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->NULLOutput: 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->NULLOutput: 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 (Min-Heap): To solve the problem using this approach follow the below idea:
This solution is based on the MIN HEAP approach. The process must start with creating a MinHeap and inserting the first element of all the k Linked Lists. Remove the root element of Minheap and put it in the output Linked List and insert the next element from the Linked List of removed element. To get the result the step must continue until there is no element left in the MinHeap.
Note: 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.
Follow the given steps to solve the problem:
- 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 Root 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 that popped out in the previous step, then insert it into the min-heap.
- Return the head node address of the merged list.
Below is the implementation of the above approach:
// 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;
}; // 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;
} // '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)
{ // Priority_queue 'pq' implemented
// 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++)
if (arr[i] != NULL)
pq.push(arr[i]);
// Handles the case when k = 0
// or lists have no elements in them
if (pq.empty())
return NULL;
struct Node *dummy = newNode(0);
struct Node *last = dummy;
// Loop till 'pq' is not empty
while (!pq.empty())
{
// Get the top element of 'pq'
struct Node* curr = pq.top();
pq.pop();
// Add the top element of 'pq'
// to the resultant merged list
last->next = curr;
last = last->next;
// Check if there is a node
// next to the 'top' node
// in the list of which 'top'
// node is a member
if (curr->next != NULL)
// Push the next node of top node
// in 'pq'
pq.push(curr->next);
}
// Address of head node of the required
// merged list
return dummy->next;
} // Function to print the singly // linked list void printList( struct Node* head)
{ while (head != NULL)
{
cout << head->data << " " ;
head = head->next;
}
} // Driver code int main()
{ // Number of linked lists
int N = 3;
// Number of elements in each list
int K = 4;
// An array of pointers storing the
// head nodes of the linked lists
Node* arr[N];
// 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, N);
// Print the merged list
printList(head);
return 0;
} |
// Java code for the above approach import java.io.*;
import java.util.*;
class Node {
int data;
Node next;
Node( int key)
{
data = key;
next = null ;
}
} // Class implements Comparator to compare Node data class NodeComparator implements Comparator<Node> {
public int compare(Node k1, Node k2)
{
if (k1.data > k2.data)
return 1 ;
else if (k1.data < k2.data)
return - 1 ;
return 0 ;
}
} class GFG {
// Function to merge k sorted linked lists
static Node mergeKList(Node[] arr, int K)
{
// Priority_queue 'queue' implemented
// as min heap with the help of
// 'compare' function
PriorityQueue<Node> queue
= new PriorityQueue<>( new NodeComparator());
Node at[] = new Node[K];
Node head = new Node( 0 );
Node last = head;
// Push the head nodes of all
// the k lists in 'queue'
for ( int i = 0 ; i < K; i++) {
if (arr[i] != null ) {
queue.add(arr[i]);
}
}
// Handles the case when k = 0
// or lists have no elements in them
if (queue.isEmpty())
return null ;
// Loop till 'queue' is not empty
while (!queue.isEmpty()) {
// Get the top element of 'queue'
Node curr = queue.poll();
// Add the top element of 'queue'
// to the resultant merged list
last.next = curr;
last = last.next;
// Check if there is a node
// next to the 'top' node
// in the list of which 'top'
// node is a member
if (curr.next != null ) {
// Push the next node of top node
// in 'queue'
queue.add(curr.next);
}
}
// Address of head node of the required
// merged list
return head.next;
}
// Print linked list
public static void printList(Node node)
{
while (node != null ) {
System.out.print(node.data + " " );
node = node.next;
}
}
public static void main(String[] args)
