Construct a linked list from 2D matrix (Iterative Approach) Last Updated : 11 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Given a Matrix mat of n*n size, the task is to construct a 2D linked list representation of the given matrix. Example: Input : mat = [[1 2 3] [4 5 6] [7 8 9]]Output : Input : mat = [[23 28] [23 28]]Output : Approach:To Construct a linked list from 2D matrix iteratively follow the steps below: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. Construct a linked list from 2D matrixBelow is the implementation of the above approach: C++ // C++ to implement linked matrix // from a 2D matrix iteratively #include <bits/stdc++.h> using namespace std; class Node { public: int data; Node *right, *down; Node(int x) { data = x; right = down = nullptr; } }; // Function to construct the linked list from the given /// matrix and return the head pointer Node* constructLinkedMatrix(vector<vector<int>>& mat) { int m = mat.size(); int n = mat[0].size(); // Stores the head of the linked list Node* mainhead = nullptr; // Stores the head of linked lists of each row vector<Node*> head(m); Node *rightcurr, *newptr; // Create m linked lists by setting all // the right nodes of every row for (int i = 0; i < m; i++) { head[i] = nullptr; for (int j = 0; j < n; j++) { newptr = new Node(mat[i][j]); // Assign the first node as the mainhead if (!mainhead) { mainhead = newptr; } // Set the head for the row or link // the current node if (!head[i]) { head[i] = newptr; } else { rightcurr->right = newptr; } rightcurr = newptr; } } // Set the down pointers for nodes in consecutive rows for (int i = 0; i < m - 1; i++) { Node *curr1 = head[i], *curr2 = head[i + 1]; // Link corresponding nodes in consecutive rows while (curr1 && curr2) { curr1->down = curr2; curr1 = curr1->right; curr2 = curr2->right; } } // Return the mainhead of the constructed // linked list return mainhead; } void printList(Node *head) { Node *currRow = head; while (currRow != nullptr) { Node *currCol = currRow; while (currCol != nullptr) { cout << currCol->data << " "; currCol = currCol->right; } cout << endl; currRow = currRow->down; } } int main() { vector<vector<int>> mat = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } }; Node* head = constructLinkedMatrix(mat); printList(head); return 0; } Java // Java to implement linked matrix // from a 2D matrix iteratively import java.util.ArrayList; class Node { int data; Node right, down; Node(int data) { this.data = data; this.right = null; this.down = null; } } class GfG { // Function to construct the linked // matrix from a 2D array static Node construct(int arr[][]) { int m = arr.length; int n = arr[0].length; // Stores the head of the linked list Node mainhead = null; // ArrayList to store the heads of // linked lists for each row ArrayList<Node> head = new ArrayList<>(m); Node rightcurr = null; // Create m linked lists by setting all // the right nodes of every row for (int i = 0; i < m; i++) { head.add(null); for (int j = 0; j < n; j++) { Node newptr = new Node(arr[i][j]); // Assign the first node as the mainhead if (mainhead == null) { mainhead = newptr; } // Set the head for the row or link // the current node if (head.get(i) == null) { head.set(i, newptr); } else { rightcurr.right = newptr; } rightcurr = newptr; } } // Set the down pointers for nodes in // consecutive rows for (int i = 0; i < m - 1; i++) { Node curr1 = head.get(i); Node curr2 = head.get(i + 1); // Link corresponding nodes in consecutive rows while (curr1 != null && curr2 != null) { curr1.down = curr2; curr1 = curr1.right; curr2 = curr2.right; } } // Return the mainhead of the constructed // linked list return mainhead; } static void printList(Node head) { Node currRow = head; while (currRow != null) { Node currCol = currRow; while (currCol != null) { System.out.print(currCol.data + " "); currCol = currCol.right; } System.out.println(); currRow = currRow.down; } } public static void main(String[] args) { int arr[][] = { {1, 2, 3}, {4, 5, 6}, {7, 8, 9} }; Node head = construct(arr); printList(head); } } Python # Python to implement linked matrix # from a 2D matrix iteratively class Node: def __init__(self, data): self.data = data self.right = None self.down = None def constructLinkedMatrix(mat): m = len(mat) n = len(mat[0]) # Stores the head of the linked list mainhead = None # Stores the head of linked lists of each row head = [None] * m rightcurr = None # Create m linked lists by setting all the # right nodes of every row for i in range(m): head[i] = None for j in range(n): newptr = Node(mat[i][j]) # Assign the first node as the mainhead if not mainhead: mainhead = newptr # Set the head for the row or link the # current node if not head[i]: head[i] = newptr else: rightcurr.right = newptr rightcurr = newptr # Set the down pointers for nodes in consecutive rows for i in range(m - 1): curr1 = head[i] curr2 = head[i + 1] # Link corresponding nodes in consecutive rows while curr1 and curr2: curr1.down = curr2 curr1 = curr1.right curr2 = curr2.right # Return the mainhead of the constructed linked list return mainhead def printList(head): currRow = head while currRow: currCol = currRow while currCol: print(currCol.data, end=" ") currCol = currCol.right print() currRow = currRow.down if __name__ == "__main__": mat = [ [1, 2, 3], [4, 5, 6], [7, 8, 9] ] head = constructLinkedMatrix(mat) printList(head) C# // C# to implement linked matrix // from a 2D matrix iteratively using System; using System.Collections.Generic; class Node { public int data; public Node right, down; public Node(int x) { data = x; right = down = null; } } class GfG { // Function to construct the linked matrix // from a List of Lists static Node Construct(List<List<int>> arr) { int m = arr.Count; int n = arr[0].Count; // Stores the head of the linked list Node mainhead = null; // List to store the heads of linked // lists for each row List<Node> head = new List<Node>(m); Node rightcurr = null; // Create m linked lists by setting all the // right nodes of every row for (int i = 0; i < m; i++) { head.Add(null); for (int j = 0; j < n; j++) { Node newptr = new Node(arr[i][j]); // Assign the first node as the mainhead if (mainhead == null) { mainhead = newptr; } // Set the head for the row or link // the current node if (head[i] == null) { head[i] = newptr; } else { rightcurr.right = newptr; } rightcurr = newptr; } } // Set the down pointers for nodes // in consecutive rows for (int i = 0; i < m - 1; i++) { Node curr1 = head[i]; Node curr2 = head[i + 1]; // Link corresponding nodes in // consecutive rows while (curr1 != null && curr2 != null) { curr1.down = curr2; curr1 = curr1.right; curr2 = curr2.right; } } // Return the mainhead of the // constructed linked list return mainhead; } static void PrintList(Node head) { Node currRow = head; while (currRow != null) { Node currCol = currRow; while (currCol != null) { Console.Write(currCol.data + " "); currCol = currCol.right; } Console.WriteLine(); currRow = currRow.down; } } static void Main(string[] args) { List<List<int>> arr = new List<List<int>> { new List<int> { 1, 2, 3 }, new List<int> { 4, 5, 6 }, new List<int> { 7, 8, 9 } }; Node head = Construct(arr); PrintList(head); } } JavaScript // Javascript to implement linked matrix // from a 2D matrix iteratively class Node { constructor(data) { this.data = data; this.right = null; this.down = null; } } // Function to construct the linked matrix // from a 2D array function constructLinkedMatrix(arr) { const m = arr.length; const n = arr[0].length; // Stores the head of the linked list let mainHead = null; // Array to store the heads of linked // lists for each row const head = new Array(m).fill(null); let rightCurr = null; // Create m linked lists by setting all the // right nodes of every row for (let i = 0; i < m; i++) { for (let j = 0; j < n; j++) { const newPtr = new Node(arr[i][j]); // Assign the first node as the mainHead if (mainHead === null) { mainHead = newPtr; } // Set the head for the row or // link the current node if (head[i] === null) { head[i] = newPtr; } else { rightCurr.right = newPtr; } rightCurr = newPtr; } } // Set the down pointers for nodes in // consecutive rows for (let i = 0; i < m - 1; i++) { let curr1 = head[i]; let curr2 = head[i + 1]; // Link corresponding nodes in consecutive rows while (curr1 && curr2) { curr1.down = curr2; curr1 = curr1.right; curr2 = curr2.right; } } // Return the mainHead of the constructed // linked list return mainHead; } function printList(head) { let currRow = head; while (currRow !== null) { let currCol = currRow; while (currCol !== null) { console.log(currCol.data + " "); currCol = currCol.right; } console.log(); currRow = currRow.down; } } const arr = [ [1, 2, 3], [4, 5, 6], [7, 8, 9] ]; const head = constructLinkedMatrix(arr); printList(head); Output1 2 3 4 5 6 7 8 9 Time Complexity: O(n^2), where n is the number of rows and columns in the matrix. This is because we need to traverse each element of the matrix exactly once to create the linked list nodes.Space Complexity: O(n^2), for storing the linked list nodes since we are creating a new node for each element in the matrix.Related articles: Construct a linked list from 2D matrix Comment More infoAdvertise with us Next Article Types of Asymptotic Notations in Complexity Analysis of Algorithms S souravdutta123 Follow Improve Article Tags : Linked List DSA Practice Tags : Linked List Similar Reads Basics & PrerequisitesTime Complexity and Space ComplexityMany times there are more than one ways to solve a problem with different algorithms and we need a way to compare multiple ways. Also, there are situations where we would like to know how much time and resources an algorithm might take when implemented. 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