Construct a linked list from 2D matrix (Iterative Approach)

Given a matrix, the task is to construct a linked list matrix in which each node is connected to its right and down node.

Example:

Input: [1 2 3
         4 5 6
         7 8 9]

Output:
1 -> 2 -> 3 -> NULL
|    |    |
v    v    v
4 -> 5 -> 6 -> NULL
|    |    |
v    v    v
7 -> 8 -> 9 -> NULL
|    |    |
v    v    v
NULL NULL NULL

A recursive solution for this problem has been already discussed in this post. Below is an iterative approach for the problem:  

• The idea is to create m linked lists (m = number of rows) whose each node stores its right node. The head pointers of each m linked lists are stored in an array of nodes.
• Then, traverse m lists, for every ith and (i+1)^th list, set the down pointers of each node of i^th list to its corresponding node of (i+1)^th list. 

Below is the implementation of the above approach: 

C++

// C++ program to construct a linked
// list from 2D matrix | Iterative Approach
#include <bits/stdc++.h>
using namespace std;

// struct node of linked list
struct node {
    int data;
    node *right, *down;
};

// utility function to create a new node with given data
node* newNode(int d)
{
    node* temp = new node;
    temp->data = d;
    temp->right = temp->down = NULL;
    return temp;
}

// utility function to print the linked list pointed to by head pointer
void display(node* head)
{
    node *rp, *dp = head;

    // loop until the down pointer is not NULL
    while (dp) {
        rp = dp;

        // loop until the right pointer is not NULL
        while (rp) {
            cout << rp->data << " ";
            rp = rp->right;
        }
        cout << endl;
        dp = dp->down;
    }
}

// function which constructs the linked list
// from the given matrix of size m * n
// and returns the head pointer of the linked list
node* constructLinkedMatrix(int mat[][3], int m, int n)
{
    // stores the head of the linked list
    node* mainhead = NULL;

    // stores the head of linked lists of each row
    node* head[m];
    node *righttemp, *newptr;

    // Firstly, we create m linked lists
    // by setting all the right nodes of every row
    for (int i = 0; i < m; i++) {

        // initially set the head of ith row as NULL
        head[i] = NULL;
        for (int j = 0; j < n; j++) {
            newptr = newNode(mat[i][j]);

            // stores the mat[0][0] node as
            // the mainhead of the linked list
            if (!mainhead)
                mainhead = newptr;

            if (!head[i])
                head[i] = newptr;
            else
                righttemp->right = newptr;

            righttemp = newptr;
        }
    }

    // Then, for every ith and (i+1)th list,
    // we set the down pointers of
    // every node of ith list
    // with its corresponding
    // node of (i+1)th list
    for (int i = 0; i < m - 1; i++) {

        node *temp1 = head[i], *temp2 = head[i + 1];

        while (temp1 && temp2) {

            temp1->down = temp2;
            temp1 = temp1->right;
            temp2 = temp2->right;
        }
    }

    // return the mainhead pointer of the linked list
    return mainhead;
}

// Driver program to test the above function
int main()
{
    int m, n; // m = rows and n = columns
    m = 3, n = 3;
    // 2D matrix
    int mat[][3] = { { 1, 2, 3 },
                     { 4, 5, 6 },
                     { 7, 8, 9 } };

    node* head = constructLinkedMatrix(mat, m, n);
    display(head);

    return 0;
}

Java

// Java implementation for above approach

// Construct a Linked List from 2-D Matrix
class LinkedMatrix
{
    class Node
    {
        int data;
        Node right, down;

        // Default Constructor
        Node()
        {
            this.right = null;
            this.down = null;
        }

        Node(int d)
        {
            this.data = d;
            this.right = null;
            this.down = null;
        }
    }

    /*
    A function to construct a Linked List
    from the given matrix of size m * n
    and returns the head pointer of the linked list
    */
    Node constructLinkedMatrix(int[][] arr,
                               int m, int n)
    {

        // stores the head of the linked list
        Node mainHead = null;

        // stores the head of linked lists
        // of each row
        Node[] head = new Node[m];
        Node rightTemp = new Node(), newptr;

        // Firstly, we create m linked lists
        // by setting all the right nodes of every row
        for(int i = 0; i < m; i++)
        {

            // initially set the head of
            // ith row as NULL
            head[i] = null;

            for(int j = 0; j < n; j++)
            {
                newptr = new Node(arr[i][j]);

