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

Output: 
0->1->2->3->4->5->6->7->8->9->10->11

Source: Merge K sorted Linked Lists | Method 2

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Approach: An efficient solution for the problem has been dicussed in Method 3 of this post. Here another solution has been provided which the uses the MIN HEAP data structure. This solution is based on the min heap approach used to solve the problem ‘merge k sorted arrays’ which is discussed here.

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; 
}; 
  
// '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' implemeted 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; 
} 

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

Time Complexity: O(nk Logk)
Auxiliary Space: 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

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

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