Merge K sorted linked lists | Set 1

Given K sorted linked lists of size N each, merge them and print the sorted output.

Examples:

Input: k = 3, n =  4
list1 = 1->3->5->7->NULL
list2 = 2->4->6->8->NULL
list3 = 0->9->10->11->NULL

Output: 0->1->2->3->4->5->6->7->8->9->10->11
Merged lists in a sorted order 
where every element is greater 
than the previous element.

Input: k = 3, n =  3
list1 = 1->3->7->NULL
list2 = 2->4->8->NULL
list3 = 9->10->11->NULL

Output: 1->2->3->4->7->8->9->10->11
Merged lists in a sorted order 
where every element is greater 
than the previous element.

Method 1 (Simple):

Approach: A Simple Solution is to initialize the result as the first list. Now traverse all lists starting from the second list. Insert every node of the currently traversed list into result in a sorted way.  

Implementation:

C++

// C++ program to merge k sorted
// arrays of size n each
#include <bits/stdc++.h>
using namespace std;

// A Linked List node
struct Node {
    int data;
    Node* next;
};

/* Function to print nodes in
   a given linked list */
void printList(Node* node)
{
    while (node != NULL) {
        printf("%d ", node->data);
        node = node->next;
    }
}

// The main function that
// takes an array of lists
// arr[0..last] and generates
// the sorted output
Node* mergeKLists(Node* arr[], int last)
{

    // Traverse form second list to last
    for (int i = 1; i <= last; i++) {
        while (true) {
            // head of both the lists,
            // 0 and ith list.
            Node *head_0 = arr[0], *head_i = arr[i];

            // Break if list ended
            if (head_i == NULL)
                break;

            // Smaller than first element
            if (head_0->data >= head_i->data) {
                arr[i] = head_i->next;
                head_i->next = head_0;
                arr[0] = head_i;
            }
            else
                // Traverse the first list
                while (head_0->next != NULL) {
                    // Smaller than next element
                    if (head_0->next->data
                        >= head_i->data) {
                        arr[i] = head_i->next;
                        head_i->next = head_0->next;
                        head_0->next = head_i;
                        break;
                    }
                    // go to next node
                    head_0 = head_0->next;

                    // if last node
                    if (head_0->next == NULL) {
                        arr[i] = head_i->next;
                        head_i->next = NULL;
                        head_0->next = head_i;
                        head_0->next->next = NULL;
                        break;
                    }
                }
        }
    }

    return arr[0];
}

// Utility function to create a new node.
Node* newNode(int data)
{
    struct Node* temp = new Node;
    temp->data = data;
    temp->next = NULL;
    return temp;
}

// Driver program to test
// above functions
int main()
{
    // Number of linked lists
    int k = 3;

    // Number of elements in each list
    int n = 4;

    // an array of pointers storing the
    // head nodes of the linked lists
    Node* arr[k];

    arr[0] = newNode(1);
    arr[0]->next = newNode(3);
    arr[0]->next->next = newNode(5);
    arr[0]->next->next->next = newNode(7);

    arr[1] = newNode(2);
    arr[1]->next = newNode(4);
    arr[1]->next->next = newNode(6);
    arr[1]->next->next->next = newNode(8);

    arr[2] = newNode(0);
    arr[2]->next = newNode(9);
    arr[2]->next->next = newNode(10);
    arr[2]->next->next->next = newNode(11);

    // Merge all lists
    Node* head = mergeKLists(arr, k - 1);

    printList(head);

    return 0;
}

Java

// Java program to merge k sorted
// arrays of size n each
import java.io.*;

// A Linked List node
class Node
{
  int data;
  Node next;

  // Utility function to create a new node.
  Node(int key)
  {
    data = key;
    next = null;
  }
}
class GFG {

  static Node head;
  static Node temp;

  /* Function to print nodes in
   a given linked list */
  static void printList(Node node)
  {
    while(node != null)
    {
      System.out.print(node.data + " ");

      node = node.next;
    }
    System.out.println();
  }

  // The main function that
  // takes an array of lists
  // arr[0..last] and generates
  // the sorted output
  static Node mergeKLists(Node arr[], int last)
  {

    // Traverse form second list to last
    for (int i = 1; i <= last; i++)
    {
      while(true)
      {

