Given a linked list which represents an integer number where each node is a digit of the represented integer. The task is to add a given digit N to the represented integer.
Examples:
Input: LL = 9 -> 9 -> 3 -> NULL, N = 7
Output: 1 -> 0 -> 0 -> 0 -> NULL
993 + 7 = 1000
Input: LL = 2 -> 9 -> 9 -> NULL, N = 5
Output: 3 -> 0 -> 4 -> NULL
Approach: An iterative approach to solve this problem has been discussed here. In this article, a recursive approach will be discussed.
The idea is to traverse the LinkedList recursively until the last node is reached. Once the last node has been reached, add the value of N to it. After adding, if the value is more than 9 then keep the carry and set mode (digit % 10) value to the node value and add carry to the previous stack frame node, and continue until all the stack frames are cleared from the stack.
If there is a carry after all the stack frames have been cleared then create a new node with this value which will be the new head of the linked list pointing to the previous head.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include<bits/stdc++.h>using namespace std;// Node class contains value// and next node referencestruct ListNode { int value; ListNode* next;};// To store the carryint carry = 0;void addNewValue(ListNode*, int);// Function that calls the recursive method// addNewValue to add a digit to the// number represented as the linked listListNode* addValue(ListNode* head, int addValue){ // Add the digit recursively addNewValue(head, addValue); // If there is a carry after the addition if (carry != 0) { // Create a new node ListNode* newHead = new ListNode(); // Assign it with carry newHead->value = carry; // Make it point to the head of // the linked list newHead->next = head; carry = 0; // Make it the new head return newHead; } // If there's not carry then // return the previous head else { return head; }}// Recursive function to add a digit to the number// represented as the given linked listvoid addNewValue(ListNode* head, int addValue){ // If it is the last node in the list if (head->next == NULL) { // Add the digit int val = head->value + addValue; // Find the carry if any head->value = val % 10; carry = val / 10; } else { // Preserve the current node's value and call // the recursive function for the next node int val = head->value; addNewValue(head->next, addValue); val = val + carry; head->value = val % 10; carry = val / 10; }}// Utility function to print the linked listvoid printList(ListNode* node){ while (node != NULL) { cout << node->value << " -> "; node = node->next; } cout<<"NULL";}// Driver codeint main(){ // Create the linked list 9 -> 9 -> 3 -> NULL ListNode* head = new ListNode(); head->value = 9; head->next = new ListNode(); head->next->value = 9; head->next->next = new ListNode(); head->next->next->value = 3; head->next->next->next = NULL; // Digit to be added int n = 7; head = addValue(head, n); printList(head);}// This code is contributed by rutvik_56 |
Java
// Java implementation of the approach// Node class contains value// and next node referenceclass ListNode { int value; ListNode next;}class GFG { // To store the carry private static int carry = 0; // Function that calls the recursive method // addNewValue to add a digit to the // number represented as the linked list public static ListNode addValue(ListNode head, int addValue) { // Add the digit recursively addNewValue(head, addValue); // If there is a carry after the addition if (carry != 0) { // Create a new node ListNode newHead = new ListNode(); // Assign it with carry newHead.value = carry; // Make it point to the head of // the linked list newHead.next = head; carry = 0; // Make it the new head return newHead; } // If there's not carry then // return the previous head else { return head; } } // Recursive function to add a digit to the number // represented as the given linked list private static void addNewValue(ListNode head, int addValue) { // If it is the last node in the list if (head.next == null) { // Add the digit int val = head.value + addValue; // Find the carry if any head.value = val % 10; carry = val / 10; } else { // Preserve the current node's value and call // the recursive function for the next node int val = head.value; addNewValue(head.next, addValue); val = val + carry; head.value = val % 10; carry = val / 10; } } // Utility function to print the linked list private static void printList(ListNode node) { while (node != null) { System.out.print(node.value + " -> "); node = node.next; } System.out.print("NULL"); } // Driver code public static void main(String[] args) { // Create the linked list 9 -> 9 -> 3 -> NULL ListNode head = new ListNode(); head.value = 9; head.next = new ListNode(); head.next.value = 9; head.next.next = new ListNode(); head.next.next.value = 3; head.next.next.next = null; // Digit to be added int n = 7; head = addValue(head, n); printList(head); }} |
Python
