Java Program To Merge K Sorted Linked Lists Using Min Heap - Set 2
Last Updated :
03 Nov, 2022
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.
Source: Merge K sorted Linked Lists | Method 2
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.
- Create a min-heap and insert the first element of all the 'k' linked lists.
- 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.
- Return the head node address of the merged list.
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<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[])
{
// 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!