        // head of both the lists,
        // 0 and ith list.
        Node head_0 = arr[0];
        Node head_i = arr[i];

        // Break if list ended
        if (head_i == null)
          break;

        // Smaller than first element
        if(head_0.data >= head_i.data)
        {
          arr[i] = head_i.next;
          head_i.next = head_0;
          arr[0] = head_i;
        }
        else
        {

          // Traverse the first list
          while (head_0.next != null)
          {

            // Smaller than next element
            if (head_0.next.data >= head_i.data)
            {
              arr[i] = head_i.next;
              head_i.next = head_0.next;
              head_0.next = head_i;
              break;
            }

            // go to next node
            head_0 = head_0.next;

            // if last node
            if (head_0.next == null)
            {
              arr[i] = head_i.next;
              head_i.next = null;
              head_0.next = head_i;
              head_0.next.next = null;
              break;
            }
          }
        }
      }
    }
    return arr[0];
  }

  // Driver program to test
  // above functions
  public static void main (String[] args)
  {

    // Number of linked lists
    int k = 3;

    // Number of elements in each list
    int n = 4;

    // an array of pointers storing the
    // head nodes of the linked lists

    Node[] arr = new Node[k];

    arr[0] = new Node(1);
    arr[0].next = new Node(3);
    arr[0].next.next = new Node(5);
    arr[0].next.next.next = new Node(7);

    arr[1] = new Node(2);
    arr[1].next = new Node(4);
    arr[1].next.next = new Node(6);
    arr[1].next.next.next = new Node(8);

    arr[2] = new Node(0);
    arr[2].next = new Node(9);
    arr[2].next.next = new Node(10);
    arr[2].next.next.next = new Node(11);

    // Merge all lists
    head = mergeKLists(arr, k - 1);
    printList(head);

  }
}

// This code is contributed by avanitrachhadiya2155

Python3

# Python3 program to merge k
# sorted arrays of size n each

# A Linked List node
class Node:

    def __init__(self, x):

        self.data = x
        self.next = None

# Function to print nodes in
# a given linked list
def printList(node):

    while (node != None):
        print(node.data,
              end = " ")
        node = node.next

# The main function that
# takes an array of lists
# arr[0..last] and generates
# the sorted output
def mergeKLists(arr, last):

    # Traverse form second
    # list to last
    for i in range(1, last + 1):
        while (True):
            # head of both the lists,
            # 0 and ith list.
            head_0 = arr[0]
            head_i = arr[i]

            # Break if list ended
            if (head_i == None):
                break

            # Smaller than first
            # element
            if (head_0.data >=
                head_i.data):
                arr[i] = head_i.next
                head_i.next = head_0
                arr[0] = head_i
            else:
                # Traverse the first list
                while (head_0.next != None):
                    # Smaller than next
                    # element
                    if (head_0.next.data >=
                        head_i.data):
                        arr[i] = head_i.next
                        head_i.next = head_0.next
                        head_0.next = head_i
                        break
                    # go to next node
                    head_0 = head_0.next
                    # if last node
                    if (head_0.next == None):
                        arr[i] = head_i.next
                        head_i.next = None
                        head_0.next = head_i
                        head_0.next.next = None
                        break
    return arr[0]

# Driver code
if __name__ == '__main__':

    # Number of linked
    # lists
    k = 3

    # Number of elements
    # in each list
    n = 4

    # an array of pointers
    # storing the head nodes
    # of the linked lists
    arr = [None for i in range(k)]

    arr[0] = Node(1)
    arr[0].next = Node(3)
    arr[0].next.next = Node(5)
    arr[0].next.next.next = Node(7)

    arr[1] = Node(2)
    arr[1].next = Node(4)
    arr[1].next.next = Node(6)
    arr[1].next.next.next = Node(8)

    arr[2] = Node(0)
    arr[2].next = Node(9)
    arr[2].next.next = Node(10)
    arr[2].next.next.next = Node(11)

    # Merge all lists
    head = mergeKLists(arr, k - 1)

    printList(head)

# This code is contributed by Mohit Kumar 29

C#

// C# program to merge k sorted
// arrays of size n each
using System;

// A Linked List node
public class Node
{
  public int data;
  public Node next;

