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.

Create a min-heap and insert the first element of all the ‘k’ linked lists.
As long as the min-heap is not empty, perform the following steps:
1. 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.
2. 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.
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 program to test above
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];
 
    // 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

// Java implementation to merge
// k sorted linked lists
// Using MIN HEAP method
import java.util.PriorityQueue;
import java.util.Comparator;
public class MergeKLists {
 
    // function to merge k
    // sorted linked lists
    public static Node mergeKSortedLists(
        Node arr[], int k)
    {
        Node head = null, last = null;
 
        // priority_queue 'pq' implemeted
        // as min heap with the
        // help of 'compare' function
        PriorityQueue<Node> pq
            = new PriorityQueue<>(
                new Comparator<Node>() {
                    public int compare(Node a, Node b)
                    {
                        return a.data - b.data;
                    }
                });
 
        // push the head nodes of all
        // the k lists in 'pq'
        for (int i = 0; i < k; i++)
            if (arr[i] != null)
                pq.add(arr[i]);
 
        // loop till 'pq' is not empty
        while (!pq.isEmpty()) {
            // get the top element of 'pq'
            Node top = pq.peek();
            pq.remove();
 
            // check if there is a node
            // next to the 'top' node
            // in the list of which 'top'
            // node is a member
            if (top.next != null)
                // push the next node in 'pq'
                pq.add(top.next);
 
            // if final merged list is empty
            if (head == null) {
                head = top;
                // points to the last node so far of
                // the final merged list
                last = top;
            }
            else {
                // insert 'top' at the end
                // of the merged list so far
                last.next = top;
 
                // update the 'last' pointer
                last = top;
            }
        }
        // head node of the required merged list
        return head;
    }
 
    // function to print the singly linked list
    public static void printList(Node head)
    {
        while (head != null) {
            System.out.print(head.data + " ");
            head = head.next;
        }
    }
 
    // Utility function to create a new node
    public Node push(int data)
    {
        Node newNode = new Node(data);
        newNode.next = null;
        return newNode;
    }
 
    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 = mergeKSortedLists(arr, k);
        printList(head);
    }
}
 
class Node {
    int data;
    Node next;
    Node(int data)
    {
        this.data = data;
        next = null;
    }
}
// This code is contributed by Gaurav Tiwari

Output

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

Complexity Analysis:

Time Complexity: O(N * k * log k), where, ‘N’ is the total number of elements among all the linked lists and ‘k’ is the total number of lists.
Insertion and deletion in a min-heap requires log k time. So the overall time complexity is O(N * log k).
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).

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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My Personal Notes

C++

Java

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