / 2
        val l1 = divide(lists, left, mid)
        val l2 = divide(lists, mid + 1, right)

        return conquer(l1, l2)
    }

    private fun conquer(l1: ListNode?, l2: ListNode?): ListNode? {
        val dummy = ListNode(0)
        var curr = dummy
        var list1 = l1
        var list2 = l2

        while (list1 != null && list2 != null) {
            if (list1.`val` <= list2.`val`) {
                curr.next = list1
                list1 = list1.next
            } else {
                curr.next = list2
                list2 = list2.next
            }
            curr = curr.next!!
        }

        curr.next = list1 ?: list2
        return dummy.next
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init() { self.val = 0; self.next = nil; }
 *     public init(_ val: Int) { self.val = val; self.next = nil; }
 *     public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
 * }
 */
class Solution {
    func mergeKLists(_ lists: [ListNode?]) -> ListNode? {
        if lists.isEmpty {
            return nil
        }
        return divide(lists, 0, lists.count - 1)
    }

    private func divide(_ lists: [ListNode?], _ l: Int, _ r: Int) -> ListNode? {
        if l > r {
            return nil
        }
        if l == r {
            return lists[l]
        }

        let mid = l + (r - l) / 2
        let left = divide(lists, l, mid)
        let right = divide(lists, mid + 1, r)
        return conquer(left, right)
    }

    private func conquer(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
        let dummy = ListNode(0)
        var curr = dummy
        var l1 = l1, l2 = l2

        while let node1 = l1, let node2 = l2 {
            if node1.val <= node2.val {
                curr.next = node1
                l1 = node1.next
            } else {
                curr.next = node2
                l2 = node2.next
            }
            curr = curr.next!
        }

        curr.next = l1 ?? l2
        return dummy.next
    }
}

Time & Space Complexity

Time complexity: $O(n \log k)$
Space complexity: $O(\log k)$

Where $k$ is the total number of lists and $n$ is the total number of nodes across $k$ lists.

6. Divide And Conquer (Iteration)

Intuition

This is the same idea as divide and conquer, but done iteratively instead of using recursion.

We repeatedly merge the lists in pairs:

In one pass:
- Merge list 0 and list 1 -> get M0
- Merge list 2 and list 3 -> get M1
- Merge list 4 and list 5 -> get M2
- ... and so on.
After this pass, we have fewer lists (about half as many).
Repeat this process on the new list of merged lists until only one list remains.

Each pairwise merge is just the usual merge of two sorted linked lists.
By always merging lists two at a time, the total work is efficient and structured, similar to the merge step in merge sort.

Algorithm

If lists is empty, return null.
While the number of lists is greater than 1:
- Create an empty list mergedLists.
- Loop over lists in steps of 2:
  - Let l1 = lists[i].
  - Let l2 = lists[i + 1] if it exists, otherwise None.
  - Merge l1 and l2 using mergeList and append the result to mergedLists.
- Set lists = mergedLists.
When the loop ends, lists[0] is the fully merged sorted list. Return it.

Merging two lists (mergeList(l1, l2)):

Create a dummy node and a tail pointer pointing to it.
While both l1 and l2 are non-empty:
- Attach the smaller of l1 and l2 to tail.next.
- Move the chosen list's pointer forward.
- Move tail forward.
Attach any remaining nodes from l1 or l2 to tail.next.
Return dummy.next as the merged head.

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class Solution:

    def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
        if not lists or len(lists) == 0:
            return None

        while len(lists) > 1:
            mergedLists = []
            for i in range(0, len(lists), 2):
                l1 = lists[i]
                l2 = lists[i + 1] if (i + 1) < len(lists) else None
                mergedLists.append(self.mergeList(l1, l2))
            lists = mergedLists
        return lists[0]

    def mergeList(self, l1, l2):
        dummy = ListNode()
        tail = dummy

        while l1 and l2:
            if l1.val < l2.val:
                tail.next = l1
                l1 = l1.next
            else:
                tail.next = l2
                l2 = l2.next
            tail = tail.next

