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# Merge K sorted linked lists | Set 1

Given K sorted linked lists of size N each, the task is to merge them all maintaining their sorted order.

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.

Recommended Practice

Naive Approach:

A Simple Solution is to initialize the result as the first list. Now traverse all lists starting from the second list. Insert every node of the currently traversed list into the result in a sorted way.

Below is the implementation of the above approach:

## C++

 `// C++ program to merge k sorted``// arrays of size n each``#include ``using` `namespace` `std;`` ` `// A Linked List node``struct` `Node {``    ``int` `data;``    ``Node* next;``};`` ` `/* Function to print nodes in``   ``a given linked list */``void` `printList(Node* node)``{``    ``while` `(node != NULL) {``        ``printf``(``"%d "``, node->data);``        ``node = node->next;``    ``}``}`` ` `// The main function that``// takes an array of lists``// arr[0..last] and generates``// the sorted output``Node* mergeKLists(Node* arr[], ``int` `last)``{`` ` `    ``// Traverse form second list to last``    ``for` `(``int` `i = 1; i <= last; i++) {``        ``while` `(``true``) {``            ``// head of both the lists,``            ``// 0 and ith list.``            ``Node *head_0 = arr[0], *head_i = arr[i];`` ` `            ``// Break if list ended``            ``if` `(head_i == NULL)``                ``break``;`` ` `            ``// Smaller than first element``            ``if` `(head_0->data >= head_i->data) {``                ``arr[i] = head_i->next;``                ``head_i->next = head_0;``                ``arr[0] = head_i;``            ``}``            ``else``                ``// Traverse the first list``                ``while` `(head_0->next != NULL) {``                    ``// Smaller than next element``                    ``if` `(head_0->next->data``                        ``>= head_i->data) {``                        ``arr[i] = head_i->next;``                        ``head_i->next = head_0->next;``                        ``head_0->next = head_i;``                        ``break``;``                    ``}``                    ``// go to next node``                    ``head_0 = head_0->next;`` ` `                    ``// if last node``                    ``if` `(head_0->next == NULL) {``                        ``arr[i] = head_i->next;``                        ``head_i->next = NULL;``                        ``head_0->next = head_i;``                        ``head_0->next->next = NULL;``                        ``break``;``                    ``}``                ``}``        ``}``    ``}`` ` `    ``return` `arr[0];``}`` ` `// Utility function to create a new node.``Node* newNode(``int` `data)``{``    ``struct` `Node* temp = ``new` `Node;``    ``temp->data = data;``    ``temp->next = NULL;``    ``return` `temp;``}`` ` `// Driver program to test``// above functions``int` `main()``{``    ``// 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[k];`` ` `    ``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 all lists``    ``Node* head = mergeKLists(arr, k - 1);`` ` `    ``printList(head);`` ` `    ``return` `0;``}`

## Java

 `// Java program to merge k sorted``// arrays of size n each``import` `java.io.*;`` ` `// A Linked List node``class` `Node {``    ``int` `data;``    ``Node next;`` ` `    ``// Utility function to create a new node.``    ``Node(``int` `key)``    ``{``        ``data = key;``        ``next = ``null``;``    ``}``}``class` `GFG {`` ` `    ``static` `Node head;``    ``static` `Node temp;`` ` `    ``/* Function to print nodes in``     ``a given linked list */``    ``static` `void` `printList(Node node)``    ``{``        ``while` `(node != ``null``) {``            ``System.out.print(node.data + ``" "``);`` ` `            ``node = node.next;``        ``}``        ``System.out.println();``    ``}`` ` `    ``// The main function that``    ``// takes an array of lists``    ``// arr[0..last] and generates``    ``// the sorted output``    ``static` `Node mergeKLists(Node arr[], ``int` `last)``    ``{`` ` `        ``// Traverse form second list to last``        ``for` `(``int` `i = ``1``; i <= last; i++) {``            ``while` `(``true``) {`` ` `                ``// head of both the lists,``                ``// 0 and ith list.``                ``Node head_0 = arr[``0``];``                ``Node head_i = arr[i];`` ` `                ``// Break if list ended``                ``if` `(head_i == ``null``)``                    ``break``;`` ` `                ``// Smaller than first element``                ``if` `(head_0.data >= head_i.data) {``                    ``arr[i] = head_i.next;``                    ``head_i.next = head_0;``                    ``arr[``0``] = head_i;``                ``}``                ``else` `{`` ` `                    ``// Traverse the first list``                    ``while` `(head_0.next != ``null``) {`` ` `                        ``// Smaller than next element``                        ``if` `(head_0.next.data``                            ``>= head_i.data) {``                            ``arr[i] = head_i.next;``                            ``head_i.next = head_0.next;``                            ``head_0.next = head_i;``                            ``break``;``                        ``}`` ` `                        ``// go to next node``                        ``head_0 = head_0.next;`` ` `                        ``// if last node``                        ``if` `(head_0.next == ``null``) {``                            ``arr[i] = head_i.next;``                            ``head_i.next = ``null``;``                            ``head_0.next = head_i;``                            ``head_0.next.next = ``null``;``                            ``break``;``                        ``}``                    ``}``                ``}``            ``}``        ``}``        ``return` `arr[``0``];``    ``}`` ` `    ``// Driver program to test``    ``// above functions``    ``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``        ``head = mergeKLists(arr, k - ``1``);``        ``printList(head);``    ``}``}`` ` `// This code is contributed by avanitrachhadiya2155`

