# Merge K sorted linked lists | Set 1

Given K sorted linked lists of size N each, 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->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.
```

Method 1 (Simple)
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 result in a sorted way.

## 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, *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 = 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;` `}`   `// 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 = newNode(1);` `    ``arr->next = newNode(3);` `    ``arr->next->next = newNode(5);` `    ``arr->next->next->next = newNode(7);`   `    ``arr = newNode(2);` `    ``arr->next = newNode(4);` `    ``arr->next->next = newNode(6);` `    ``arr->next->next->next = newNode(8);`   `    ``arr = newNode(0);` `    ``arr->next = newNode(9);` `    ``arr->next->next = newNode(10);` `    ``arr->next->next->next = newNode(11);`   `    ``// Merge all lists` `    ``Node* head = mergeKLists(arr, k - 1);`   `    ``printList(head);`   `    ``return` `0;` `}`

## 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 prnodes 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`

Output:

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

```

Complexity Analysis:

• Time complexity: O(N2), where N is a total number of nodes, i.e., N = kn.
• Auxiliary Space: O(1).
As no extra space is required.

Method 2: Min Heap
A Better solution is to use Min Heap-based solution which is discussed here for arrays. The time complexity of this solution would be O(nk Log k)

Method 3: Divide and Conquer
In this post, Divide and Conquer approach is discussed. This approach doesn’t require extra space for heap and works in O(nk Log k)
It is known that merging of two linked lists can be done in O(n) time and O(1) space (For arrays O(n) space is required).

1. The idea is to pair up K lists and merge each pair in linear time using O(1) space.
2. 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.
3. 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(Log n) extra space for recursive calls,` `   ``but it can be easily modified to work with same time and` `   ``O(1) extra space  */` `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;` `}`   `// 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 = newNode(1);` `    ``arr->next = newNode(3);` `    ``arr->next->next = newNode(5);` `    ``arr->next->next->next = newNode(7);`   `    ``arr = newNode(2);` `    ``arr->next = newNode(4);` `    ``arr->next->next = newNode(6);` `    ``arr->next->next->next = newNode(8);`   `    ``arr = newNode(0);` `    ``arr->next = newNode(9);` `    ``arr->next->next = newNode(10);` `    ``arr->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;
}

/* 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
Node arr[] = new Node[k];

arr = new Node(1);
arr.next = new Node(3);
arr.next.next = new Node(5);
arr.next.next.next = new Node(7);

arr = new Node(2);
arr.next = new Node(4);
arr.next.next = new Node(6);
arr.next.next.next = new Node(8);

arr = new Node(0);
arr.next = new Node(9);
arr.next.next = new Node(10);
arr.next.next.next = new Node(11);

// Merge all lists
Node head = mergeKLists(arr, k - 1);
}
}

class Node {
int data;
Node next;
Node(int data)
{
this.data = data;
}
}
// This code is contributed by Gaurav Tiwari
``````

## 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;` `    ``}`   `    ``/* 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 = ``new` `Node(1);` `        ``arr.next = ``new` `Node(3);` `        ``arr.next.next = ``new` `Node(5);` `        ``arr.next.next.next = ``new` `Node(7);`   `        ``arr = ``new` `Node(2);` `        ``arr.next = ``new` `Node(4);` `        ``arr.next.next = ``new` `Node(6);` `        ``arr.next.next.next = ``new` `Node(8);`   `        ``arr = ``new` `Node(0);` `        ``arr.next = ``new` `Node(9);` `        ``arr.next.next = ``new` `Node(10);` `        ``arr.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 */`

Output:

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

```

Complexity Analysis:

• Time Complexity: O(nk logk).
As outer while loop in function mergeKLists() runs log k times and every time it processes nk elements.
• Auxiliary Space: O(1).
As no extra space is required.

Merge k sorted linked lists | Set 2 (Using Min Heap)