  // Utility function to create a new node.
  public Node(int key)
  {
    data = key;
    next = null;
  }
}

public class GFG
{
  static Node head;

  /* Function to print nodes in
   a given linked list */
  static void printList(Node node)
  {
    while(node != null)
    {
      Console.Write(node.data + " ");
      node = node.next;
    }
    Console.WriteLine();
  }

  // The main function that
  // takes an array of lists
  // arr[0..last] and generates
  // the sorted output
  static Node mergeKLists(Node[] arr, int last)
  {

    // Traverse form second list to last
    for (int i = 1; i <= last; i++)
    {
      while(true)
      {

        // head of both the lists,
        // 0 and ith list.
        Node head_0 = arr[0];
        Node head_i = arr[i];

        // Break if list ended
        if (head_i == null)
          break;

        // Smaller than first element
        if(head_0.data >= head_i.data)
        {
          arr[i] = head_i.next;
          head_i.next = head_0;
          arr[0] = head_i;
        }
        else
        {

          // Traverse the first list
          while (head_0.next != null)
          {

            // Smaller than next element
            if (head_0.next.data >= head_i.data)
            {
              arr[i] = head_i.next;
              head_i.next = head_0.next;
              head_0.next = head_i;
              break;
            }

            // go to next node
            head_0 = head_0.next;

            // if last node
            if (head_0.next == null)
            {
              arr[i] = head_i.next;
              head_i.next = null;
              head_0.next = head_i;
              head_0.next.next = null;
              break;
            }
          }
        }
      }
    }
    return arr[0];
  }
  static public void Main ()
  {

    // Number of linked lists
    int k = 3;

    // an array of pointers storing the
    // head nodes of the linked lists
    Node[] arr = new Node[k];

    arr[0] = new Node(1);
    arr[0].next = new Node(3);
    arr[0].next.next = new Node(5);
    arr[0].next.next.next = new Node(7);

    arr[1] = new Node(2);
    arr[1].next = new Node(4);
    arr[1].next.next = new Node(6);
    arr[1].next.next.next = new Node(8);

    arr[2] = new Node(0);
    arr[2].next = new Node(9);
    arr[2].next.next = new Node(10);
    arr[2].next.next.next = new Node(11);

    // Merge all lists
    head = mergeKLists(arr, k - 1);
    printList(head);
  }
}

// This code is contributed by rag2127

Javascript

<script>
// Javascript program to merge k sorted
// arrays of size n each

    // A Linked List node
    class Node
    {
        // Utility function to create a new node.
        constructor(key)
        {
            this.data=key;
            this.next=null;
        }
    }

    let head;
    let temp;

/* Function to print nodes in
a given linked list */
function printList(node)
{
    while(node != null)
    {
      document.write(node.data + " ");

      node = node.next;
    }
    document.write("<br>");

}

// The main function that
// takes an array of lists
// arr[0..last] and generates
// the sorted output
function mergeKLists(arr,last)
{
    // Traverse form second list to last
    for (let i = 1; i <= last; i++)
    {
      while(true)
      {

        // head of both the lists,
        // 0 and ith list.
        let head_0 = arr[0];
        let head_i = arr[i];

        // Break if list ended
        if (head_i == null)
          break;

        // Smaller than first element
        if(head_0.data >= head_i.data)
        {
          arr[i] = head_i.next;
          head_i.next = head_0;
          arr[0] = head_i;
        }
        else
        {

          // Traverse the first list
          while (head_0.next != null)
          {

            // Smaller than next element
            if (head_0.next.data >= head_i.data)
            {
              arr[i] = head_i.next;
              head_i.next = head_0.next;
              head_0.next = head_i;
              break;
            }

            // go to next node
            head_0 = head_0.next;

            // if last node
            if (head_0.next == null)
            {
              arr[i] = head_i.next;
              head_i.next = null;
              head_0.next = head_i;
              head_0.next.next = null;
              break;
            }
          }
        }
      }
    }
    return arr[0];
}

// Driver program to test
// above functions
// Number of linked lists
let k = 3;
// Number of elements in each list
let n = 4;
// an array of pointers storing the
// head nodes of the linked lists
let  arr = new Array(k);
arr[0] = new Node(1);
arr[0].next = new Node(3);
arr[0].next.next = new Node(5);
arr[0].next.next.next = new Node(7);

arr[1] = new Node(2);
arr[1].next = new Node(4);
arr[1].next.next = new Node(6);
arr[1].next.next.next = new Node(8);

arr[2] = new Node(0);
arr[2].next = new Node(9);
arr[2].next.next = new Node(10);
arr[2].next.next.next = new Node(11);

// Merge all lists
head = mergeKLists(arr, k - 1);
printList(head);



// This code is contributed by unknown2108

</script>

Output

0 1 2 3 4 5 6 7 8 9 10 11 

Complexity Analysis:

• Time complexity: O(nk²)
• Auxiliary Space: O(1). 
  As no extra space is required.