## Python3

 `# Python3 program to merge k``# sorted arrays of size n each`` ` `# A Linked List node`` ` ` ` `class` `Node:`` ` `    ``def` `__init__(``self``, x):`` ` `        ``self``.data ``=` `x``        ``self``.``next` `=` `None`` ` `# Function to print nodes in``# a given linked list`` ` ` ` `def` `printList(node):`` ` `    ``while` `(node !``=` `None``):``        ``print``(node.data,``              ``end``=``" "``)``        ``node ``=` `node.``next`` ` `# The main function that``# takes an array of lists``# arr[0..last] and generates``# the sorted output`` ` ` ` `def` `mergeKLists(arr, last):`` ` `    ``# Traverse form second``    ``# list to last``    ``for` `i ``in` `range``(``1``, last ``+` `1``):``        ``while` `(``True``):``            ``# head of both the lists,``            ``# 0 and ith list.``            ``head_0 ``=` `arr[``0``]``            ``head_i ``=` `arr[i]`` ` `            ``# Break if list ended``            ``if` `(head_i ``=``=` `None``):``                ``break`` ` `            ``# Smaller than first``            ``# element``            ``if` `(head_0.data >``=``                    ``head_i.data):``                ``arr[i] ``=` `head_i.``next``                ``head_i.``next` `=` `head_0``                ``arr[``0``] ``=` `head_i``            ``else``:``                ``# Traverse the first list``                ``while` `(head_0.``next` `!``=` `None``):``                    ``# Smaller than next``                    ``# element``                    ``if` `(head_0.``next``.data >``=``                            ``head_i.data):``                        ``arr[i] ``=` `head_i.``next``                        ``head_i.``next` `=` `head_0.``next``                        ``head_0.``next` `=` `head_i``                        ``break``                    ``# go to next node``                    ``head_0 ``=` `head_0.``next``                    ``# if last node``                    ``if` `(head_0.``next` `=``=` `None``):``                        ``arr[i] ``=` `head_i.``next``                        ``head_i.``next` `=` `None``                        ``head_0.``next` `=` `head_i``                        ``head_0.``next``.``next` `=` `None``                        ``break``    ``return` `arr[``0``]`` ` ` ` `# Driver code``if` `__name__ ``=``=` `'__main__'``:`` ` `    ``# Number of linked``    ``# lists``    ``k ``=` `3`` ` `    ``# Number of elements``    ``# in each list``    ``n ``=` `4`` ` `    ``# an array of pointers``    ``# storing the head nodes``    ``# of the linked lists``    ``arr ``=` `[``None` `for` `i ``in` `range``(k)]`` ` `    ``arr[``0``] ``=` `Node(``1``)``    ``arr[``0``].``next` `=` `Node(``3``)``    ``arr[``0``].``next``.``next` `=` `Node(``5``)``    ``arr[``0``].``next``.``next``.``next` `=` `Node(``7``)`` ` `    ``arr[``1``] ``=` `Node(``2``)``    ``arr[``1``].``next` `=` `Node(``4``)``    ``arr[``1``].``next``.``next` `=` `Node(``6``)``    ``arr[``1``].``next``.``next``.``next` `=` `Node(``8``)`` ` `    ``arr[``2``] ``=` `Node(``0``)``    ``arr[``2``].``next` `=` `Node(``9``)``    ``arr[``2``].``next``.``next` `=` `Node(``10``)``    ``arr[``2``].``next``.``next``.``next` `=` `Node(``11``)`` ` `    ``# Merge all lists``    ``head ``=` `mergeKLists(arr, k ``-` `1``)`` ` `    ``printList(head)`` ` `# This code is contributed by Mohit Kumar 29`

