Add 1 to a number represented as linked list

Number is represented in linked list such that each digit corresponds to a node in linked list. Add 1 to it. For example 1999 is represented as (1-> 9-> 9 -> 9) and adding 1 to it should change it to (2->0->0->0)

Below are the steps :

  1. Reverse given linked list. For example, 1-> 9-> 9 -> 9 is converted to 9-> 9 -> 9 ->1.
  2. Start traversing linked list from leftmost node and add 1 to it. If there is a carry, move to the next node. Keep moving to the next node while there is a carry.
  3. Reverse modified linked list and return head.

Below is the implementation of above steps.

C++

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// C++ program to add 1 to a linked list 
#include <bits/stdc++.h>
using namespace std;
  
/* Linked list node */
class Node 
    public:
    int data; 
    Node* next; 
}; 
  
/* Function to create a new node with given data */
Node *newNode(int data) 
    Node *new_node = new Node; 
    new_node->data = data; 
    new_node->next = NULL; 
    return new_node; 
  
/* Function to reverse the linked list */
Node *reverse(Node *head) 
    Node * prev = NULL; 
    Node * current = head; 
    Node * next; 
    while (current != NULL) 
    
        next = current->next; 
        current->next = prev; 
        prev = current; 
        current = next; 
    
    return prev; 
  
/* Adds one to a linked lists and return the head 
node of resultant list */
Node *addOneUtil(Node *head) 
    // res is head node of the resultant list 
    Node* res = head; 
    Node *temp, *prev = NULL; 
  
    int carry = 1, sum; 
  
    while (head != NULL) //while both lists exist 
    
        // Calculate value of next digit in resultant list. 
        // The next digit is sum of following things 
        // (i) Carry 
        // (ii) Next digit of head list (if there is a 
        // next digit) 
        sum = carry + head->data; 
  
        // update carry for next calulation 
        carry = (sum >= 10)? 1 : 0; 
  
        // update sum if it is greater than 10 
        sum = sum % 10; 
  
        // Create a new node with sum as data 
        head->data = sum; 
  
        // Move head and second pointers to next nodes 
        temp = head; 
        head = head->next; 
    
  
    // if some carry is still there, add a new node to 
    // result list. 
    if (carry > 0) 
        temp->next = newNode(carry); 
  
    // return head of the resultant list 
    return res; 
  
// This function mainly uses addOneUtil(). 
Node* addOne(Node *head) 
    // Reverse linked list 
    head = reverse(head); 
  
    // Add one from left to right of reversed 
    // list 
    head = addOneUtil(head); 
  
    // Reverse the modified list 
    return reverse(head); 
  
// A utility function to print a linked list 
void printList(Node *node) 
    while (node != NULL) 
    
        cout << node->data; 
        node = node->next; 
    
    cout<<endl;
  
/* Driver program to test above function */
int main(void
    Node *head = newNode(1); 
    head->next = newNode(9); 
    head->next->next = newNode(9); 
    head->next->next->next = newNode(9); 
  
    cout << "List is "
    printList(head); 
  
    head = addOne(head); 
  
    cout << "\nResultant list is "
    printList(head); 
  
    return 0; 
  
// This is code is contributed by rathbhupendra

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C

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// C program to add 1 to a linked list
#include<bits/stdc++.h>
  
/* Linked list node */
struct Node
{
    int data;
    Node* next;
};
  
/* Function to create a new node with given data */
Node *newNode(int data)
{
    Node *new_node = new Node;
    new_node->data = data;
    new_node->next = NULL;
    return new_node;
}
  
/* Function to reverse the linked list */
Node *reverse(Node *head)
{
    Node * prev   = NULL;
    Node * current = head;
    Node * next;
    while (current != NULL)
    {
        next  = current->next;
        current->next = prev;
        prev = current;
        current = next;
    }
    return prev;
}
  
/* Adds one to a linked lists and return the head
   node of resultant list */
Node *addOneUtil(Node *head)
{
    // res is head node of the resultant list
    Node* res = head;
    Node *temp, *prev = NULL;
  
    int carry = 1, sum;
  
    while (head != NULL) //while both lists exist
    {
        // Calculate value of next digit in resultant list.
        // The next digit is sum of following things
        // (i) Carry
        // (ii) Next digit of head list (if there is a
        //     next digit)
        sum = carry + head->data;
  
        // update carry for next calulation
        carry = (sum >= 10)? 1 : 0;
  
        // update sum if it is greater than 10
        sum = sum % 10;
  