Method 2: Min Heap. 

A Better solution is to use Min Heap-based solution which is discussed here for arrays. The time complexity of this solution would be O(nk Log k)

Method 3: Divide and Conquer. 

In this post, Divide and Conquer approach is discussed. This approach doesn’t require extra space for heap and works in O(nk Log k)
It is known that merging of two linked lists can be done in O(n) time and O(n) space. 

1. The idea is to pair up K lists and merge each pair in linear time using O(n) space.
2. After the first cycle, K/2 lists are left each of size 2*N. After the second cycle, K/4 lists are left each of size 4*N and so on.
3. Repeat the procedure until we have only one list left.

Below is the implementation of the above idea. 

C++

// C++ program to merge k sorted
// arrays of size n each
#include <bits/stdc++.h>
using namespace std;

// A Linked List node
struct Node {
    int data;
    Node* next;
};

/* Function to print nodes in
   a given linked list */
void printList(Node* node)
{
    while (node != NULL) {
        printf("%d ", node->data);
        node = node->next;
    }
}

/* Takes two lists sorted in increasing order, and merge
   their nodes together to make one big sorted list. Below
   function takes O(n) extra space for recursive calls,
    */
Node* SortedMerge(Node* a, Node* b)
{
    Node* result = NULL;

    /* Base cases */
    if (a == NULL)
        return (b);
    else if (b == NULL)
        return (a);

    /* Pick either a or b, and recur */
    if (a->data <= b->data) {
        result = a;
        result->next = SortedMerge(a->next, b);
    }
    else {
        result = b;
        result->next = SortedMerge(a, b->next);
    }

    return result;
}

// The main function that takes an array of lists
// arr[0..last] and generates the sorted output
Node* mergeKLists(Node* arr[], int last)
{
    // repeat until only one list is left
    while (last != 0) {
        int i = 0, j = last;

        // (i, j) forms a pair
        while (i < j) {
            // merge List i with List j and store
            // merged list in List i
            arr[i] = SortedMerge(arr[i], arr[j]);

            // consider next pair
            i++, j--;

            // If all pairs are merged, update last
            if (i >= j)
                last = j;
        }
    }

    return arr[0];
}

// Utility function to create a new node.
Node* newNode(int data)
{
    struct Node* temp = new Node;
    temp->data = data;
    temp->next = NULL;
    return temp;
}

// Driver program to test above functions
int main()
{
    int k = 3; // Number of linked lists
    int n = 4; // Number of elements in each list

    // an array of pointers storing the head nodes
    // of the linked lists
    Node* arr[k];

    arr[0] = newNode(1);
    arr[0]->next = newNode(3);
    arr[0]->next->next = newNode(5);
    arr[0]->next->next->next = newNode(7);

    arr[1] = newNode(2);
    arr[1]->next = newNode(4);
    arr[1]->next->next = newNode(6);
    arr[1]->next->next->next = newNode(8);

    arr[2] = newNode(0);
    arr[2]->next = newNode(9);
    arr[2]->next->next = newNode(10);
    arr[2]->next->next->next = newNode(11);

    // Merge all lists
    Node* head = mergeKLists(arr, k - 1);

    printList(head);

    return 0;
}

Java

// Java program to merge k sorted arrays of size n each
public class MergeKSortedLists {

    /* Takes two lists sorted in increasing order, and merge
    their nodes together to make one big sorted list. Below
    function takes O(Log n) extra space for recursive calls,
    but it can be easily modified to work with same time and
    O(1) extra space  */
    public static Node SortedMerge(Node a, Node b)
    {
        Node result = null;
        /* Base cases */
        if (a == null)
            return b;
        else if (b == null)
            return a;