## C#

 `// C# program to merge k sorted``// arrays of size n each``using` `System;`` ` `// A Linked List node``public` `class` `Node {``    ``public` `int` `data;``    ``public` `Node next;`` ` `    ``// Utility function to create a new node.``    ``public` `Node(``int` `key)``    ``{``        ``data = key;``        ``next = ``null``;``    ``}``}`` ` `public` `class` `GFG {``    ``static` `Node head;`` ` `    ``/* Function to print nodes in``     ``a given linked list */``    ``static` `void` `printList(Node node)``    ``{``        ``while` `(node != ``null``) {``            ``Console.Write(node.data + ``" "``);``            ``node = node.next;``        ``}``        ``Console.WriteLine();``    ``}`` ` `    ``// The main function that``    ``// takes an array of lists``    ``// arr[0..last] and generates``    ``// the sorted output``    ``static` `Node mergeKLists(Node[] arr, ``int` `last)``    ``{`` ` `        ``// Traverse form second list to last``        ``for` `(``int` `i = 1; i <= last; i++) {``            ``while` `(``true``) {`` ` `                ``// head of both the lists,``                ``// 0 and ith list.``                ``Node head_0 = arr[0];``                ``Node head_i = arr[i];`` ` `                ``// Break if list ended``                ``if` `(head_i == ``null``)``                    ``break``;`` ` `                ``// Smaller than first element``                ``if` `(head_0.data >= head_i.data) {``                    ``arr[i] = head_i.next;``                    ``head_i.next = head_0;``                    ``arr[0] = head_i;``                ``}``                ``else` `{`` ` `                    ``// Traverse the first list``                    ``while` `(head_0.next != ``null``) {`` ` `                        ``// Smaller than next element``                        ``if` `(head_0.next.data``                            ``>= head_i.data) {``                            ``arr[i] = head_i.next;``                            ``head_i.next = head_0.next;``                            ``head_0.next = head_i;``                            ``break``;``                        ``}`` ` `                        ``// go to next node``                        ``head_0 = head_0.next;`` ` `                        ``// if last node``                        ``if` `(head_0.next == ``null``) {``                            ``arr[i] = head_i.next;``                            ``head_i.next = ``null``;``                            ``head_0.next = head_i;``                            ``head_0.next.next = ``null``;``                            ``break``;``                        ``}``                    ``}``                ``}``            ``}``        ``}``        ``return` `arr[0];``    ``}``    ``static` `public` `void` `Main()``    ``{`` ` `        ``// Number of linked lists``        ``int` `k = 3;`` ` `        ``// 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``        ``head = mergeKLists(arr, k - 1);``        ``printList(head);``    ``}``}`` ` `// This code is contributed by rag2127`

## Javascript

 ``

Output

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

Time complexity: O(NK-1), Traversing N times on each of the K lists.
Auxiliary Space: O(1).

## Merge K sorted linked lists using Min Heap:

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 the removed element. To get the result the step must continue until there is no element left in the MinHeap.

For a more detailed solution and code checkout, this article Merge k sorted linked lists | Set 2 (Using Min Heap).

Time Complexity: O(N*K*LogK)
Auxiliary Space: O(K)

## Merge K sorted linked lists usingDivide and Conquer:

The idea is to pair up a sorted list after which K/2 list will be left to be merged and repeat this till all the lists gets merged.

Follow the steps below to solve the problem:

• Pair up K lists and merge each pair in linear time using O(N) space.
• After the first cycle, K/2 lists are left each of size 2*N. After the second cycle, K/4 lists are left each of size 4*N and so on.
• Repeat the procedure until we have only one list left.

