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# Java Program To Merge K Sorted Linked Lists Using Min Heap – Set 2

Given k linked lists each of size n and each list is sorted in non-decreasing order, merge them into a single sorted (non-decreasing order) linked list and print the sorted linked list as 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.```

An efficient solution for the problem has been discussed 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.

1. Create a min-heap and insert the first element of all the ‘k’ linked lists.
2. As long as the min-heap is not empty, perform the following steps:
• Remove the top element of the min-heap (which is the current minimum among all the elements in the min-heap) and add it to the result list.
• If there exists an element (in the same linked list) next to the element popped out in previous step, insert it into the min-heap.

Below is the implementation of the above approach:

## 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' implemented``        ``// as min heap with the``        ``// help of 'compare' function``        ``PriorityQueue pq =``                ``new` `PriorityQueue<>(``new` `Comparator()``        ``{``            ``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[])``    ``{``        ``// Number of linked lists``        ``int` `k = ``3``;``    ` `        ``// Number of elements in each list``        ``int` `n = ``4``;` `        ``// 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 `

Complexity Analysis:

• Time Complexity: O(N * log k) or O(n * k * log k), where, ‘N’ is the total number of elements among all the linked lists, ‘k’ is the total number of lists, and ‘n’ is the size of each linked list.
Insertion and deletion operation will be performed in min-heap for all N nodes.
Insertion and deletion in a min-heap require log k time.
• Auxiliary Space: O(k).
The priority queue will have atmost ‘k’ number of elements at any point of time, hence the additional space required for our algorithm is O(k).

Please refer complete article on Merge k sorted linked lists | Set 2 (Using Min Heap) for more details!

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