        if l1:
            tail.next = l1
        if l2:
            tail.next = l2

        return dummy.next

/**
 * 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; }
 * }
 */

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

        while (lists.length > 1) {
            List<ListNode> mergedLists = new ArrayList<>();
            for (int i = 0; i < lists.length; i += 2) {
                ListNode l1 = lists[i];
                ListNode l2 = (i + 1) < lists.length ? lists[i + 1] : null;
                mergedLists.add(mergeList(l1, l2));
            }
            lists = mergedLists.toArray(new ListNode[0]);
        }
        return lists[0];
    }

    private ListNode mergeList(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode();
        ListNode tail = dummy;

        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                tail.next = l1;
                l1 = l1.next;
            } else {
                tail.next = l2;
                l2 = l2.next;
            }
            tail = tail.next;
        }

        if (l1 != null) {
            tail.next = l1;
        }
        if (l2 != null) {
            tail.next = l2;
        }

        return dummy.next;
    }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */

class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        if (lists.empty()) {
            return nullptr;
        }

        while (lists.size() > 1) {
            vector<ListNode*> mergedLists;
            for (int i = 0; i < lists.size(); i += 2) {
                ListNode* l1 = lists[i];
                ListNode* l2 = (i + 1) < lists.size() ? lists[i + 1] : nullptr;
                mergedLists.push_back(mergeList(l1, l2));
            }
            lists = mergedLists;
        }
        return lists[0];
    }

private:
    ListNode* mergeList(ListNode* l1, ListNode* l2) {
        ListNode dummy;
        ListNode* tail = &dummy;

        while (l1 && l2) {
            if (l1->val < l2->val) {
                tail->next = l1;
                l1 = l1->next;
            } else {
                tail->next = l2;
                l2 = l2->next;
            }
            tail = tail->next;
        }

        if (l1) {
            tail->next = l1;
        }
        if (l2) {
            tail->next = l2;
        }

        return dummy.next;
    }
};

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     constructor(val = 0, next = null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */

class Solution {
    /**
     * @param {ListNode[]} lists
     * @return {ListNode}
     */
    mergeKLists(lists) {
        if (!lists || lists.length === 0) {
            return null;
        }

        while (lists.length > 1) {
            const mergedLists = [];
            for (let i = 0; i < lists.length; i += 2) {
                const l1 = lists[i];
                const l2 = i + 1 < lists.length ? lists[i + 1] : null;
                mergedLists.push(this.mergeList(l1, l2));
            }
            lists = mergedLists;
        }
        return lists[0];
    }

    /**
     * @param {ListNode} l1
     * @param  {ListNode} l2
     * @return {ListNode}
     */
    mergeList(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;
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */

public class Solution {
    public ListNode MergeKLists(ListNode[] lists) {
        if (lists == null || lists.Length == 0) {
            return null;
        }

        while (lists.Length > 1) {
            List<ListNode> mergedLists = new List<ListNode>();
            for (int i = 0; i < lists.Length; i += 2) {
                ListNode l1 = lists[i];
                ListNode l2 = (i + 1) < lists.Length ? lists[i + 1] : null;
                mergedLists.Add(MergeList(l1, l2));
            }
            lists = mergedLists.ToArray();
        }
        return lists[0];
    }

    private ListNode MergeList(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode();
        ListNode tail = dummy;

        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                tail.next = l1;
                l1 = l1.next;
            } else {
                tail.next = l2;
                l2 = l2.next;
            }
            tail = tail.next;
        }

        if (l1 != null) {
            tail.next = l1;
        }
        if (l2 != null) {
            tail.next = l2;
        }

        return dummy.next;
    }
}

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeKLists(lists []*ListNode) *ListNode {
    if len(lists) == 0 {
        return nil
    }

    for len(lists) > 1 {
        var mergedLists []*ListNode
        for i := 0; i < len(lists); i += 2 {
            l1 := lists[i]
            var l2 *ListNode
            if i+1 < len(lists) {
                l2 = lists[i+1]
            }
            mergedLists = append(mergedLists, mergeList(l1, l2))
        }
        lists = mergedLists
    }
    return lists[0]
}

func mergeList(l1, l2 *ListNode) *ListNode {
    dummy := &ListNode{}
    tail := dummy

