Construct a linked list from 2D matrix

Given a matrix. Convert it into a linked list matrix such that each node is connected to its next right and down node.

Input : 2D matrix 
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

Question Source: Factset Interview Experience | Set 9

Approach: The problem can be solved based on the following idea:

Connect every cell to its right cell of the same row and to its bottom cell in the same column and also for each cell and keep track of those created node. 

Follow the steps mentioned below to solve this problem:

• First create a variable of Node type, which will store address of its right and bottom Node corresponding to cell in the matrix.
• Recursively do the following steps for any cell in the matrix:
  • If Node is not created for any corresponding cell in the matrix, then create a new Node and store it.
  • Else we reach at some cell which has already been created for its corresponding cell in the matrix then return that stored Node.
  • Attach Node to its right and bottom cell which is created and return the current Node.
• Finally return the root Node.

Below is the implementation of the above approach:

C++

// CPP program to construct a linked list
// from given 2D matrix
#include <bits/stdc++.h>
using namespace std;
// struct node of linked list
struct Node {
int data;
Node* right, *down;
};
// returns head pointer of linked list
// constructed from 2D matrix
Node* construct(int arr[][4], int i, int j,
int m, int n, vector<vector<Node*>> &visited)
{
// return if i or j is out of bounds
if (i > m - 1 || j > n - 1)
return NULL;
// Check if node is previously created then,
// don't need to create new/
if(visited[i][j]){
return visited[i][j];
}
// create a new node for current i and j
// and recursively allocate its down and
// right pointers
Node* temp = new Node();
visited[i][j] = temp;
temp->data = arr[i][j];
temp->right = construct(arr, i, j + 1, m, n, visited);
temp->down = construct(arr, i + 1, j, m, n, visited);
return temp;
}
// utility function for displaying
// linked list data
void display(Node* head)
{
// pointer to move right
Node* Rp;
// pointer to move down
Node* Dp = head;
// loop till node->down is not NULL
while (Dp) {
Rp = Dp;
// loop till node->right is not NULL
while (Rp) {
cout << Rp->data << " ";
Rp = Rp->right;
}
cout << "\n";
Dp = Dp->down;
}
}
// driver program
int main()
{
// 2D matrix
int arr[][4] = {
{ 1, 2, 3, 0},
{ 4, 5, 6 , 1},
{ 7, 8, 9 , 2},
{ 7, 8, 9 , 2}
};
int m = 4, n = 4;
vector<vector<Node*>> visited(m, vector<Node*>(n));
Node* head = construct(arr, 0, 0, m, n, visited);
display(head);
return 0;
}

Java

// Java program to construct a linked list
// from given 2D matrix
public class Linked_list_2D_Matrix {
// node of linked list
static class Node {
int data;
Node right;
Node down;
};
// returns head pointer of linked list
// constructed from 2D matrix
static Node construct(int arr[][], int i, int j,
int m, int n, Node[][] visited) {
// return if i or j is out of bounds
if (i > m - 1 || j > n - 1)
return null;
// Check if node is previously created then,
// don't need to create new/
if(visited[i][j] != null){
return visited[i][j];
}
// create a new node for current i and j
// and recursively allocate its down and
// right pointers
Node temp = new Node();
visited[i][j] = temp;
temp.data = arr[i][j];
temp.right = construct(arr, i, j + 1, m, n, visited);
temp.down = construct(arr, i + 1, j, m, n, visited);
return temp;
}
// utility function for displaying
// linked list data
static void display(Node head) {
// pointer to move right
Node Rp;
// pointer to move down
Node Dp = head;
// loop till node->down is not NULL
while (Dp != null) {
Rp = Dp;
// loop till node->right is not NULL
while (Rp != null) {
System.out.print(Rp.data + " ");
Rp = Rp.right;
}
System.out.println();
Dp = Dp.down;
}
}
// driver program
public static void main(String args[]) {
// 2D matrix
int arr[][] = {
{ 1, 2, 3, 0},
{ 4, 5, 6 , 1},
{ 7, 8, 9 , 2},
{ 7, 8, 9 , 2}
};
int m = 4, n = 4;
// List<List<Node>> arr = new ArrayList<List<Node>>();
Node[][] visited = new Node[m][n];
Node head = construct(arr, 0, 0, m, n, visited);
display(head);
}
}
// This code is contributed by Sumit Ghosh

Output

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

Time complexity: O(N*M), where N is the number of row and M is the number of column.
Auxiliary space: O(N*M)

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