        // Create a new node with sum as data
        head->data = sum;
  
        // Move head and second pointers to next nodes
        temp = head;
        head = head->next;
    }
  
    // if some carry is still there, add a new node to
    // result list.
    if (carry > 0)
        temp->next = newNode(carry);
  
    // return head of the resultant list
    return res;
}
  
// This function mainly uses addOneUtil().
Node* addOne(Node *head)
{
    // Reverse linked list
    head = reverse(head);
  
    // Add one from left to right of reversed
    // list
    head = addOneUtil(head);
  
    // Reverse the modified list
    return reverse(head);
}
  
// A utility function to print a linked list
void printList(Node *node)
{
    while (node != NULL)
    {
        printf("%d", node->data);
        node = node->next;
    }
    printf("\n");
}
  
/* Driver program to test above function */
int main(void)
{
    Node *head = newNode(1);
    head->next = newNode(9);
    head->next->next = newNode(9);
    head->next->next->next = newNode(9);
  
    printf("List is ");
    printList(head);
  
    head = addOne(head);
  
    printf("\nResultant list is ");
    printList(head);
  
    return 0;
}

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Java

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// Java program to add 1 to a linked list 
class GfG 
{
  
/* Linked list node */
static class Node 
    int data; 
    Node next; 
}
  
/* Function to create a new node with given data */
static Node newNode(int data) 
    Node new_node = new Node(); 
    new_node.data = data; 
    new_node.next = null
    return new_node; 
  
/* Function to reverse the linked list */
static Node reverse(Node head) 
    Node prev = null
    Node current = head; 
    Node next = null
    while (current != null
    
        next = current.next; 
        current.next = prev; 
        prev = current; 
        current = next; 
    
    return prev; 
  
/* Adds one to a linked lists and return the head 
node of resultant list */
static Node addOneUtil(Node head) 
    // res is head node of the resultant list 
    Node res = head; 
    Node temp = null, prev = null
  
    int carry = 1, sum; 
  
    while (head != null) //while both lists exist 
    
        // Calculate value of next digit in resultant list. 
        // The next digit is sum of following things 
        // (i) Carry 
        // (ii) Next digit of head list (if there is a 
        // next digit) 
        sum = carry + head.data; 
  
        // update carry for next calulation 
        carry = (sum >= 10)? 1 : 0
  
        // update sum if it is greater than 10 
        sum = sum % 10
  
        // Create a new node with sum as data 
        head.data = sum; 
  
        // Move head and second pointers to next nodes 
        temp = head; 
        head = head.next; 
    
  
    // if some carry is still there, add a new node to 
    // result list. 
    if (carry > 0
        temp.next = newNode(carry); 
  
    // return head of the resultant list 
    return res; 
  
// This function mainly uses addOneUtil(). 
static Node addOne(Node head) 
    // Reverse linked list 
    head = reverse(head); 
  
    // Add one from left to right of reversed 
    // list 
    head = addOneUtil(head); 
  
    // Reverse the modified list 
    return reverse(head); 
  
// A utility function to print a linked list 
static void printList(Node node) 
    while (node != null
    
        System.out.print(node.data); 
        node = node.next; 
    
    System.out.println(); 
  
/* Driver code */
public static void main(String[] args) 
    Node head = newNode(1); 
    head.next = newNode(9); 
    head.next.next = newNode(9); 
    head.next.next.next = newNode(9); 
  
    System.out.print("List is "); 
    printList(head); 
  
    head = addOne(head); 
    System.out.println();
    System.out.print("Resultant list is "); 
    printList(head); 
}
  
// This code is contributed by prerna saini

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Python3

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# Python3 program to add 1 to a linked list 
import sys
import math
  
# Linked list node
class Node:
    def __init__(self,data):
        self.data = data
        self.next = None
  
# Function to create a new node with given data */
def newNode(data):
    return Node(data)
  
# Function to reverse the linked list */
def reverseList(head):
    if not head:
        return
    curNode = head
    prevNode = head
    nextNode = head.next
    curNode.next = None
  
    while(nextNode):
        curNode = nextNode
        nextNode = nextNode.next
        curNode.next = prevNode
        prevNode = curNode
      
    return curNode
  
# Adds one to a linked lists and return the head 
#node of resultant list
def addOne(head):
  
    # Reverse linked list 
    head = reverseList(head)
    temp = head
    carry = 1
    while(temp):
        temp.data += carry
        if temp.data % 10 == 0:
            # update carry for next calulation 
            carry = 1
            temp.data = 0
        else:
            # update carry for next calulation 
            carry = 0
        temp = temp.next
  