        /* Pick either a or b, and recur */
        if (a.data <= b.data) {
            result = a;
            result.next = SortedMerge(a.next, b);
        }
        else {
            result = b;
            result.next = SortedMerge(a, b.next);
        }

        return result;
    }

    // The main function that takes an array of lists
    // arr[0..last] and generates the sorted output
    public static Node mergeKLists(Node arr[], int last)
    {
        // repeat until only one list is left
        while (last != 0) {
            int i = 0, j = last;

            // (i, j) forms a pair
            while (i < j) {
                // merge List i with List j and store
                // merged list in List i
                arr[i] = SortedMerge(arr[i], arr[j]);

                // consider next pair
                i++;
                j--;

                // If all pairs are merged, update last
                if (i >= j)
                    last = j;
            }
        }

        return arr[0];
    }

    /* Function to print nodes in a given linked list */
    public static void printList(Node node)
    {
        while (node != null) {
            System.out.print(node.data + " ");
            node = node.next;
        }
    }

    public static void main(String args[])
    {
        int k = 3; // Number of linked lists
        int n = 4; // Number of elements in each list

        // an array of pointers storing the head nodes
        // of the linked lists
        Node arr[] = new Node[k];

        arr[0] = new Node(1);
        arr[0].next = new Node(3);
        arr[0].next.next = new Node(5);
        arr[0].next.next.next = new Node(7);

        arr[1] = new Node(2);
        arr[1].next = new Node(4);
        arr[1].next.next = new Node(6);
        arr[1].next.next.next = new Node(8);

        arr[2] = new Node(0);
        arr[2].next = new Node(9);
        arr[2].next.next = new Node(10);
        arr[2].next.next.next = new Node(11);

        // Merge all lists
        Node head = mergeKLists(arr, k - 1);
        printList(head);
    }
}

class Node {
    int data;
    Node next;
    Node(int data)
    {
        this.data = data;
    }
}
// This code is contributed by Gaurav Tiwari

Python3

# Python3 program to merge k sorted
# arrays of size n each

# A Linked List node
class Node:

    def __init__(self):

        self.data = 0
        self.next = None

# Function to print nodes in a
# given linked list
def printList(node):

    while (node != None):
        print(node.data, end = ' ')
        node = node.next

# Takes two lists sorted in increasing order,
# and merge their nodes together to make one
# big sorted list. Below function takes
# O(Log n) extra space for recursive calls,
# but it can be easily modified to work with
# same time and O(1) extra space
def SortedMerge(a, b):

    result = None

    # Base cases
    if (a == None):
        return(b)
    elif (b == None):
        return(a)

    # Pick either a or b, and recur
    if (a.data <= b.data):
        result = a
        result.next = SortedMerge(a.next, b)
    else:
        result = b
        result.next = SortedMerge(a, b.next)

    return result

# The main function that takes an array of lists
# arr[0..last] and generates the sorted output
def mergeKLists(arr, last):

    # Repeat until only one list is left
    while (last != 0):
        i = 0
        j = last

        # (i, j) forms a pair
        while (i < j):

            # Merge List i with List j and store
            # merged list in List i
            arr[i] = SortedMerge(arr[i], arr[j])

            # Consider next pair
            i += 1
            j -= 1

            # If all pairs are merged, update last
            if (i >= j):
                last = j

    return arr[0]

# Utility function to create a new node.
def newNode(data):

    temp = Node()
    temp.data = data
    temp.next = None
    return temp

# Driver code
if __name__=='__main__':

    # Number of linked lists
    k = 3

    # Number of elements in each list
    n = 4

    # An array of pointers storing the
    # head nodes of the linked lists
    arr = [0 for i in range(k)]

    arr[0] = newNode(1)
    arr[0].next = newNode(3)
    arr[0].next.next = newNode(5)
    arr[0].next.next.next = newNode(7)

    arr[1] = newNode(2)
    arr[1].next = newNode(4)
    arr[1].next.next = newNode(6)
    arr[1].next.next.next = newNode(8)

    arr[2] = newNode(0)
    arr[2].next = newNode(9)
    arr[2].next.next = newNode(10)
    arr[2].next.next.next = newNode(11)

    # Merge all lists
    head = mergeKLists(arr, k - 1)

    printList(head)

