Flattening a linked list | Set 2

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 below code)
  • Pointer to a linked list where this node is head (we call it ‘down’ pointer in the below code).

All linked lists are sorted. See the following example
Examples: 

Input: 
5 -> 10 -> 19 -> 28
|    |     |     |
V    V     V     V
7    20    22    35
|          |     |
V          V     V
8          50    40
|                |
V                V
30               45

Output: 5->7->8->10->19->20->22->28->30->35->40->45->50

Input: 
5 -> 10 -> 19 -> 28
|          |   
V          V    
7          22   
|          |   
V          V    
8          50 
|              
V               
30              

Output: 5->7->8->10->19->20->22->30->50

In the previous post, we have to use the merge() process of merge sort for linked lists to flatten the linked list. 
In this post, we will solve it using Heap.

Approach: The idea is to observe that from each top node there are N nodes that are connected in a downward direction but observe that all the downward nodes are in sorted order. So the task is to sort this entire thing in increasing order(or decreasing order).

  1. Push the head of all the linked lists in the downward list in the priority queue.
  2. Pop the smallest node from the priority queue.
  3. Check the location of the node so that the next node pointed by the current node can be pushed into the priority queue.
  4. Again pop out the smallest element and insert the next node pointed by the current node till the heap becomes empty.
  5. Keep on adding the data of nodes in a new linked list that are pooped out to the new list.
  6. Print the linked list formed above.

Below is the implementation of the above approach:
 



C++

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// C++ program for Flattening
// a linked list using Heaps
#include <bits/stdc++.h>
using namespace std;
 
// Structure of given Linked list
struct Node {
    int data;
    struct Node* right;
    struct Node* down;
 
    Node(int x)
    {
        data = x;
        right = NULL;
        down = NULL;
    }
};
 
// Function to print the
// linked list
void printList(Node* Node)
{
    while (Node != NULL) {
        printf("%d ", Node->data);
        Node = Node->down;
    }
}
 
// Function that compares the value
// pointed by node and returns true
// if first data is greater
struct compare {
    bool operator()(Node* a, Node* b)
    {
        return a->data > b->data;
    }
};
 
// Function which returns the root
// of the flattened linked list
Node* flatten(Node* root)
{
    Node* ptr = root;
    Node* head = NULL;
 
    // Min Heap which will return
    // smallest element currently
    // present in heap
    priority_queue<Node*,
            vector<Node*>,
             compare> pq;
 
    // Push the head nodes of each
    // downward linked list
    while (ptr != NULL) {
        pq.push(ptr);
        ptr = ptr->right;
    }
 
    // This loop will execute
    // till the map becomes empty
    while (!pq.empty()) {
 
        // Pop out the node that
        // contains element
        // currently in heap
        Node* temp = pq.top();
        pq.pop();
 
        // Push the next node pointed by
        // the current node into heap
        // if it is not null
        if (temp->down != NULL) {
            pq.push(temp->down);
        }
 
        // Create new linked list
        // that is to be returned
        if (head == NULL) {
            head = temp;
            ptr = temp;
            ptr->right = NULL;
        }
        else {
            ptr->down = temp;
            ptr = temp;
            ptr->right = NULL;
        }
    }
 
    // Pointer to head node
    // in the linked list
    return head;
}
 
// Create and push new nodes
void push(Node** head_ref, int new_data)
{
    Node* new_node = (Node*)malloc(sizeof(Node));
    new_node->right = NULL;
    new_node->data = new_data;
    new_node->down = (*head_ref);
 
    (*head_ref) = new_node;
}
 
// Driver Code
int main()
{
    Node* root = NULL;
 
    // Given Linked List
    push(&root, 30);
    push(&root, 8);
    push(&root, 7);
    push(&root, 5);
 
    push(&(root->right), 20);
    push(&(root->right), 10);
 
    push(&(root->right->right), 50);
    push(&(root->right->right), 22);
    push(&(root->right->right), 19);
 
    push(&(root->right->right->right), 45);
    push(&(root->right->right->right), 40);
    push(&(root->right->right->right), 35);
    push(&(root->right->right->right), 20);
 
