Add 1 to a number represented as linked list

Last Updated : 10 Sep, 2024

A number is represented in linked list such that each digit corresponds to a node in linked list. The task is to add 1 to it.

Examples:

Input: head: 4 -> 5 -> 6
Output: head: 4 -> 5 -> 7
Explanation: Adding 1 to number represented by Linked List = 456 + 1 = 457

Input: head: 2 -> 1 -> 6 -> 9
Output: head: 2 -> 1 -> 7 -> 0
Explanation: Adding 1 to number represented by Linked List = 2169 + 1 = 2170

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Table of Content

[Expected Approach - 1] Using Recursive Method - O(n) Time and O(n) Space:

The idea is to recursively traverse the linked list, reaching the last node first. This ensures that the least significant digit is processed before others. Add 1 to the value of the last node and compute any carry resulting from this addition. While backtracking, update each node's value based on the carry propagated from the subsequent node. After traversing, if the carry is not equals to 0, create new node with the data as carry and insert it at head.

Below is the implementation of above approach.

C++ 
// C++ program to add one to a linked list

#include <iostream>
using namespace std;

class Node {
  public:
    int data;
    Node *next;

    Node(int x) {
       	data = x;
        next = nullptr;
    }
};

// Recursively add 1 from end to beginning and return
// carry after all nodes are processed.
int addWithCarry(Node *head) {
  
    // If linked list is empty, return carry
    if (head == nullptr) {
        return 1;
    }

    // Add carry returned by the next node call
    int res = head->data + addWithCarry(head->next);

    // Update data and return new carry
    head->data = res % 10;
    return res / 10;
}

Node *addOne(Node *head) {
  
    // Add 1 to linked list from end to beginning
    int carry = addWithCarry(head);

    // If there is carry after updating all nodes,
    // then we need to add a new node to the linked list
    if (carry) {
        Node *newNode = new Node(carry);
        newNode->next = head;

      	// New node becomes head now
      	return newNode; 
    }

    return head;
}

void printList(Node *head) {
    Node *curr = head;
    while (curr != nullptr) {
        cout << curr->data << " ";
        curr = curr->next;
    }
    cout << endl;
}

int main() {
  
  	// Create a hard-coded linked list:
    // 1 -> 9 -> 9 -> 9
    Node *head = new Node(1);
    head->next = new Node(9);
    head->next->next = new Node(9);
    head->next->next->next = new Node(9);
  
    head = addOne(head);

    printList(head);

    return 0;
}
C 
// C program to add one to a linked list

#include <stdio.h>
#include <stdlib.h>

struct Node {
    int data;
    struct Node* next;
};

struct Node* createNode(int data);

// Recursively add 1 from end to beginning and return
// carry after all nodes are processed.
int addWithCarry(struct Node* head) {

    // If linked list is empty, return carry
    if (head == NULL) {
        return 1;
    }

    // Add carry returned by the next node call
    int res = head->data + addWithCarry(head->next);

    // Update data and return new carry
    head->data = res % 10;
    return res / 10;
}

struct Node* addOne(struct Node* head) {

    // Add 1 to linked list from end to beginning
    int carry = addWithCarry(head);

    // If there is carry after updating all nodes,
    // then we need to add a new node to the linked list
    if (carry) {
        struct Node* newNode = createNode(carry);
        newNode->next = head;

        // New node becomes head now
        return newNode;
    }

    return head;
}

void printList(struct Node* head) {
    struct Node* curr = head;
    while (curr != NULL) {
        printf("%d ", curr->data);
        curr = curr->next;
    }
    printf("\n");
}

struct Node* createNode(int data) {
    struct Node* newNode = 
      (struct Node*)malloc(sizeof(struct Node));
    newNode->data = data;
    newNode->next = NULL;
    return newNode;
}

int main() {

    // Create a hard-coded linked list:
    // 1 -> 9 -> 9 -> 9
    struct Node* head = createNode(1);
    head->next = createNode(9);
    head->next->next = createNode(9);
    head->next->next->next = createNode(9);

    head = addOne(head);

    printList(head);

    return 0;
}
Java 
// Java program to add one to a linked list

class Node {
    int data;
    Node next;