Below is the implementation of the above idea.

## C++

 `// C++ program to merge k sorted``// arrays of size n each``#include ``using` `namespace` `std;`` ` `// A Linked List node``struct` `Node {``    ``int` `data;``    ``Node* next;``};`` ` `/* Function to print nodes in``   ``a given linked list */``void` `printList(Node* node)``{``    ``while` `(node != NULL) {``        ``printf``(``"%d "``, node->data);``        ``node = node->next;``    ``}``}`` ` `/* Takes two lists sorted in increasing order, and merge``   ``their nodes together to make one big sorted list. Below``   ``function takes O(n) extra space for recursive calls,``    ``*/``Node* SortedMerge(Node* a, Node* b)``{``    ``Node* result = NULL;`` ` `    ``/* Base cases */``    ``if` `(a == NULL)``        ``return` `(b);``    ``else` `if` `(b == NULL)``        ``return` `(a);`` ` `    ``/* Pick either a or b, and recur */``    ``if` `(a->data <= b->data) {``        ``result = a;``        ``result->next = SortedMerge(a->next, b);``    ``}``    ``else` `{``        ``result = b;``        ``result->next = SortedMerge(a, b->next);``    ``}`` ` `    ``return` `result;``}`` ` `// The main function that takes an array of lists``// arr[0..last] and generates the sorted output``Node* mergeKLists(Node* arr[], ``int` `last)``{``    ``// repeat until only one list is left``    ``while` `(last != 0) {``        ``int` `i = 0, j = last;`` ` `        ``// (i, j) forms a pair``        ``while` `(i < j) {``            ``// merge List i with List j and store``            ``// merged list in List i``            ``arr[i] = SortedMerge(arr[i], arr[j]);`` ` `            ``// consider next pair``            ``i++, j--;`` ` `            ``// If all pairs are merged, update last``            ``if` `(i >= j)``                ``last = j;``        ``}``    ``}`` ` `    ``return` `arr[0];``}`` ` `// Utility function to create a new node.``Node* newNode(``int` `data)``{``    ``struct` `Node* temp = ``new` `Node;``    ``temp->data = data;``    ``temp->next = NULL;``    ``return` `temp;``}`` ` `// Driver program to test above functions``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];`` ` `    ``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 all lists``    ``Node* head = mergeKLists(arr, k - 1);`` ` `    ``printList(head);`` ` `    ``return` `0;``}`

## Java

 `// Java program to merge k sorted arrays of size n each``public` `class` `MergeKSortedLists {`` ` `    ``/* Takes two lists sorted in increasing order, and merge``    ``their nodes together to make one big sorted list. Below``    ``function takes O(Log n) extra space for recursive calls,``    ``but it can be easily modified to work with same time and``    ``O(1) extra space  */``    ``public` `static` `Node SortedMerge(Node a, Node b)``    ``{``        ``Node result = ``null``;``        ``/* Base cases */``        ``if` `(a == ``null``)``            ``return` `b;``        ``else` `if` `(b == ``null``)``            ``return` `a;`` ` `        ``/* Pick either a or b, and recur */``        ``if` `(a.data <= b.data) {``            ``result = a;``            ``result.next = SortedMerge(a.next, b);``        ``}``        ``else` `{``            ``result = b;``            ``result.next = SortedMerge(a, b.next);``        ``}`` ` `        ``return` `result;``    ``}`` ` `    ``// The main function that takes an array of lists``    ``// arr[0..last] and generates the sorted output``    ``public` `static` `Node mergeKLists(Node arr[], ``int` `last)``    ``{``        ``// repeat until only one list is left``        ``while` `(last != ``0``) {``            ``int` `i = ``0``, j = last;`` ` `            ``// (i, j) forms a pair``            ``while` `(i < j) {``                ``// merge List i with List j and store``                ``// merged list in List i``                ``arr[i] = SortedMerge(arr[i], arr[j]);`` ` `                ``// consider next pair``                ``i++;``                ``j--;`` ` `                ``// If all pairs are merged, update last``                ``if` `(i >= j)``                    ``last = j;``            ``}``        ``}`` ` `        ``return` `arr[``0``];``    ``}`` ` `    ``/* Function to print nodes in a given 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` `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[] = ``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 = mergeKLists(arr, k - ``1``);``        ``printList(head);``    ``}``}`` ` `class` `Node {``    ``int` `data;``    ``Node next;``    ``Node(``int` `data) { ``this``.data = data; }``}``// This code is contributed by Gaurav Tiwari`