                // stores the mat[0][0] node as
                // the mainhead of the linked list
                if(mainHead == null)
                    mainHead = newptr;
                if(head[i] == null)
                    head[i] = newptr;
                else
                    rightTemp.right = newptr;

                rightTemp = newptr;
            }
        }

        // Then, for every ith and (i+1)th list,
        // we set the down pointers of
        // every node of ith list
        // with its corresponding
        // node of (i+1)th list
        for(int i = 0; i < m - 1; i++)
        {
            Node temp1 = head[i], temp2 = head[i + 1];

            while(temp1 != null && temp2 != null)
            {
                temp1.down = temp2;
                temp1 = temp1.right;
                temp2 = temp2.right;
            }
        }

        // return the mainhead pointer
        // of the linked list
        return mainHead;
    }

    // utility function to print the
    // linked list pointed to by head pointer
    void display(Node head)
    {
        Node rp, dp = head;

        // loop until the down pointer
        // is not NULL
        while(dp != null)
        {
            rp = dp;

            // loop until the right pointer
            // is not NULL
            while(rp != null)
            {
                System.out.print(rp.data + " ");
                rp = rp.right;
            }
            System.out.println();
            dp = dp.down;
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        LinkedMatrix Obj = new LinkedMatrix();
        int m = 3, n = 3; // m = rows and n = columns

        // 2-D Matrix
        int[][] arr = {{ 1, 2, 3 },
                       { 4, 5, 6 },
                       { 7, 8, 9 }};

        Node head = Obj.constructLinkedMatrix(arr, m, n);
        Obj.display(head);
    }
}

// This code is contributed by Rhythem

Python3

# Python3 program to construct a linked
# list from 2D matrix | Iterative Approach

# struct node of linked list
class node:
    def __init__(self, data):
        self.data = data
        self.right = None
        self.down = None

# utility function to create a new node with given data
def newNode(d):
    temp = node(d)
    return temp;

# utility function to print the linked list pointed to by head pointer
def display(head):
    rp = None
    dp = head;

    # loop until the down pointer is not None
    while (dp != None):
        rp = dp;

        # loop until the right pointer is not None
        while rp != None:
            print(rp.data, end = ' ')
            rp = rp.right;
        print()
        dp = dp.down;

# function which constructs the linked list
# from the given matrix of size m * n
# and returns the head pointer of the linked list
def constructLinkedMatrix(mat, m, n):

    # stores the head of the linked list
    mainhead = None;

    # stores the head of linked lists of each row
    head = [0 for i in range(m)];
    righttemp = None
    newptr = None

    # Firstly, we create m linked lists
    # by setting all the right nodes of every row
    for i in range(m):

        # initially set the head of ith row as None
        head[i] = None;
        for j in range(n):
            newptr = newNode(mat[i][j]);

            # stores the mat[0][0] node as
            # the mainhead of the linked list
            if (not mainhead):
                mainhead = newptr;

            if (not head[i]):
                head[i] = newptr;
            else:
                righttemp.right = newptr;
            righttemp = newptr;

    # Then, for every ith and (i+1)th list,
    # we set the down pointers of
    # every node of ith list
    # with its corresponding
    # node of (i+1)th list
    for i in range(m - 1):
        temp1 = head[i]
        temp2 = head[i + 1];

        while (temp1 and temp2):
            temp1.down = temp2;
            temp1 = temp1.right;
            temp2 = temp2.right;

    # return the mainhead pointer of the linked list
    return mainhead;

# Driver code
if __name__ == '__main__':
    m = 3
    n = 3;

    # 2D matrix
    mat= [ [ 1, 2, 3 ],
          [ 4, 5, 6 ],
          [ 7, 8, 9 ] ];

    head = constructLinkedMatrix(mat, m, n);
    display(head);

# This code is contributed by rutvik_56

C#

// C# implementation for above approach
using System;

// Construct a Linked List from 2-D Matrix
class GFG
{
    class Node
    {
        public int data;
        public Node right, down;

        // Default Constructor
        public Node()
        {
            this.right = null;
            this.down = null;
        }

        public Node(int d)
        {
            this.data = d;
            this.right = null;
            this.down = null;
        }
    }

    /*
    A function to construct a Linked List
    from the given matrix of size m * n
    and returns the head pointer of the linked list
    */
    Node constructLinkedMatrix(int[,] arr,
                               int m, int n)
    {

        // stores the head of the linked list
        Node mainHead = null;

