Java Program To 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.  

Java

// Java program to merge k sorted
// linked lists 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 code
  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

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. 

Java

// Java program to merge k sorted
// linked lists 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[])
    {
        // 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
        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

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.

Please refer complete article on Merge K sorted linked lists | Set 1 for more details!