# 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
```

## 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++ 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++)
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

// 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
}

// function to print the singly linked list
{
cout << head->data << " ";
}
}

// 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
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;
}
```

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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