# Python implementation of the approach# Node class contains value# and next node referenceclass ListNode: def __init__(self, new_data): self.value = new_data self.next = None# To store the carrycarry = 0# Function that calls the recursive method# addNewValue to add a digit to the# number represented as the linked listdef addValue(head, addValue): global carry # Add the digit recursively addNewValue(head, addValue) # If there is a carry after the addition if (carry != 0) : # Create a node newHead = ListNode(0) # Assign it with carry newHead.value = carry # Make it point to the head of # the linked list newHead.next = head carry = 0 # Make it the head return newHead # If there's not carry then # return the previous head else : return head # Recursive function to add a digit to the number# represented as the given linked listdef addNewValue(head,addValue): global carry # If it is the last node in the list if (head.next == None) : # Add the digit val = head.value + addValue # Find the carry if any head.value = val % 10 carry = int(val / 10) else : # Preserve the current node's value and call # the recursive function for the next node val = head.value addNewValue(head.next, addValue) val = val + carry head.value = val % 10 carry = int(val / 10) # Utility function to print the linked listdef printList(node): while (node != None) : print(node.value ,end= " -> ") node = node.next print("None") # Driver code# Create the linked list 9 -> 9 -> 3 -> Nonehead = ListNode(0)head.value = 9head.next = ListNode(0)head.next.value = 9head.next.next = ListNode(0)head.next.next.value = 3head.next.next.next = None# Digit to be addedn = 7head = addValue(head, n)printList(head)# This code is contributed by Arnab Kundu |
C#
// C# implementation of the approach using System;// Node class contains value// and next node referencepublic class ListNode { public int value; public ListNode next;}class GFG{ // To store the carry private static int carry = 0; // Function that calls the recursive method // addNewValue to add a digit to the // number represented as the linked list public static ListNode addValue(ListNode head, int addValue) { // Add the digit recursively addNewValue(head, addValue); // If there is a carry after the addition if (carry != 0) { // Create a new node ListNode newHead = new ListNode(); // Assign it with carry newHead.value = carry; // Make it point to the head of // the linked list newHead.next = head; carry = 0; // Make it the new head return newHead; } // If there's not carry then // return the previous head else { return head; } } // Recursive function to add a digit to the number // represented as the given linked list private static void addNewValue(ListNode head, int addValue) { // If it is the last node in the list if (head.next == null) { // Add the digit int val = head.value + addValue; // Find the carry if any head.value = val % 10; carry = val / 10; } else { // Preserve the current node's value and call // the recursive function for the next node int val = head.value; addNewValue(head.next, addValue); val = val + carry; head.value = val % 10; carry = val / 10; } } // Utility function to print the linked list private static void printList(ListNode node) { while (node != null) { Console.Write(node.value + " -> "); node = node.next; } Console.Write("NULL"); } // Driver code public static void Main(String[] args) { // Create the linked list 9 -> 9 -> 3 -> NULL ListNode head = new ListNode(); head.value = 9; head.next = new ListNode(); head.next.value = 9; head.next.next = new ListNode(); head.next.next.value = 3; head.next.next.next = null; // Digit to be added int n = 7; head = addValue(head, n); printList(head); }}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// JavaScript implementation of the approach// Node class contains value// and next node referenceclass ListNode { constructor() { this.value = 0; this.next = null; } } // To store the carrylet carry = 0;// Function that calls the recursive method// addNewValue to add a digit to the// number represented as the linked listfunction addValue( head, addValue){ // Add the digit recursively addNewValue(head, addValue); // If there is a carry after the addition if (carry != 0) { // Create a new node var newHead = new ListNode(); // Assign it with carry newHead.value = carry; // Make it point to the head of // the linked list newHead.next = head; carry = 0; // Make it the new head return newHead; } // If there's not carry then // return the previous head else { return head; }}// Recursive function to add a digit to the number// represented as the given linked listfunction addNewValue( head, addValue){ // If it is the last node in the list if (head.next == null) { // Add the digit let val = head.value + addValue; // Find the carry if any head.value = val % 10; carry = Math.floor(val / 10); } else { // Preserve the current node's value and call // the recursive function for the next node let val = head.value; addNewValue(head.next, addValue); val = val + carry; head.value = val % 10; carry = Math.floor(val / 10); } }// Utility function to print the linked listfunction printList( node){ while (node != null) { document.write(node.value + " -> "); node = node.next; } document.write("NULL");}// Driver Code// Create the linked list 9 -> 9 -> 3 -> NULLvar head = new ListNode();head.value = 9;head.next = new ListNode();head.next.value = 9;head.next.next = new ListNode();head.next.next.value = 3;head.next.next.next = null;// Digit to be addedlet n = 7;head = addValue(head, n);printList(head); </script> |
Output:
1 -> 0 -> 0 -> 0 -> NULL
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