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

**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 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; }

Output:

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

Time Complexity: O(nk Logk)

Auxiliary Space: O(k)

### Asked in: Amazon

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