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

Given k linked lists each of size n and each list is sorted in non-decreasing order, merge them into a single sorted (non-decreasing order) linked list and print the sorted linked list as 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.

Source: Merge K sorted Linked Lists | Method 2

An efficient solution for the problem has been discussed in Method 3 of this post.

Approach: This solution is based on the MIN HEAP approach used to solve the problem ‘merge k sorted arrays’ which is discussed here.
MinHeap: A Min-Heap is a complete binary tree in which the value in each internal node is smaller than or equal to the values in the children of that node. Mapping the elements of a heap into an array is trivial: if a node is stored at index k, then its left child is stored at index 2k + 1 and its right child at index 2k + 2.

1. Create a min-heap and insert the first element of all the ‘k’ linked lists.
2. As long as the min-heap is not empty, perform the following steps: 
   • Remove the top element of the min-heap (which is the current minimum among all the elements in the min-heap) and add it to the result list.
   • If there exists an element (in the same linked list) next to the element popped out in previous step, insert it into the min-heap.
3. Return the head node address of the merged list.

Below is the implementation of the above approach:

C++

// C++ implementation to merge k
// sorted linked lists
// | Using MIN HEAP method
#include <bits/stdc++.h>
using namespace std;

struct Node
{
    int data;
    struct Node* next;
};

// Utility function to create
// a new node
struct Node* newNode(int data)
{
    // Allocate node
    struct Node* new_node = new Node();

    // Put in the data
    new_node->data = data;
    new_node->next = NULL;

    return new_node;
}

// 'compare' function used to build
// up the priority queue
struct compare
{
    bool operator()(
         struct Node* a, struct Node* b)
    {
        return a->data > b->data;
    }
};

// Function to merge k sorted linked lists
struct Node* mergeKSortedLists(
             struct Node* arr[], int k)
{
    // Priority_queue 'pq' implemented
    // as min heap with the help of
    // 'compare' function
    priority_queue<Node*, vector<Node*>, compare> pq;

    // Push the head nodes of all
    // the k lists in 'pq'
    for (int i = 0; i < k; i++)
        if (arr[i] != NULL)
            pq.push(arr[i]);

      // Handles the case when k = 0
      // or lists have no elements in them
      if (pq.empty())
        return NULL;

      struct Node *dummy = newNode(0);
      struct Node *last = dummy;

    // Loop till 'pq' is not empty
    while (!pq.empty())
    {
        // Get the top element of 'pq'
        struct Node* curr = pq.top();
        pq.pop();

        // Add the top element of 'pq'
        // to the resultant merged list
        last->next = curr;
        last = last->next;

        // Check if there is a node
         // next to the 'top' node
        // in the list of which 'top'
        // node is a member
        if (curr->next != NULL)

        // Push the next node of top node
        // in 'pq'
        pq.push(curr->next);
    }

    // Address of head node of the required
    // merged list
    return dummy->next;
}

// Function to print the singly
// linked list
void printList(struct Node* head)
{
    while (head != NULL)
    {
        cout << head->data << " ";
        head = head->next;
    }
}

// Driver code
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];

    // Creating k = 3 sorted lists
    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 the k sorted lists
    struct Node* head = mergeKSortedLists(arr, k);

    // Print the merged list
    printList(head);

    return 0;
}

Java

/*package whatever //do not write package name here */

import java.io.*;
import java.util.*;

class Node {
    int data;
    Node next;

    Node(int key)
    {
        data = key;
        next = null;
    }
}
// Class implements Comparator to compare Node data
class NodeComparator implements Comparator<Node> {

    public int compare(Node k1, Node k2)
    {
        if (k1.data > k2.data)
            return 1;
        else if (k1.data < k2.data)
            return -1;
        return 0;
    }
}
class GFG {
    // Function to merge k sorted linked lists
    static Node mergeKList(Node[] arr, int K)
    {
        // Priority_queue 'queue' implemented
        // as min heap with the help of
        // 'compare' function
        PriorityQueue<Node> queue
            = new PriorityQueue<>(new NodeComparator());
        Node at[] = new Node[K];
        Node head = new Node(0);
        Node last = head;
        // Push the head nodes of all
        // the k lists in 'queue'
        for (int i = 0; i < K; i++) {
            if (arr[i] != null) {
                queue.add(arr[i]);
            }
        }
        // Handles the case when k = 0
        // or lists have no elements in them
        if (queue.isEmpty())
            return null;
        // Loop till 'queue' is not empty
        while (!queue.isEmpty()) {
            // Get the top element of 'queue'
            Node curr = queue.poll();

            // Add the top element of 'queue'
            // to the resultant merged list
            last.next = curr;
            last = last.next;
            // Check if there is a node
            // next to the 'top' node
            // in the list of which 'top'
            // node is a member
            if (curr.next != null) {
                // Push the next node of top node
                // in 'queue'
                queue.add(curr.next);
            }
        }
        // Address of head node of the required
        // merged list
        return head.next;
    }
    // Print 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 N = 4;
        // array to store head of linkedlist
        Node[] a = new Node[N];
        // Linkedlist1
        Node head1 = new Node(1);
        a[0] = head1;
        head1.next = new Node(2);
        head1.next.next = new Node(3);
        // Limkedlist2
        Node head2 = new Node(4);
        a[1] = head2;
        head2.next = new Node(5);
        // Linkedlist3
        Node head3 = new Node(5);
        a[2] = head3;
        head3.next = new Node(6);
        // Linkedlist4
        Node head4 = new Node(7);
        a[3] = head4;
        head4.next = new Node(8);

        Node res = mergeKList(a, N);

        if (res != null)
            printList(res);
        System.out.println();
    }
}

Output:

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

Complexity Analysis:

• Time Complexity: O(N * log k) or O(n * k * log k), where, ‘N’ is the total number of elements among all the linked lists, ‘k’ is the total number of lists, and ‘n’ is the size of each linked list.
  Insertion and deletion operation will be performed in min-heap for all N nodes.
  Insertion and deletion in a min-heap require log k time.
• Auxiliary Space: O(k).
  The priority queue will have atmost ‘k’ number of elements at any point of time, hence the additional space required for our algorithm is O(k).

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