    // Flatten the list
    root = flatten(root);
 
    // Print the flatened linked list
    printList(root);
 
    return 0;
}

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Java

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// Java program for Flattening
// a linked list using Heaps
import java.util.*;
 
// Linked list Node
class Node {
    int data;
    Node right, down;
    Node(int data)
    {
        this.data = data;
        right = null;
        down = null;
    }
}
 
class pair {
    int val;
    Node head;
 
    pair(Node head, int val)
    {
        this.val = val;
        this.head = head;
    }
}
 
// Class that compares the value
// pointed by node and make
// LinkedList sorted
class pairComp implements Comparator<pair> {
    public int compare(pair p1, pair p2)
    {
        return p1.val - p2.val;
    }
}
 
class GFG {
 
    // Function which returns the root
    // of the flattened linked list
    public static Node flatten(Node root)
    {
        Node ptr = root;
        Node h = null;
 
        // Min Heap which will return
        // smallest element currently
        // present in heap
        PriorityQueue<pair> pq
            = new PriorityQueue<pair>(
                        new pairComp());
 
        // Push the head nodes of each
        // downward linked list
        while (ptr != null) {
            pq.add(new pair(ptr, ptr.data));
            ptr = ptr.right;
        }
 
        // This loop will execute
        // till the pq becomes empty
        while (!pq.isEmpty()) {
 
            // Pop out the node that
            // contains element
            // currently in heap
            Node temp = pq.poll().head;
 
            // Push the next node pointed by
            // the current node into heap
            // if it is not null
            if (temp.down != null) {
                pq.add(new pair(temp.down,
                                temp.down.data));
            }
 
            // Create new linked list
            // that is to be returned
            if (h == null) {
                h = temp;
                ptr = temp;
                ptr.right = null;
            }
            else {
                ptr.down = temp;
                ptr = temp;
                ptr.right = null;
            }
        }
 
        // Pointer to head node
        // in the linked list
        return h;
    }
 
    // Create and push new nodes
    public static Node push(Node head_ref,
                            int data)
    {
 
        // Allocate the Node &
        // Put in the data
        Node new_node = new Node(data);
 
        // Make next of new Node as head
        new_node.down = head_ref;
 
        // Move the head to point to new Node
        head_ref = new_node;
 
        // return to link it back
        return head_ref;
    }
 
    // Function to print the
    // linked list
    public static void printList(Node h)
    {
        while (h != null) {
            System.out.print(h.data + " ");
            h = h.down;
        }
    }
 
    // Driver code
    public static void main(String args[])
    {
        Node head = null;
 
        head = push(head, 30);
        head = push(head, 8);
        head = push(head, 7);
        head = push(head, 5);
 
        head.right = push(head.right, 20);
        head.right = push(head.right, 10);
 
        head.right.right = push(
                  head.right.right, 50);
        head.right.right = push(
                  head.right.right, 22);
        head.right.right = push(
                  head.right.right, 19);
 
        head.right.right.right
            = push(
          head.right.right.right, 45);
        head.right.right.right
            = push(
              head.right.right.right, 40);
        head.right.right.right
            = push(
              head.right.right.right, 35);
        head.right.right.right
            = push(head.right.right.right, 20);
 
        // Flatten the list
        head = flatten(head);
 
        printList(head);
    }
}
 
// This code is contributed by Naresh Saharan
// and Sagar Jangra and Tridib Samanta

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Output

5 7 8 10 19 20 20 22 30 35 40 45 50 

Time Complexity: O(k * log k) + O((N-k) * log k) = O(N * log k), where ‘k‘ is the number of nodes in the topmost horizontal linked list and ‘N‘ is the total number of nodes among all the linked lists. ‘log k‘ time is taken for the min-heapify procedure.
Auxiliary Space: O(k) for the min-heap, where ‘k‘ is the number of nodes in the topmost horizontal linked list. The min-heap will have at the most ‘k‘ number of nodes at any time.
 

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