Add one to a number represented as linked list | Set 2

Given a singly linked list which represents a number where each node contains only one digit [0 – 9]. The task is to add 1 to the number represented by the given linked list and print the new linked list.

Examples:

Input: 1 -> 2 -> 9 -> 9 -> NULL
Output:
Original list is : 1 2 9 9
Resultant list is : 1 3 0 0



Input: 9 -> 9 -> 9 -> 9 -> NULL
Output:
Original list is : 9 9 9 9
Resultant list is : 1 0 0 0 0

Approach: A previous implementation of the above problem problem was discussed in this post. However, one of the implementations requires the linked list to be reversed and the other makes use of recursion. An O(1) space complexity solution is been discussed here which doesn’t require the linked list to be reversed.
The main focus in this question is on the digit 9 which creates all the changes otherwise for other digits we have to just increment their value by 1 but if we change node’s value with value 9 it makes a carry which then has to be passed through the linked list.

Find the last node in the linked list which is not equal to 9. Now there are three cases:

  1. If there is no such node i.e. the value of every node is 9 then the new linked list will contain all 0s and a single 1 inserted at the head of the linked list.
  2. If the rightmost node which is not equal to 9 is the last node in the linked list then add 1 to this node and return the head of the linked list.
  3. If the node is other than the last node i.e. every node after it is equal to 9 then add 1 to the current node and change all the nodes after it to 0.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Node of the linked list
struct Node {
    int data;
    Node* next;
};
  
// Function to create a new node
Node* create_Node(int data)
{
    Node* temp = new Node();
    temp->data = data;
    temp->next = NULL;
    return temp;
}
  
// Function to print the linked list
void print(Node* head)
{
  
    Node* temp = head;
    while (temp != NULL) {
        cout << temp->data << " ";
        temp = temp->next;
    }
    cout << endl;
}
  
// Function to add one to a number
// represented as linked list
Node* addOne(Node* head)
{
  
    // To store the last node in the linked
    // list which is not equal to 9
    Node* last = NULL;
    Node* cur = head;
  
    // Iterate till the last node
    while (cur->next != NULL) {
  
        if (cur->data != 9) {
            last = cur;
        }
        cur = cur->next;
    }
  
    // If last node is not equal to 9
    // add 1 to it and return the head
    if (cur->data != 9) {
        cur->data++;
        return head;
    }
  
    // If list is of the type 9 -> 9 -> 9 ...
    if (last == NULL) {
        last = new Node();
        last->data = 0;
        last->next = head;
        head = last;
    }
  
    // For cases when the righmost node which
    // is not equal to 9 is not the last
    // node in the linked list
    last->data++;
    last = last->next;
  
    while (last != NULL) {
        last->data = 0;
        last = last->next;
    }
  
    return head;
}
  
// Driver code
int main()
{
    Node* head = create_Node(1);
    head->next = create_Node(2);
    head->next->next = create_Node(9);
    head->next->next->next = create_Node(9);
  
    cout << "Original list is : ";
    print(head);
  
    head = addOne(head);
  
    cout << "Resultant list is : ";
    print(head);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG
{
  
// Node of the linked list
static class Node 
{
    int data;
    Node next;
};
  
// Function to create a new node
static Node create_Node(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.next = null;
    return temp;
}
  
// Function to print the linked list
static void print(Node head)
{
    Node temp = head;
    while (temp != null
    {
        System.out.print(temp.data + " ");
        temp = temp.next;
    }
    System.out.println();
}
  
// Function to add one to a number
// represented as linked list
static Node addOne(Node head)
{
  
    // To store the last node in the linked
    // list which is not equal to 9
    Node last = null;
    Node cur = head;
  
    // Iterate till the last node
    while (cur.next != null)
    {
        if (cur.data != 9)
        {
            last = cur;
        }
        cur = cur.next;
    }
  
    // If last node is not equal to 9
    // add 1 to it and return the head
    if (cur.data != 9)
    {
        cur.data++;
        return head;
    }
  
    // If list is of the type 9 . 9 . 9 ...
    if (last == null)
    {
        last = new Node();
        last.data = 0;
        last.next = head;
        head = last;
    }
  
    // For cases when the righmost node which
    // is not equal to 9 is not the last
    // node in the linked list
    last.data++;
    last = last.next;
  
    while (last != null
    {
        last.data = 0;
        last = last.next;
    }
    return head;
}
  
// Driver code
public static void main(String[] args) 
{
    Node head = create_Node(1);
    head.next = create_Node(2);
    head.next.next = create_Node(9);
    head.next.next.next = create_Node(9);
  
    System.out.print("Original list is : ");
    print(head);
  
    head = addOne(head);
  
    System.out.print("Resultant list is : ");
    print(head);
}
}
  
// This code is contributed 
// by PrinciRaj1992 

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C#

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// C# implementation of the approach
using System;
      
class GFG
{
  
// Node of the linked list
public class Node 
{
    public int data;
    public Node next;
};
  
// Function to create a new node
static Node create_Node(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.next = null;
    return temp;
}
  
// Function to print the linked list
static void print(Node head)
{
    Node temp = head;
    while (temp != null
    {
        Console.Write(temp.data + " ");
        temp = temp.next;
    }
    Console.WriteLine();
}
  
// Function to add one to a number
// represented as linked list
static Node addOne(Node head)
{
  
    // To store the last node in the linked
    // list which is not equal to 9
    Node last = null;
    Node cur = head;
  
    // Iterate till the last node
    while (cur.next != null)
    {
        if (cur.data != 9)
        {
            last = cur;
        }
        cur = cur.next;
    }
  
    // If last node is not equal to 9
    // add 1 to it and return the head
    if (cur.data != 9)
    {
        cur.data++;
        return head;
    }
  
    // If list is of the type 9 . 9 . 9 ...
    if (last == null)
    {
        last = new Node();
        last.data = 0;
        last.next = head;
        head = last;
    }
  
    // For cases when the righmost node which
    // is not equal to 9 is not the last
    // node in the linked list
    last.data++;
    last = last.next;
  
    while (last != null
    {
        last.data = 0;
        last = last.next;
    }
    return head;
}
  
// Driver code
public static void Main(String[] args) 
{
    Node head = create_Node(1);
    head.next = create_Node(2);
    head.next.next = create_Node(9);
    head.next.next.next = create_Node(9);
  
    Console.Write("Original list is : ");
    print(head);
  
    head = addOne(head);
  
    Console.Write("Resultant list is : ");
    print(head);
}
}
  
// This code is contributed by 29AjayKumar

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Output:

Original list is : 1 2 9 9 
Resultant list is : 1 3 0 0


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Improved By : princiraj1992, 29AjayKumar



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