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++ 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; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

1 2 3 4 5 6 7 8 9

**Time complexity: ** O(M * N)

## Recommended Posts:

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- Recursive Approach to find nth node from the end in the linked list
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- Check if linked list is sorted (Iterative and Recursive)
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