    # Reverse the modified list             
    return reverseList(head)
  
# A utility function to print a linked list 
def printList(head):
    if not head:
        return
    while(head):
        print("{}".format(head.data),end="")
        head=head.next
  
# Driver code
if __name__=='__main__':
    head = newNode(1)
    head.next = newNode(9)
    head.next.next = newNode(9)
    head.next.next.next = newNode(9)
  
    print("List is: ",end = "")
    printList(head)
  
    head = addOne(head)
  
    print("\nResultant list is: ",end="")
    printList(head)
  
  
# This code is contributed by Vikash Kumar 37

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C#

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// C# program to add 1 to a linked list
using System;
  
class GfG 
{
  
/* Linked list node */
public class Node 
    public int data; 
    public Node next; 
}
  
/* Function to create a new node with given data */
static Node newNode(int data) 
    Node new_node = new Node(); 
    new_node.data = data; 
    new_node.next = null
    return new_node; 
  
/* Function to reverse the linked list */
static Node reverse(Node head) 
    Node prev = null
    Node current = head; 
    Node next = null
    while (current != null
    
        next = current.next; 
        current.next = prev; 
        prev = current; 
        current = next; 
    
    return prev; 
  
/* Adds one to a linked lists and return the head 
node of resultant list */
static Node addOneUtil(Node head) 
    // res is head node of the resultant list 
    Node res = head; 
    Node temp = null, prev = null
  
    int carry = 1, sum; 
  
    while (head != null) //while both lists exist 
    
        // Calculate value of next digit in resultant list. 
        // The next digit is sum of following things 
        // (i) Carry 
        // (ii) Next digit of head list (if there is a 
        // next digit) 
        sum = carry + head.data; 
  
        // update carry for next calulation 
        carry = (sum >= 10)? 1 : 0; 
  
        // update sum if it is greater than 10 
        sum = sum % 10; 
  
        // Create a new node with sum as data 
        head.data = sum; 
  
        // Move head and second pointers to next nodes 
        temp = head; 
        head = head.next; 
    
  
    // if some carry is still there, add a new node to 
    // result list. 
    if (carry > 0) 
        temp.next = newNode(carry); 
  
    // return head of the resultant list 
    return res; 
  
// This function mainly uses addOneUtil(). 
static Node addOne(Node head) 
    // Reverse linked list 
    head = reverse(head); 
  
    // Add one from left to right of reversed 
    // list 
    head = addOneUtil(head); 
  
    // Reverse the modified list 
    return reverse(head); 
  
// A utility function to print a linked list 
static void printList(Node node) 
    while (node != null
    
        Console.Write(node.data); 
        node = node.next; 
    
    Console.WriteLine(); 
  
/* Driver code */
public static void Main(String[] args) 
    Node head = newNode(1); 
    head.next = newNode(9); 
    head.next.next = newNode(9); 
    head.next.next.next = newNode(9); 
  
    Console.Write("List is "); 
    printList(head); 
  
    head = addOne(head); 
    Console.WriteLine();
    Console.Write("Resultant list is "); 
    printList(head); 
}
}
  
// This code contributed by Rajput-Ji

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Output:

List is 1999

Resultant list is 2000

 
Recursive Implementation:
We can recursively reach the last node and forward carry to previous nodes. Recursive solution doesn’t require reversing of linked list. We can also use a stack in place of recursion to temporarily hold nodes.

Below is the implementation of recursive solution.

C++

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// Recursive C++ program to add 1 to a linked list
#include<bits/stdc++.h>
  
/* Linked list node */
struct Node
{
    int data;
    Node* next;
};
  
/* Function to create a new node with given data */
Node *newNode(int data)
{
    Node *new_node = new Node;
    new_node->data = data;
    new_node->next = NULL;
    return new_node;
}
  
// Recursively add 1 from end to beginning and returns
// carry after all nodes are processed.
int addWithCarry(Node *head)
{
    // If linked list is empty, then
    // return carry
    if (head == NULL)
        return 1;
  
    // Add carry returned be next node call
    int res = head->data + addWithCarry(head->next);
  
    // Update data and return new carry
    head->data = (res) % 10;
    return (res) / 10;
}
  
// This function mainly uses addWithCarry().
Node* addOne(Node *head)
{
    // Add 1 to linked list from end to beginning
    int carry = addWithCarry(head);
  
