Given a linked list where every node represents a linked list and contains two pointers of its type:
- Pointer to next node in the main list (we call it ‘right’ pointer in the code below).
- Pointer to a linked list where this node is headed (we call it the ‘down’ pointer in the code below).
All linked lists are sorted. See the following example
5 -> 10 -> 19 -> 28
| | | |
V V V V
7 20 22 35
| | |
V V V
8 50 40
| |
V V
30 45
Write a function flatten() to flatten the lists into a single linked list. The flattened linked list should also be sorted. For example, for the above input list, output list should be 5->7->8->10->19->20->22->28->30->35->40->45->50.
The idea is to use the Merge() process of merge sort for linked lists. We use merge() to merge lists one by one. We recursively merge() the current list with the already flattened list.
The down pointer is used to link nodes of the flattened list.
Below is the implementation of the above approach:
Java
// Java program for flattening // a Linked Listclass LinkedList{ // Head of list Node head; // Linked list Node class Node { int data; Node right, down; Node(int data) { this.data = data; right = null; down = null; } } // An utility function to merge // two sorted linked lists Node merge(Node a, Node b) { // If first linked list is empty // then second is the answer if (a == null) return b; // If second linked list is empty // then first is the result if (b == null) return a; // Compare the data members of the // two linked lists and put the // larger one in the result Node result; if (a.data < b.data) { result = a; result.down = merge(a.down, b); } else { result = b; result.down = merge(a, b.down); } result.right = null; return result; } Node flatten(Node root) { // Base Cases if (root == null || root.right == null) return root; // Recur for list on right root.right = flatten(root.right); // Now merge root = merge(root, root.right); // Return the root // it will be in turn merged with // its left return root; } /* Utility function to insert a node at beginning of the linked list */ Node push(Node head_ref, int data) { /* 1 & 2: Allocate the Node & Put in the data */ Node new_node = new Node(data); // 3. Make next of new Node as head new_node.down = head_ref; // 4. Move the head to point to // new Node head_ref = new_node; // 5. return to link it back return head_ref; } void printList() { Node temp = head; while (temp != null) { System.out.print(temp.data + " "); temp = temp.down; } System.out.println(); } // Driver code public static void main(String args[]) { LinkedList L = new LinkedList(); /* Create the following linked list 5 -> 10 -> 19 -> 28 | | | | V V V V 7 20 22 35 | | | V V V 8 50 40 | | V V 30 45 */ L.head = L.push(L.head, 30); L.head = L.push(L.head, 8); L.head = L.push(L.head, 7); L.head = L.push(L.head, 5); L.head.right = L.push(L.head.right, 20); L.head.right = L.push(L.head.right, 10); L.head.right.right = L.push(L.head.right.right, 50); L.head.right.right = L.push(L.head.right.right, 22); L.head.right.right = L.push(L.head.right.right, 19); L.head.right.right.right = L.push(L.head.right.right.right, 45); L.head.right.right.right = L.push(L.head.right.right.right, 40); L.head.right.right.right = L.push(L.head.right.right.right, 35); L.head.right.right.right = L.push(L.head.right.right.right, 20); // Flatten the list L.head = L.flatten(L.head); L.printList(); }} // This code is contributed by Rajat Mishra |
Output:
5 7 8 10 19 20 20 22 30 35 40 45 50
Time Complexity: O(N*N*M) – where N is the no of nodes in main linked list (reachable using right pointer) and M is the no of node in a single sub linked list (reachable using down pointer).
Space Complexity: O(N*M) as the recursive functions will use recursive stack of size equivalent to total number of elements in the lists.
Please refer complete article on Flattening a Linked List for more details!
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