## Python3

 `# Python3 program to merge k sorted``# arrays of size n each`` ` `# A Linked List node``class` `Node:`` ` `    ``def` `__init__(``self``):`` ` `        ``self``.data ``=` `0``        ``self``.``next` `=` `None`` ` `# Function to print nodes in a``# given linked list``def` `printList(node):`` ` `    ``while` `(node !``=` `None``):``        ``print``(node.data, end``=``' '``)``        ``node ``=` `node.``next`` ` `# Takes two lists sorted in increasing order,``# and merge their nodes together to make one``# big sorted list. Below function takes``# O(Log n) extra space for recursive calls,``# but it can be easily modified to work with``# same time and O(1) extra space``def` `SortedMerge(a, b):`` ` `    ``result ``=` `None`` ` `    ``# Base cases``    ``if` `(a ``=``=` `None``):``        ``return``(b)``    ``elif` `(b ``=``=` `None``):``        ``return``(a)`` ` `    ``# Pick either a or b, and recur``    ``if` `(a.data <``=` `b.data):``        ``result ``=` `a``        ``result.``next` `=` `SortedMerge(a.``next``, b)``    ``else``:``        ``result ``=` `b``        ``result.``next` `=` `SortedMerge(a, b.``next``)`` ` `    ``return` `result`` ` `# The main function that takes an array of lists``# arr[0..last] and generates the sorted output``def` `mergeKLists(arr, last):`` ` `    ``# Repeat until only one list is left``    ``while` `(last !``=` `0``):``        ``i ``=` `0``        ``j ``=` `last`` ` `        ``# (i, j) forms a pair``        ``while` `(i < j):`` ` `            ``# Merge List i with List j and store``            ``# merged list in List i``            ``arr[i] ``=` `SortedMerge(arr[i], arr[j])`` ` `            ``# Consider next pair``            ``i ``+``=` `1``            ``j ``-``=` `1`` ` `            ``# If all pairs are merged, update last``            ``if` `(i >``=` `j):``                ``last ``=` `j`` ` `    ``return` `arr[``0``]`` ` `# Utility function to create a new node.``def` `newNode(data):`` ` `    ``temp ``=` `Node()``    ``temp.data ``=` `data``    ``temp.``next` `=` `None``    ``return` `temp`` ` ` ` `# Driver code``if` `__name__ ``=``=` `'__main__'``:`` ` `    ``# Number of linked lists``    ``k ``=` `3`` ` `    ``# Number of elements in each list``    ``n ``=` `4`` ` `    ``# An array of pointers storing the``    ``# head nodes of the linked lists``    ``arr ``=` `[``0` `for` `i ``in` `range``(k)]`` ` `    ``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 all lists``    ``head ``=` `mergeKLists(arr, k ``-` `1``)`` ` `    ``printList(head)`` ` `# This code is contributed by rutvik_56`