{
int N = 3 ;
// array to store head of linkedlist
Node[] a = new Node[N];
// Linkedlist1
Node head1 = new Node( 1 );
a[ 0 ] = head1;
head1.next = new Node( 3 );
head1.next.next = new Node( 5 );
head1.next.next.next = new Node( 7 );
// Limkedlist2
Node head2 = new Node( 2 );
a[ 1 ] = head2;
head2.next = new Node( 4 );
head2.next.next = new Node( 6 );
head2.next.next.next = new Node( 8 );
// Linkedlist3
Node head3 = new Node( 0 );
a[ 2 ] = head3;
head3.next = new Node( 9 );
head3.next.next = new Node( 10 );
head3.next.next.next = new Node( 11 );
Node res = mergeKList(a, N);
if (res != null )
printList(res);
System.out.println();
}
} |
# Python implementation to merge k sorted linked lists # Using MIN HEAP method import heapq
class Node:
def __init__( self , data):
self .data = data
self . next = None
def mergeKList(arr, K):
# Priority_queue 'queue' implemented
# as min heap with the help of
# 'compare' function
queue = []
for i in range (K):
if arr[i] ! = None :
heapq.heappush(queue, (arr[i].data, arr[i]))
dummy = Node( 0 )
last = dummy
while queue:
curr = heapq.heappop(queue)[ 1 ]
last. next = curr
last = last. next
if curr. next ! = None :
heapq.heappush(queue, (curr. next .data, curr. next ))
return dummy. next
def printList(node):
while node ! = None :
print (node.data, end = " " )
node = node. next
print ()
if __name__ = = "__main__" :
N = 3
a = [ None ] * N
# Linkedlist1
head1 = Node( 1 )
a[ 0 ] = head1
head1. next = Node( 3 )
head1. next . next = Node( 5 )
head1. next . next . next = Node( 7 )
# Limkedlist2
head2 = Node( 2 )
a[ 1 ] = head2
head2. next = Node( 4 )
head2. next . next = Node( 6 )
head2. next . next . next = Node( 8 )
# Linkedlist3
head3 = Node( 0 )
a[ 2 ] = head3
head3. next = Node( 9 )
head3. next . next = Node( 10 )
head3. next . next . next = Node( 11 )
res = mergeKList(a, N)
if res ! = None :
printList(res)
# This code is contributed by lokesh. |
using System;
using System.Collections.Generic;
class Node
{ public int data;
public Node next;
public Node( int key)
{
data = key;
next = null ;
}
} // Custom comparer to compare nodes based on data class NodeComparer : IComparer<Node>
{ public int Compare(Node k1, Node k2)
{
if (k1.data > k2.data)
return 1;
else if (k1.data < k2.data)
return -1;
return 0;
}
} class GFG
{ // Function to merge k sorted linked lists
static Node MergeKList(Node[] arr, int K)
{
// Priority queue implemented as a min-heap using custom comparer
PriorityQueue<Node> queue = new PriorityQueue<Node>( new NodeComparer());
Node head = new Node(0);
Node last = head;
// Enqueue the head nodes of all k lists in the queue
for ( int i = 0; i < K; i++)
{
if (arr[i] != null )
{
queue.Enqueue(arr[i]);
}
}
// If the queue is empty, return null
if (queue.Count == 0)
return null ;
// Loop until the queue is not empty
while (queue.Count > 0)
{
// Get the node with the smallest data from the queue
Node curr = queue.Dequeue();
// Add the node to the resultant merged list
last.next = curr;
last = last.next;
// Check if there is a node next to the current node in its list
if (curr.next != null )
{
// Enqueue the next node in the list
queue.Enqueue(curr.next);
}
}
// Return the head of the merged list
return head.next;
}
// Print linked list
public static void PrintList(Node node)
{
while (node != null )
{
Console.Write(node.data + " " );
node = node.next;
}
}
static void Main( string [] args)
{
int N = 3;
Node[] a = new Node[N];
// Linked list 1
Node head1 = new Node(1);
a[0] = head1;
head1.next = new Node(3);
head1.next.next = new Node(5);
head1.next.next.next = new Node(7);
// Linked list 2
Node head2 = new Node(2);
a[1] = head2;
head2.next = new Node(4);
head2.next.next = new Node(6);
head2.next.next.next = new Node(8);
// Linked list 3
Node head3 = new Node(0);
a[2] = head3;
head3.next = new Node(9);
head3.next.next = new Node(10);
head3.next.next.next = new Node(11);
Node res = MergeKList(a, N);
if (res != null )
PrintList(res);
Console.WriteLine();
}
} // Priority queue implementation class PriorityQueue<T>
{ private List<T> heap;
private IComparer<T> comparer;
public PriorityQueue(IComparer<T> comparer = null )
