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

Last Updated : 06 Jul, 2021

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 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 the node's value with the 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:

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.
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.
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++

// 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 rightmost 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; }

Java

// 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 rightmost 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

Python3

# A Python3 implementation of the approach  # Node of the linked list class Node():     def __init__(self):         self.data = None         self.next = None  # Function to create a new node def create_Node(data):     temp = Node()     temp.data = data     temp.next = None     return temp  # Function to print the linked list def printList(head):     temp = head     while (temp != None):         print(temp.data, end = ' ')         temp = temp.next     print()  # Function to add one to a number # represented as linked list def addOne(head):      # To store the last node in the     # linked list which is not equal to 9     last = None     cur = head      # Iterate till the last node     while(cur.next != None):         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 += 1         return head      # If list is of the type 9 -> 9 -> 9 ...     if(last == None):         last = Node()         last.data = 0         last.next = head         head = last      # For cases when the rightmost node which     # is not equal to 9 is not the last     # node in the linked list     last.data += 1     last = last.next      while(last != None):         last.data = 0         last = last.next      return head  # Driver code if __name__=='__main__':     head = create_Node(1)     head.next = create_Node(2)     head.next.next = create_Node(9)     head.next.next.next = create_Node(9)      print("Original list is : ", end = "")     printList(head)      head = addOne(head)      print("Resultant list is : ", end = "")     printList(head)  # This code is contributed by Yashyasvi Agarwal

// 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 rightmost 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

JavaScript

<script>      // Javascript implementation of the approach  // Node of the linked list class Node {      constructor()     {         this.data =0;         this.next = null;     } };  // Function to create a new node function create_Node(data) {     var temp = new Node();     temp.data = data;     temp.next = null;     return temp; }  // Function to print the linked list function print(head) {      var temp = head;     while (temp != null) {         document.write( temp.data + " ");         temp = temp.next;     }     document.write("<br>"); }  // Function to add one to a number // represented as linked list function addOne(head) {      // To store the last node in the linked     // list which is not equal to 9     var last = null;     var 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 rightmost 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 var head = create_Node(1); head.next = create_Node(2); head.next.next = create_Node(9); head.next.next.next = create_Node(9); document.write( "Original list is : "); print(head); head = addOne(head); document.write( "Resultant list is : "); print(head);  // This code is contributed by rrrtnx. </script>

Output:

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

Time Complexity: O(N)

Space Complexity: O(1)

Add Two Numbers Represented as Linked List

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

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