Add the given digit to a number stored in a linked list

Given a linked list which represents an integer number where every node is a digit if the represented integer. The task is to add a given digit N to the represented integer.

Examples:

Input: 9 -> 9 -> 3 -> NULL, N = 7
Output:
9 -> 9 -> 3 -> NULL
1 -> 0 -> 0 -> 0 -> NULL

Input: 2 -> 9 -> 9 -> NULL, N = 5
Output:
2 -> 9 -> 9 -> NULL
3 -> 0 -> 4 -> NULL

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: We have already discussed the approach for adding 1 to a number stored in linked list int this article but the code requires reversal of the linked list.
In this post, we have extended the problem to adding any digit to the number stored in a linked list and achieving the same without reversal or recursion.
The idea is to traverse the list and while traversing maintain a pointer to the last node whose value is less than 9. This is because we are adding a single digit to the number stored in the linked list. So, the maximum value of carry (if present) can be 1. Suppose we start propagating the carry from the least significant digit towards most significant digit, then the propagation will stop as soon as it finds a number less than 9.
After the complete traversal of the list in this manner, we have finally reached the last node of the linked list and also maintained a pointer to the latest node whose value is less than 9.
Two cases can arise:

There can be overflow after adding the number in the last digit i.e. value at the node is greater than 9.
No overflow i.e. after adding the value at the node is less than 10.

In the first case, we have to propagate the carry from the latest node whose value is less than 9 to the last node.

In the second case, we don’t have to do anything else.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach 
#include <iostream> 
using namespace std; 
  
// Node structure containing data 
// and pointer to the next Node 
struct node { 
  
    int key; 
    node* next; 
  
    node(int n) 
    { 
        key = n; 
        next = NULL; 
    } 
}; 
  
// Linked list class 
class LinkedList { 
  
    node* head; 
  
public: 
    // Deafault constructor for 
    // creating empty list 
    LinkedList(); 
  
    // Insert a node in linked list 
    void insert(node* n); 
  
    // Adding a single digit to the list 
    void addDigit(int n); 
  
    // Print the linked list 
    void printList(); 
}; 
  
LinkedList::LinkedList() 
{ 
    // Empty List 
    head = NULL; 
} 
  
// Function to insert a node at the 
// head of the linked list 
void LinkedList::insert(node* n) 
{ 
    // Empty List 
    if (head == NULL) 
        head = n; 
  
    // Insert in the beginning of the list 
    else { 
        n->next = head; 
        head = n; 
    } 
} 
  
// Function to print the linked list 
void LinkedList::printList() 
{ 
    node* ptr = head; 
  
    while (ptr) { 
        cout << ptr->key << " -> "; 
        ptr = ptr->next; 
    } 
    cout << "NULL" << endl; 
} 
  
// Function to add a digit to the integer 
// represented as a linked list 
void LinkedList::addDigit(int n) 
{ 
  
    // To keep track of the last node 
    // whose value is less than 9 
    node* lastNode = NULL; 
    node* curr = head; 
  
    while (curr->next) { 
  
        // If found a node with value 
        // less than 9 
        if (curr->key < 9) 
            lastNode = curr; 
  
        // Otherwise keep traversing 
        // the list till end 
        curr = curr->next; 
    } 
  
    // Add the given digit to the last node 
    curr->key = curr->key + n; 
  
    // In case of overflow in the last node 
    if (curr->key > 9) { 
  
        curr->key = curr->key % 10; 
  
        // If the list is of the 
        // form 9 -> 9 -> 9 -> ... 
        if (lastNode == NULL) { 
  
            // Insert a node at the beginnig as 
            // there would be overflow in the 
            // head in this case 
            insert(new node(1)); 
  
            // Adjust the lastNode pointer to 
            // propagate the carry effect to 
            // all the nodes of the list 
            lastNode = head->next; 
        } 
  
        // Forward propagate carry effect 
        while (lastNode != curr) { 
            lastNode->key = (lastNode->key + 1) % 10; 
            lastNode = lastNode->next; 
        } 
    } 
} 
  
// Driver code 
int main() 
{ 
    // Creating the linked list 
    LinkedList* l1 = new LinkedList(); 
  
    // Adding elements to the linked list 
    l1->insert(new node(9)); 
    l1->insert(new node(9)); 
    l1->insert(new node(1)); 
  
    // Printing the original list 
    l1->printList(); 
  
    // Adding the digit 
    l1->addDigit(5); 
  
    // Printing the modified list 
    l1->printList(); 
  
    return 0; 
} 

Output:

1 -> 9 -> 9 -> NULL
2 -> 0 -> 4 -> NULL

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

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