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

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

    return dummy.Next
}

/**
 * Example:
 * var li = ListNode(5)
 * var v = li.`val`
 * Definition for singly-linked list.
 * class ListNode(var `val`: Int) {
 *     var next: ListNode? = null
 * }
 */
class Solution {
    fun mergeKLists(lists: Array<ListNode?>): ListNode? {
        if (lists.isEmpty()) return null

        var currentLists = lists.toList()
        while (currentLists.size > 1) {
            val mergedLists = mutableListOf<ListNode?>()
            for (i in currentLists.indices step 2) {
                val l1 = currentLists[i]
                val l2 = if (i + 1 < currentLists.size) currentLists[i + 1] else null
                mergedLists.add(mergeList(l1, l2))
            }
            currentLists = mergedLists
        }
        return currentLists[0]
    }

    private fun mergeList(l1: ListNode?, l2: ListNode?): ListNode? {
        val dummy = ListNode(0)
        var tail = dummy
        var list1 = l1
        var list2 = l2

        while (list1 != null && list2 != null) {
            if (list1.`val` < list2.`val`) {
                tail.next = list1
                list1 = list1.next
            } else {
                tail.next = list2
                list2 = list2.next
            }
            tail = tail.next!!
        }

        tail.next = list1 ?: list2
        return dummy.next
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init() { self.val = 0; self.next = nil; }
 *     public init(_ val: Int) { self.val = val; self.next = nil; }
 *     public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
 * }
 */
class Solution {
    func mergeKLists(_ lists: [ListNode?]) -> ListNode? {
        if lists.isEmpty {
            return nil
        }

        var lists = lists

        while lists.count > 1 {
            var mergedLists: [ListNode?] = []

            for i in stride(from: 0, to: lists.count, by: 2) {
                let l1 = lists[i]
                let l2 = i + 1 < lists.count ? lists[i + 1] : nil
                mergedLists.append(mergeList(l1, l2))
            }

            lists = mergedLists
        }

        return lists[0]
    }

    private func mergeList(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
        let dummy = ListNode(0)
        var tail = dummy
        var l1 = l1, l2 = l2

        while let node1 = l1, let node2 = l2 {
            if node1.val < node2.val {
                tail.next = node1
                l1 = node1.next
            } else {
                tail.next = node2
                l2 = node2.next
            }
            tail = tail.next!
        }

        tail.next = l1 ?? l2
        return dummy.next
    }
}

Time & Space Complexity

Time complexity: $O(n \log k)$
Space complexity: $O(k)$

Where $k$ is the total number of lists and $n$ is the total number of nodes across $k$ lists.

Common Pitfalls

Not Handling Empty Lists in the Input Array

The input array lists may contain null or empty linked lists. Failing to check for these before accessing node values will cause null pointer exceptions. Always verify that a list is non-empty before processing it.

Forgetting to Advance the Chosen List's Pointer

After selecting the smallest node from among the k lists, you must move that list's head pointer to the next node. Forgetting this step causes an infinite loop where the same node is repeatedly selected.

Incorrect Comparator for Min-Heap

When using a priority queue or min-heap, the comparator must compare node values correctly. In some languages, the default heap is a max-heap (like Python's heapq with negative values or C++'s priority_queue). Using the wrong comparison direction results in a max-heap instead of a min-heap, producing an incorrectly sorted output.

Not Using a Dummy Node for the Result List

Building the result list without a dummy head node requires special handling for the first node and complicates the logic. Using a dummy node simplifies the code by allowing uniform treatment of all nodes, then returning dummy.next as the result.

Modifying the Input Lists Array During Iteration

When merging lists one by one or in the divide-and-conquer approach, be careful about how you update the lists array. Overwriting elements while still iterating can lead to incorrect merges or skipped lists. Either use a separate array for merged results or iterate carefully with proper indexing.