Construya una lista enlazada a partir de una array 2D (enfoque iterativo)

Dada una array, la tarea es construir una array de lista enlazada en la que cada Node esté conectado a su Node derecho e inferior.
Ejemplo: 
 

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

Ya se ha discutido una solución recursiva para este problema en esta publicación . A continuación se muestra un enfoque iterativo para el problema: 
 

• La idea es crear m listas enlazadas (m = número de filas) en las que cada Node almacene su Node derecho. Los punteros principales de cada m listas enlazadas se almacenan en una array de Nodes.
• Luego, recorra m listas, para cada i-ésima y (i+1) ^-ésima lista, establezca los punteros hacia abajo de cada Node de la i ^-ésima lista a su Node correspondiente de (i+1) ^-ésima lista.

A continuación se muestra la implementación del enfoque anterior: 
 

// 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 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 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# 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

1 2 3 
4 5 6 
7 8 9

Complejidad del tiempo: O(M * N)