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.


Input: k = 3, n =  4
list1 = 1->3->5->7->NULL
list2 = 2->4->6->8->NULL
list3 = 0->9->10->11


Source: Merge K sorted Linked Lists | Method 2

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++ 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++)
    // loop till 'pq' is not empty
    while (!pq.empty()) {
        // get the top element of 'pq'
        struct Node* top =;
        // 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'
        // 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
    return 0;



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 or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

My Personal Notes arrow_drop_up