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Run Length Decoding in Linked List
  • Last Updated : 24 Dec, 2020

Given an encoded Linked List which is encoded using the Run Length Encoding algorithm. The task is to decode the given linked list and generate the input string.
Run Length Encoding: In run length encoding, the input string is encoded by replacing a substring of repeated character in the string by the character followed by its count. If the character is single and is non-repeating than it’s count is not added. For Example, if the input string is “wwwwaaadexxxxxx”, then the function should return “w4a3dex6”
Examples: 
 

Input : List = a->5->b->r->3->NULL 
Output : string = “aaaaabrrr” 
Explanation : 
From the linked list, the character is ‘a’ and it’s count is 5 so the character is repeated 5 times
The next character is ‘b’ and the next character to it is not a number hence the character ‘b’ is repeated only once. 
The next character is ‘r’ and the count is 3 hence the character is repeated 3 times.
Input : List = a->b->r->3->a->3->NULL 
Output : string = “abrrraaa” 
 

 

Approach
 

  • Traverse through the linked list.
  • Store the current character in a variable c.
  • Check if the next node is a number and store the number in count else count is 1.
  • Append the character c to the list count times.

Below is the implementation of the above approach: 
 

C++




// C++ program to decode a linked list
#include <bits/stdc++.h>
using namespace std;
 
// Linked list node
struct Node {
    char data;
    struct Node* next;
};
 
// Utility function to create a new Node
Node* newNode(char data)
{
    Node* temp = new Node;
    temp->data = data;
    temp->next = NULL;
 
    return temp;
}
 
// Function to append nodes to a list
void append(struct Node* head_ref, char new_data)
{
    struct Node* new_node = new Node;
 
    struct Node* last = head_ref;
 
    new_node->data = new_data;
 
    new_node->next = NULL;
 
    if (head_ref == NULL) {
        head_ref = new_node;
        return;
    }
 
    while (last->next != NULL)
        last = last->next;
 
    last->next = new_node;
    return;
}
 
// Function to print list
void printList(Node* node)
{
    while (node != NULL) {
        cout << node->data << " ";
 
        node = node->next;
    }
}
 
// Function to decode the linked list
string decodeList(Node* head)
{
    Node* p = head;
    string res = "";
    int count;
 
    // While there are nodes left
    while (p) {
 
        // To store the count by which the current
        // character needs to be repeated
        count = 0;
 
        // Get the current character
        char c = p->data;
        if (p->next) {
            Node* temp = p->next;
 
            // If current node is a digit
            if (temp && temp->data >= '0'
                && temp->data <= '9') {
 
                // Generate the integer from
                // the consecutive digits
                while (temp && temp->data >= '0'
                       && temp->data <= '9') {
                    count = count * 10 + (temp->data - '0');
                    temp = temp->next;
                }
                p = temp;
            }
            else {
                count = 1;
                p = p->next;
            }
        }
        else {
            count = 1;
            p = p->next;
        }
 
        // Repeat the character count times
        for (int i = 0; i < count; i++) {
            res += c;
        }
    }
 
    return res;
}
 
// Driver code
int main()
{
    // Creating the linked list
    Node* head = newNode('a');
    head->next = newNode('5');
    head->next->next = newNode('b');
    head->next->next->next = newNode('r');
 
    cout << decodeList(head);
 
    return 0;
}

Java




// Java program to decode a linked list
class GFG {
 
    // Linked list node
    static class Node {
        char data;
        Node next;
    };
 
    // Utility function to create a new Node
    static Node newNode(char data)
    {
        Node temp = new Node();
        temp.data = data;
        temp.next = null;
 
        return temp;
    }
 
    // Function to append nodes to a list
    static void append(Node head_ref, char new_data)
    {
        Node new_node = new Node();
 
        Node last = head_ref;
 
        new_node.data = new_data;
 
        new_node.next = null;
 
        if (head_ref == null) {
            head_ref = new_node;
            return;
        }
 
        while (last.next != null)
            last = last.next;
 
        last.next = new_node;
        return;
    }
 
    // Function to print list
    static void printList(Node node)
    {
        while (node != null) {
            System.out.print(node.data + " ");
 
            node = node.next;
        }
    }
 
    // Function to decode the linked list
    static String decodeList(Node head)
    {
        // Pointer used to traverse through all
        // the nodes in the list
        Node p = head;
 
        // String to store the decoded message
        String res = "";
 
        int count;
 
        // While there are nodes left
        while (p != null) {
 
            // To store the count by which the current
            // character needs to be repeated
            count = 0;
 
            // Get the current character
            char c = p.data;
            if (p.next != null) {
                Node temp = p.next;
 
                // If current node is a digit
                if (temp != null && temp.data >= '0'
                    && temp.data <= '9') {
 