## C#

 `// C# program to merge k sorted arrays of size n each``using` `System;`` ` `public` `class` `MergeKSortedLists {`` ` `    ``/* Takes two lists sorted in``    ``increasing order, and merge``    ``their nodes together to make``    ``one big sorted list. Below``    ``function takes O(Log n) extra``    ``space for recursive calls,``    ``but it can be easily modified``    ``to work with same time and``    ``O(1) extra space */``    ``public` `static` `Node SortedMerge(Node a, Node b)``    ``{``        ``Node result = ``null``;`` ` `        ``/* Base cases */``        ``if` `(a == ``null``)``            ``return` `b;``        ``else` `if` `(b == ``null``)``            ``return` `a;`` ` `        ``/* Pick either a or b, and recur */``        ``if` `(a.data <= b.data) {``            ``result = a;``            ``result.next = SortedMerge(a.next, b);``        ``}``        ``else` `{``            ``result = b;``            ``result.next = SortedMerge(a, b.next);``        ``}`` ` `        ``return` `result;``    ``}`` ` `    ``// The main function that takes``    ``// an array of lists arr[0..last]``    ``// and generates the sorted output``    ``public` `static` `Node mergeKLists(Node[] arr, ``int` `last)``    ``{``        ``// repeat until only one list is left``        ``while` `(last != 0) {``            ``int` `i = 0, j = last;`` ` `            ``// (i, j) forms a pair``            ``while` `(i < j) {``                ``// merge List i with List j and store``                ``// merged list in List i``                ``arr[i] = SortedMerge(arr[i], arr[j]);`` ` `                ``// consider next pair``                ``i++;``                ``j--;`` ` `                ``// If all pairs are merged, update last``                ``if` `(i >= j)``                    ``last = j;``            ``}``        ``}`` ` `        ``return` `arr[0];``    ``}`` ` `    ``/* Function to print nodes in a given linked list */``    ``public` `static` `void` `printList(Node node)``    ``{``        ``while` `(node != ``null``) {``            ``Console.Write(node.data + ``" "``);``            ``node = node.next;``        ``}``    ``}`` ` `    ``public` `static` `void` `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 = ``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 = mergeKLists(arr, k - 1);``        ``printList(head);``    ``}``}`` ` `public` `class` `Node {``    ``public` `int` `data;``    ``public` `Node next;``    ``public` `Node(``int` `data) { ``this``.data = data; }``}`` ` `/* This code contributed by PrinciRaj1992 */`

## Javascript

 ``

Output

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

Time Complexity: O(N * K * log K), As outer while loop in function mergeKLists() runs log K times and every time it processes N*K elements.
Auxiliary Space: O(N * K), Because recursion is used in SortedMerge() and to merge the final 2 linked lists of size N/2, N recursive calls will be made.

## Merge K sorted linked lists by Selecting min of top element:

The idea is to select the minimum of top elements iteratively store that in a new node and increment the pointer of the minimum element.

Follow the steps below to solve the problem:

• Find the node with the smallest value in all the K lists and
• Increment the current pointer to the next node of the list where the smallest node is found.
• Now make a new node and append the node to the head node of the resultant list and point the head list with this new node
• Repeat these steps till all nodes have been used.

Below is the implementation of the above approach:

## C++

 `// C++ program to merge k sorted arrays of size n each``#include ``using` `namespace` `std;`` ` `// A Linked List node``struct` `Node {``    ``int` `data;``    ``Node* next;``    ``Node(``int` `x)``    ``{``        ``data = x;``        ``next = NULL;``    ``}``};`` ` `/* Function to print nodes in a given linked list */``void` `printList(Node* node)``{``    ``while` `(node != NULL) {``        ``printf``(``"%d "``, node->data);``        ``node = node->next;``    ``}``    ``cout << endl;``}`` ` `/*Linked list Node structure`` ` `struct Node``{``int data;``Node* next;``Node(int x){``    ``data = x;``    ``next = NULL;``}``};``*/`` ` `    ``// Function to merge K sorted linked list.``    ``Node* mergeKLists(Node* arr[], ``int` `K)``    ``{``        ``Node* head = NULL;``        ``while` `(1) {``            ``int` `a = 0;``            ``int` `z;``            ``Node* curr;``            ``int` `min = INT_MAX;``            ``for` `(``int` `i = 0; i < K; i++) {``                ``if` `(arr[i] != NULL) {``                    ``a++;``                    ``if` `(arr[i]->data < min) {``                        ``min = arr[i]->data;``                        ``z = i;``                    ``}``                ``}``            ``}``            ``if` `(a != 0) {``                ``arr[z] = arr[z]->next;``                ``Node* temp = ``new` `Node(min);``                ``if` `(head == NULL) {``                    ``head = temp;``                    ``curr = temp;``                ``}``                ``else` `{``                    ``curr->next = temp;``                    ``curr = temp;``                ``}``            ``}``            ``else` `{``                ``return` `head;``            ``}``        ``}``    ``}`` ` `// { Driver Code Starts.`` ` `// Driver program to test above functions``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];`` ` `    ``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);``     ` `    ``Node* head = mergeKLists(arr, k);`` ` `    ``printList(head);``    ``return` `0;``}`` ` `// } Driver Code Ends`

Output

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

Time complexity: O(N*K2), There are N*K nodes in total and to find the smallest node it takes K times so for the N*K nodes it will take N*K*K time.
Auxiliary Space: O(N)

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