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 result as first list. Now traverse all lists starting from second list. Insert every node of currently traversed list into result in a sorted way.

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to merge k sorted
// arrays of size n each
#include <bits/stdc++.h>
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;
}

chevron_right


Output:

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

Complexity Analysis:



  • Time complexity: O(N2), where N is 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. 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 first cycle, K/2 lists are left each of size 2*N. After 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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to merge k sorted
// arrays of size n each
#include <bits/stdc++.h>
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[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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 */

chevron_right


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)

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up