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

Given k sorted linked lists each of size n, 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.

Source: Merge K sorted Linked Lists | Method 2

An efficient solution for the problem has been dicussed 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.



The process must start with creating a MinHeap and inserting the first element of all the k linked list. Remove the root element of Minheap and insert it in the output list and insert the next element from the linked list of removed element. To get the result the step must continue until there is no element in the MinHeap left.

C++

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// 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;
};
  
// '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)
{
    struct Node *head = NULL, *last;
  
    // 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]);
  
    // loop till 'pq' is not empty
    while (!pq.empty()) {
  
        // get the top element of 'pq'
        struct Node* top = pq.top();
        pq.pop();
  
        // 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.push(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
void printList(struct Node* head)
{
    while (head != NULL) {
        cout << head->data << " ";
        head = head->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;
}
  
// 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;
}

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Java

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// 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

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Output:

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

Complexity Analysis:

  • Time Complexity: O( n * k * log k).
    Insertion and deletion in a Min Heap requires log k time. So the Overall time complexity is O( n * k * log k)
  • Auxiliary Space: O(k).
    k is the space required to store the priority queue.

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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