# This code is contributed by rutvik_56

C#

// C# program to merge k sorted arrays of size n each
using System;

public class MergeKSortedLists {

    /* Takes two lists sorted in
    increasing order, and merge
    their nodes together to make
    one big sorted list. Below
    function takes O(Log n) extra
    space for recursive calls,
    but it can be easily modified
    to work with same time and
    O(1) extra space */
    public static Node SortedMerge(Node a, Node b)
    {
        Node result = null;

        /* Base cases */
        if (a == null)
            return b;
        else if (b == null)
            return a;

        /* Pick either a or b, and recur */
        if (a.data <= b.data) {
            result = a;
            result.next = SortedMerge(a.next, b);
        }
        else {
            result = b;
            result.next = SortedMerge(a, b.next);
        }

        return result;
    }

    // The main function that takes
    // an array of lists arr[0..last]
    // and generates the sorted output
    public static Node mergeKLists(Node[] arr, int last)
    {
        // repeat until only one list is left
        while (last != 0) {
            int i = 0, j = last;

            // (i, j) forms a pair
            while (i < j) {
                // merge List i with List j and store
                // merged list in List i
                arr[i] = SortedMerge(arr[i], arr[j]);

                // consider next pair
                i++;
                j--;

                // If all pairs are merged, update last
                if (i >= j)
                    last = j;
            }
        }

        return arr[0];
    }

    /* Function to print nodes in a given linked list */
    public static void printList(Node node)
    {
        while (node != null) {
            Console.Write(node.data + " ");
            node = node.next;
        }
    }

    public static void Main()
    {
        int k = 3; // Number of linked lists
        //int n = 4; // Number of elements in each list

        // An array of pointers storing the head nodes
        // of the linked lists
        Node[] arr = new Node[k];

        arr[0] = new Node(1);
        arr[0].next = new Node(3);
        arr[0].next.next = new Node(5);
        arr[0].next.next.next = new Node(7);

        arr[1] = new Node(2);
        arr[1].next = new Node(4);
        arr[1].next.next = new Node(6);
        arr[1].next.next.next = new Node(8);

        arr[2] = new Node(0);
        arr[2].next = new Node(9);
        arr[2].next.next = new Node(10);
        arr[2].next.next.next = new Node(11);

        // Merge all lists
        Node head = mergeKLists(arr, k - 1);
        printList(head);
    }
}

public class Node {
    public int data;
    public Node next;
    public Node(int data)
    {
        this.data = data;
    }
}

/* This code contributed by PrinciRaj1992 */

Javascript

<script>

// JavaScript program to merge k sorted
// arrays of size n each
class Node {
    constructor(val) {
        this.data = val;
        this.next = null;
    }
}

    /*
      Takes two lists sorted in increasing order,
     and merge their nodes together to
     make one big sorted list.
     Below function takes O(Log n) extra space for
     * recursive calls, but it can be easily modified
     to work with same time and
     * O(1) extra space
     */
    function SortedMerge(a,  b) {
var result = null;
        /* Base cases */
        if (a == null)
            return b;
        else if (b == null)
            return a;

        /* Pick either a or b, and recur */
        if (a.data <= b.data) {
            result = a;
            result.next = SortedMerge(a.next, b);
        } else {
            result = b;
            result.next = SortedMerge(a, b.next);
        }

        return result;
    }

    // The main function that takes an array of lists
    // arr[0..last] and generates the sorted output
    function mergeKLists(arr , last) {
        // repeat until only one list is left
        while (last != 0) {
            var i = 0, j = last;

            // (i, j) forms a pair
            while (i < j) {
                // merge List i with List j and store
                // merged list in List i
                arr[i] = SortedMerge(arr[i], arr[j]);

                // consider next pair
                i++;
                j--;

                // If all pairs are merged, update last
                if (i >= j)
                    last = j;
            }
        }

        return arr[0];
    }

    /* Function to print nodes in a given linked list */
    function printList(node) {
        while (node != null) {
            document.write(node.data + " ");
            node = node.next;
        }
    }


        var k = 3; // Number of linked lists
        var n = 4; // Number of elements in each list