{
this .heap = new List<T>();
this .comparer = comparer ?? Comparer<T>.Default;
}
public int Count => heap.Count;
// Enqueue operation
public void Enqueue(T item)
{
heap.Add(item);
int i = heap.Count - 1;
while (i > 0)
{
int parent = (i - 1) / 2;
if (comparer.Compare(heap[parent], heap[i]) <= 0)
break ;
Swap(parent, i);
i = parent;
}
}
// Dequeue operation
public T Dequeue()
{
if (heap.Count == 0)
throw new InvalidOperationException( "Priority queue is empty." );
T result = heap[0];
int last = heap.Count - 1;
heap[0] = heap[last];
heap.RemoveAt(last);
last--;
int i = 0;
while ( true )
{
int left = 2 * i + 1;
if (left > last)
break ;
int right = left + 1;
int minChild = left;
if (right <= last && comparer.Compare(heap[right], heap[left]) < 0)
minChild = right;
if (comparer.Compare(heap[i], heap[minChild]) <= 0)
break ;
Swap(i, minChild);
i = minChild;
}
return result;
}
// Swap two elements in the heap
private void Swap( int i, int j)
{
T temp = heap[i];
heap[i] = heap[j];
heap[j] = temp;
}
} |
class Node { constructor(key) {
this .data = key;
this .next = null ;
}
} // Comparator function to compare Node data function nodeComparator(k1, k2) {
if (k1.data < k2.data) return -1;
if (k1.data > k2.data) return 1;
return 0;
} class PriorityQueue { constructor(compare) {
this .heap = [];
this .compare = compare;
}
enqueue(value) {
this .heap.push(value);
this .bubbleUp();
}
bubbleUp() {
let index = this .heap.length - 1;
while (index > 0) {
let element = this .heap[index],
parentIndex = Math.floor((index - 1) / 2),
parent = this .heap[parentIndex];
if ( this .compare(element, parent) < 0) break ;
this .heap[index] = parent;
this .heap[parentIndex] = element;
index = parentIndex;
}
}
dequeue() {
let max = this .heap[0];
let end = this .heap.pop();
if ( this .heap.length > 0) {
this .heap[0] = end;
this .sinkDown(0);
}
return max;
}
sinkDown(index) {
let left = 2 * index + 1,
right = 2 * index + 2,
largest = index;
if (
left < this .heap.length &&
this .compare( this .heap[left], this .heap[largest]) > 0
) {
largest = left;
}
if (
right < this .heap.length &&
this .compare( this .heap[right], this .heap[largest]) > 0
) {
largest = right;
}
if (largest !== index) {
[ this .heap[largest], this .heap[index]] = [
this .heap[index],
this .heap[largest],
];
this .sinkDown(largest);
}
}
isEmpty() {
return this .heap.length === 0;
}
} function mergeKLists(arr, K) {
const queue = new PriorityQueue(nodeComparator);
const at = new Array(K);
const head = new Node(0);
let last = head;
// Push the head nodes of all the k lists in 'queue'
for (let i = 0; i < K; i++) {
if (arr[i] !== null ) {
queue.enqueue(arr[i]);
}
}
// Handles the case when k = 0
// or lists have no elements in them
if (queue.isEmpty()) return null ;
// Loop till 'queue' is not empty
while (!queue.isEmpty()) {
// Get the top element of 'queue'
const curr = queue.dequeue();
// Add the top element of 'queue' to the resultant merged list
last.next = curr;
last = last.next;
// Check if there is a node next to the 'top' node
// in the list of which 'top' node is a member
if (curr.next !== null ) {
// Push the next node of top node in 'queue'
queue.enqueue(curr.next);
}
}
// Address of head node of the required merged list
return head.next;
} // Print linked list function printList(node) {
let str = "" ;
while (node !== null ) {
str += `${node.data} `;
node = node.next;
}
console.log(str);
} // Testing const N = 3; // array to store head of linkedlist const a = new Array(N);
// Linkedlist1 const head1 = new Node(1);
a[0] = head1; head1.next = new Node(3);
head1.next.next = new Node(5);
head1.next.next.next = new Node(7);
// Limkedlist2 const head2 = new Node(2);
a[1] = head2; head2.next = new Node(4);
head2.next.next = new Node(6);
head2.next.next.next = new Node(8);
// Linkedlist3 const head3 = new Node(0);
a[2] = head3; head3.next = new Node(9);
head3.next.next = new Node(10);
head3.next.next.next = new Node(11);
const res = mergeKLists(a, N); if (res !== null ) printList(res);
console.log(); |
0 1 2 3 4 5 6 7 8 9 10 11
Time Complexity: O(N * K * log K)
Auxiliary Space: O(K)