# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ith list to its corresponding node of (i+1)th list. Below is the implementation of the above approach:

 `// C++ program to construct a linked ` `// list from 2D matrix | Iterative Approach ` `#include ` `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[], ``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 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[] = { { 1, 2, 3 }, ` `                     ``{ 4, 5, 6 }, ` `                     ``{ 7, 8, 9 } }; ` ` `  `    ``node* head = constructLinkedMatrix(mat, m, n); ` `    ``display(head); ` ` `  `    ``return` `0; ` `} `

Output:

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

Time complexity: O(M * N)

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