        // an array of pointers storing the head nodes
        // of the linked lists
var arr = Array(k);

        arr[0] = new Node(1);
        arr[0].next = new Node(3);
        arr[0].next.next = new Node(5);
        arr[0].next.next.next = new Node(7);

        arr[1] = new Node(2);
        arr[1].next = new Node(4);
        arr[1].next.next = new Node(6);
        arr[1].next.next.next = new Node(8);

        arr[2] = new Node(0);
        arr[2].next = new Node(9);
        arr[2].next.next = new Node(10);
        arr[2].next.next.next = new Node(11);

        // Merge all lists
var head = mergeKLists(arr, k - 1);
        printList(head);

// This code contributed by gauravrajput1

</script>

Output

0 1 2 3 4 5 6 7 8 9 10 11 

Complexity Analysis:

     Assuming N(n*k) is the total number of nodes, n is the size of each linked list, and k is the total number of linked lists.

• Time Complexity: O(N*log k) or O(n*k*log k)
  As outer while loop in function mergeKLists() runs log k times and every time it processes n*k elements.
• Auxiliary Space: O(N) or O(n*k)
  Because recursion is used in SortedMerge() and to merge the final 2 linked lists of size N/2, N recursive calls will be made.

Merge k sorted linked lists | Set 2 (Using Min Heap)

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

 Method 4:   Selecting min of top element iteratively

Approach : Select the minimum of top elements iteratively store that in a new node and increment the pointer of the minimum element.

Implementation:

C++

// C++ program to merge k sorted arrays of size n each
#include <bits/stdc++.h>
using namespace std;

// A Linked List node
struct Node
{
    int data;
    Node* next;
    Node(int x)
    {
      data = x;
        next = NULL;
    }
};

/* Function to print nodes in a given linked list */
void printList(Node* node)
{
    while (node != NULL)
    {
        printf("%d ", node->data);
        node = node->next;
    }
    cout << endl;
}

/*Linked list Node structure

struct Node
{
int data;
Node* next;
Node(int x){
    data = x;
    next = NULL;
}
};
*/

class Solution
{
public:

    // Function to merge K sorted linked list.
    Node* mergeKLists(Node* arr[], int K)
    {
        Node* head = NULL;
        while (1)
        {
            int a = 0;
            int z;
            Node* curr;
            int min = INT_MAX;
            for (int i = 0; i < K; i++)
            {
                if (arr[i] != NULL)
                {
                    a++;
                    if (arr[i]->data < min)
                    {
                        min = arr[i]->data;
                        z = i;
                    }
                }
            }
            if (a != 0)
            {
                arr[z] = arr[z]->next;
                Node* temp = new Node(min);
                if (head == NULL)
                {
                    head = temp;
                    curr = temp;
                }
                else
                {
                    curr->next = temp;
                    curr = temp;
                }
            }
            else
            {
                return head;
            }
        }
    }
};

// { Driver Code Starts.

// Driver program to test above functions
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int N;
        cin >> N;
        struct Node* arr[N];
        for (int j = 0; j < N; j++)
        {
            int n;
            cin >> n;
            int x;
            cin >> x;
            arr[j] = new Node(x);
            Node* curr = arr[j];
            n--;
            for (int i = 0; i < n; i++)
            {
                cin >> x;
                Node* temp = new Node(x);
                curr->next = temp;
                curr = temp;
            }
        }
        Solution obj;
        Node* res = obj.mergeKLists(arr, N);
        printList(res);
    }
    return 0;
}

// } Driver Code Ends

Output

Time complexity: O(n*k²) 
Auxiliary Space: O(n)

Let’s take an example to understand the Time Complexity

arr[0] = 0->1->2->NULL

arr[1] = 3->4->5->NULL

arr[2] = 6->7->8->NULL

N = 3 K = 3

I have taken this example to make you understand the complexity in an easy way.

According to the approach, we find 0 as the minimum element in the first k nodes, to find this we will take K time. 

Similarly, we found 1 as a minimum in K time, this process takes runs for N nodes in the first list, So the time taken will be O(NK) for arr[0], after this, our minimum element come in the remaining two lists so our loop works for K-1 and for 2nd list time taken will be N*K-1. This process will continue till we merge all nodes of all lists.

So the equation can be 

NK + N*K-1 + N*K-2 + …… + N * 1

N(K+ K-1 + K-2 + … + 1)

Which is equal to O (N*K^2)