    // If there is carry after processing all nodes,
    // then we need to add a new node to linked list
    if (carry)
    {
        Node *newNode = new Node;
        newNode->data = carry;
        newNode->next = head;
        return newNode; // New node becomes head now
    }
  
    return head;
}
  
// A utility function to print a linked list
void printList(Node *node)
{
    while (node != NULL)
    {
        printf("%d", node->data);
        node = node->next;
    }
    printf("\n");
}
  
/* Driver program to test above function */
int main(void)
{
    Node *head = newNode(1);
    head->next = newNode(9);
    head->next->next = newNode(9);
    head->next->next->next = newNode(9);
  
    printf("List is ");
    printList(head);
  
    head = addOne(head);
  
    printf("\nResultant list is ");
    printList(head);
  
    return 0;
}

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Java

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// Recursive Java program to add 1 to a linked list
class GfG {
  
    /* Linked list node */
    static class Node
    {
        int data;
        Node next;
    }
  
    /* Function to create a new node with given data */
    static Node newNode(int data) 
    {
        Node new_node = new Node();
        new_node.data = data;
        new_node.next = null;
        return new_node;
    }
  
    // Recursively add 1 from end to beginning and returns
    // carry after all nodes are processed.
    static int addWithCarry(Node head)
    {
  
        // If linked list is empty, then
        // return carry
        if (head == null)
            return 1;
  
        // Add carry returned be next node call
        int res = head.data + addWithCarry(head.next);
  
        // Update data and return new carry
        head.data = (res) % 10;
        return (res) / 10;
    }
  
    // This function mainly uses addWithCarry(). 
    static Node addOne(Node head)
    {
  
        // Add 1 to linked list from end to beginning
        int carry = addWithCarry(head);
  
        // If there is carry after processing all nodes,
        // then we need to add a new node to linked list
        if (carry > 0)
        {
            Node newNode = newNode(carry);
            newNode.next = head;
            return newNode; // New node becomes head now
        }
  
        return head;
    }
  
    // A utility function to print a linked list
    static void printList(Node node)
    {
        while (node != null)
        {
            System.out.print(node.data);
            node = node.next;
        }
        System.out.println();
    }
  
    /* Driver code */
    public static void main(String[] args)
    {
        Node head = newNode(1);
        head.next = newNode(9);
        head.next.next = newNode(9);
        head.next.next.next = newNode(9);
  
        System.out.print("List is ");
        printList(head);
  
        head = addOne(head);
        System.out.println();
        System.out.print("Resultant list is ");
        printList(head);
    }
}
  
// This code is contributed by shubham96301

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C#

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// Recursive C# program to add 1 to a linked list 
using System; 
  
class GfG 
  
    /* Linked list node */
    public class Node 
    
        public int data; 
        public Node next; 
    
  
    /* Function to create a new node with given data */
    public static Node newNode(int data) 
    
        Node new_node = new Node(); 
        new_node.data = data; 
        new_node.next = null
        return new_node; 
    
  
    // Recursively add 1 from end to beginning and returns 
    // carry after all nodes are processed. 
    public static int addWithCarry(Node head) 
    
  
        // If linked list is empty, then 
        // return carry 
        if (head == null
            return 1; 
  
        // Add carry returned be next node call 
        int res = head.data + addWithCarry(head.next); 
  
        // Update data and return new carry 
        head.data = (res) % 10; 
        return (res) / 10; 
    
  
    // This function mainly uses addWithCarry(). 
    public static Node addOne(Node head) 
    
  
        // Add 1 to linked list from end to beginning 
        int carry = addWithCarry(head); 
        Node newNodes = null;
          
        // If there is carry after processing all nodes, 
        // then we need to add a new node to linked list 
        if (carry > 0) 
        
            newNodes = newNode(carry); 
            newNodes.next = head; 
            return newNodes; // New node becomes head now 
        
  
        return head; 
    
  
    // A utility function to print a linked list 
    public static void printList(Node node) 
    
        while (node != null
        
            Console.Write(node.data); 
            node = node.next; 
        
        Console.WriteLine(); 
    
  
    /* Driver code */
    public static void Main(String[] args) 
    
        Node head = newNode(1); 
        head.next = newNode(9); 
        head.next.next = newNode(9); 
        head.next.next.next = newNode(9); 
  
        Console.Write("List is "); 
        printList(head); 
  
        head = addOne(head); 
        Console.WriteLine(); 
        Console.Write("Resultant list is "); 
        printList(head); 
    
  
/* This code contributed by PrinciRaj1992 */

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Output:

List is 1999

Resultant list is 2000

This article is contributed by Aditya Goel. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



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