        // stores the head of linked lists
        // of each row
        Node[] head = new Node[m];
        Node rightTemp = new Node(), newptr;

        // Firstly, we create m linked lists
        // by setting all the right nodes of every row
        for(int i = 0; i < m; i++)
        {

            // initially set the head of
            // ith row as NULL
            head[i] = null;

            for(int j = 0; j < n; j++)
            {
                newptr = new Node(arr[i, j]);

                // stores the mat[0][0] node as
                // the mainhead of the linked list
                if(mainHead == null)
                    mainHead = newptr;
                if(head[i] == null)
                    head[i] = newptr;
                else
                    rightTemp.right = newptr;

                rightTemp = newptr;
            }
        }

        // Then, for every ith and (i+1)th list,
        // we set the down pointers of
        // every node of ith list
        // with its corresponding
        // node of (i+1)th list
        for(int i = 0; i < m - 1; i++)
        {
            Node temp1 = head[i],
                 temp2 = head[i + 1];

            while(temp1 != null && temp2 != null)
            {
                temp1.down = temp2;
                temp1 = temp1.right;
                temp2 = temp2.right;
            }
        }

        // return the mainhead pointer
        // of the linked list
        return mainHead;
    }

    // utility function to print the
    // linked list pointed to by head pointer
    void display(Node head)
    {
        Node rp, dp = head;

        // loop until the down pointer
        // is not NULL
        while(dp != null)
        {
            rp = dp;

            // loop until the right pointer
            // is not NULL
            while(rp != null)
            {
                Console.Write(rp.data + " ");
                rp = rp.right;
            }
            Console.WriteLine();
            dp = dp.down;
        }
    }

    // Driver Code
    public static void Main(String[] args)
    {
        GFG Obj = new GFG();
        int m = 3, n = 3; // m = rows and n = columns

        // 2-D Matrix
        int[,] arr = {{ 1, 2, 3 },
                      { 4, 5, 6 },
                      { 7, 8, 9 }};

        Node head = Obj.constructLinkedMatrix(arr, m, n);
        Obj.display(head);
    }
}

// This code is contributed by PrinciRaj1992

Javascript

// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)
//  JavaScript program to construct a linked
//  list from 2D matrix  Iterative approach
//  node of linked list
class node{
    constructor(data){
        this.data = data;
        this.right = null;
        this.down = null;
    }
}

//  utility function to create a new node with given data
function newNode(d){
    return new node(d);
}

//  utility function to print the linked list pointed to by head pointer
function display(head){
    let rp = head;
    let dp = head;

    //  loop until the down pointer is not null
    while(dp != null){
        rp = dp;

        //  loop until the right pointer is not null
        while(rp != null){
            document.write(rp.data +  " ");
            rp = rp.right;
        }
        document.write("<br>");
        dp = dp.down;
    }
}

//  function which constructs the linked list
//  from the given matrix of size m  n
//  and returns the head pointer of the linked list
function constructLinkedMatrix(mat, m, n){
    //  stores the head of the linked list
    let mainhead = null;

    //  stores the head of linked lists of each row
    let head = [];
    let righttemp;
    let newptr;

    //  Firstly, we create m linked lists
    //  by setting all the right nodes of every row
    for(let i = 0; i < m; i++){
        //  initially set the head of ith row as NULL
        head[i] = null;
        for(let j = 0; j<n; j++){
            newptr = newNode(mat[i][j]);

            //  stores the mat[0][0] node as
            //  the mainhead of the linked list
            if(mainhead == null){
                mainhead = newptr;
            }
            if(head[i] == null){
                head[i] = newptr;
            }
            else{
                righttemp.right = newptr;
            }
            righttemp = newptr;
        }
    }

    //  Then, for every ith and (i+1)th list,
    //  we set the down pointers of
    //  every node of ith list
    //  with its corresponding
    //  node of (i+1)th list
    for(let i = 0; i<m-1; i++){
        let temp1 = head[i];
        let temp2 = head[i+1];

        while(temp1 != null && temp2 != null){
            temp1.down = temp2;
            temp1 = temp1.right;
            temp2 = temp2.right;
        }
    }

    //  return the mainhead pointer of the linked list
    return mainhead;
}

//  driver program
let m = 3;
let n = 3;
let mat = [[1, 2, 3], [4, 5, 6], [7, 8, 9]];
let head = constructLinkedMatrix(mat, m, n);
display(head);

Output

1 2 3 
4 5 6 
7 8 9 

Time complexity: O(M * N)