                    // Generate the integer from
                    // the consecutive digits
                    while (temp != null && temp.data >= '0'
                           && temp.data <= '9') {
                        count = count * 10 + (temp.data - '0');
                        temp = temp.next;
                    }
                    p = temp;
                }
                else {
                    count = 1;
                    p = p.next;
                }
            }
            else {
                count = 1;
                p = p.next;
            }
 
            // Repeat the character count times
            for (int i = 0; i < count; i++) {
                res += c;
            }
        }
 
        return res;
    }
 
    // Driver code
    public static void main(String args[])
    {
        // Creating the linked list
        Node head = newNode('a');
        head.next = newNode('5');
        head.next.next = newNode('b');
        head.next.next.next = newNode('r');
 
        System.out.println(decodeList(head));
    }
}
 
/* This code contributed by PrinciRaj1992 */

Python3




# Python3 program to decode a linked list
  
# Linked list node
class Node:  
    def __init__(self, data):       
        self.data = data
        self.next = None
  
# Utility function to create a Node
def newNode(data):
    temp = Node(data);
    temp.next = None;
  
    return temp;
 
# Function to append nodes to a list
def append(head_ref, new_data):
    new_node = Node;
    last = head_ref;
    new_node.data = new_data;
    new_node.next = None;
    if (head_ref == None):
        head_ref = new_node;
        return;
    while (last.next != None):
        last = last.next;
    last.next = new_node;
    return;
  
# Function to print list
def printList(node):
    while (node != None):       
        print(node.data, end = ' ')
        node = node.next;
      
# Function to decode the linked list
def decodeList(head):
    p = head;
    res = "";
    count = 0
  
    # While there are nodes left
    while (p != None):
  
        # To store the count by which the current
        # character needs to be repeated
        count = 0;
  
        # Get the current character
        c = p.data;
        if (p.next != None):
             
            temp = p.next;
  
            # If current node is a digit
            if (temp != None and ord(temp.data) >= ord('0')
                and ord(temp.data) <= ord('9')):
  
                # Generate the integer from
                # the consecutive digits
                while (temp != None and ord(temp.data) >= ord('0')
                       and ord(temp.data) <= ord('9')):
                    count = count * 10 + ord(temp.data) - ord('0')
                    temp = temp.next;
             
                p = temp;
             
            else:
                count = 1;
                p = p.next;           
        else:
            count = 1;
            p = p.next;
          
        # Repeat the character count times
        for i in range(0, count):      
            res += c;
    return res;
  
# Driver code
if __name__=='__main__':
     
    # Creating the linked list
    head = newNode('a');
    head.next = newNode('5');
    head.next.next = newNode('b');
    head.next.next.next = newNode('r');
  
    print(decodeList(head))
 
  # This code is contributed by rutvik_56

C#




// C# program to decode a linked list
using System;
 
public class GFG {
 
    // Linked list node
    public class Node {
        public char data;
        public Node next;
    };
 
    // Utility function to create a new Node
    static Node newNode(char data)
    {
        Node temp = new Node();
        temp.data = data;
        temp.next = null;
 
        return temp;
    }
 
    // Function to append nodes to a list
    static void append(Node head_ref, char new_data)
    {
        Node new_node = new Node();
 
        Node last = head_ref;
 
        new_node.data = new_data;
 
        new_node.next = null;
 
        if (head_ref == null) {
            head_ref = new_node;
            return;
        }
 
        while (last.next != null)
            last = last.next;
 
        last.next = new_node;
        return;
    }
 
    // Function to print list
    static void printList(Node node)
    {
        while (node != null) {
            Console.Write(node.data + " ");
 
            node = node.next;
        }
    }
 
    // Function to decode the linked list
    static String decodeList(Node head)
    {
        // Pointer used to traverse through all
        // the nodes in the list
        Node p = head;
 
        // String to store the decoded message
        String res = "";
 
        int count;
 
        // While there are nodes left
        while (p != null) {
 
            // To store the count by which the current
            // character needs to be repeated
            count = 0;
 
            // Get the current character
            char c = p.data;
            if (p.next != null) {
                Node temp = p.next;
 
                // If current node is a digit
                if (temp != null && temp.data >= '0'
                    && temp.data <= '9') {
 
                    // Generate the integer from
                    // the consecutive digits
                    while (temp != null && temp.data >= '0'
                           && temp.data <= '9') {
                        count = count * 10 + (temp.data - '0');
                        temp = temp.next;
                    }
                    p = temp;
                }
                else {
                    count = 1;
                    p = p.next;
                }
            }
            else {
                count = 1;
                p = p.next;
            }
 
            // Repeat the character count times
            for (int i = 0; i < count; i++) {
                res += c;
            }
        }
 
        return res;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        // Creating the linked list
        Node head = newNode('a');
        head.next = newNode('5');
        head.next.next = newNode('b');
        head.next.next.next = newNode('r');
 
        Console.WriteLine(decodeList(head));
    }
}
 
// This code is contributed by 29AjayKumar
Output: 
aaaaabr

 




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