    Node(int data) {
        this.data = data;
        this.next = null;
    }
}

// Recursively add 1 from end to beginning and return
// carry after all nodes are processed.
class GfG {
    
    static int addWithCarry(Node head) {
        
        // If linked list is empty, return carry
        if (head == null) {
            return 1;
        }

        // Add carry returned by the next node call
        int res = head.data + addWithCarry(head.next);

        // Update data and return new carry
        head.data = res % 10;
        return res / 10;
    }

    static Node addOne(Node head) {
        
        // Add 1 to linked list from end to beginning
        int carry = addWithCarry(head);

        // If there is carry after updating all nodes,
        // then we need to add a new node to the linked list
        if (carry > 0) {
            Node newNode = new Node(carry);
            newNode.next = head;

            // New node becomes head now
            return newNode;
        }

        return head;
    }

    static void printList(Node head) {
        Node curr = head;
        while (curr != null) {
            System.out.print(curr.data + " ");
            curr = curr.next;
        }
        System.out.println();
    }

    public static void main(String[] args) {

        // Create a hard-coded linked list:
        // 1 -> 9 -> 9 -> 9
        Node head = new Node(1);
        head.next = new Node(9);
        head.next.next = new Node(9);
        head.next.next.next = new Node(9);

        head = addOne(head);

        printList(head);
    }
}
Python 
# Python program to add one to a linked list

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

# Recursively add 1 from end to beginning and return
# carry after all nodes are processed.
def addWithCarry(head):
    
    # If linked list is empty, return carry
    if head is None:
        return 1
    
    # Add carry returned by the next node call
    res = head.data + addWithCarry(head.next)
    
    # Update data and return new carry
    head.data = res % 10
    return res // 10

def addOne(head):
    
    # Add 1 to linked list from end to beginning
    carry = addWithCarry(head)
    
    # If there is carry after updating all nodes,
    # then we need to add a new node to the linked list
    if carry:
        newNode = Node(carry)
        newNode.next = head
        
        # New node becomes head now
        return newNode
    
    return head

def printList(head):
    curr = head
    while curr:
        print(curr.data, end=" ")
        curr = curr.next
    print()

if __name__ == "__main__":
    
    # Create a hard-coded linked list:
    # 1 -> 9 -> 9 -> 9
    head = Node(1)
    head.next = Node(9)
    head.next.next = Node(9)
    head.next.next.next = Node(9)
    
    head = addOne(head)
    
    printList(head)
C# 
// C# program to add 1 to a linked list 

using System;

class Node {
    public int data;
    public Node next;

    public Node(int data) {
        this.data = data;
        this.next = null;
    }
}

class GfG {

    // Recursively add 1 from end to beginning and return
    // carry after all nodes are processed.
    static int addWithCarry(Node head) {
        
        // If linked list is empty, return carry
        if (head == null) {
            return 1;
        }

        // Add carry returned by the next node call
        int res = head.data + addWithCarry(head.next);

        // Update data and return new carry
        head.data = res % 10;
        return res / 10;
    }

    static Node addOne(Node head) {
        
        // Add 1 to linked list from end to beginning
        int carry = addWithCarry(head);

        // If there is carry after updating all nodes,
        // then we need to add a new node to the linked list
        if (carry > 0) {
            Node newNode = new Node(carry);
            newNode.next = head;

            // New node becomes head now
            return newNode;
        }

        return head;
    }

    static void printList(Node head) {
        Node curr = head;
        while (curr != null) {
            Console.Write(curr.data + " ");
            curr = curr.next;
        }
        Console.WriteLine();
    }

    static void Main() {

        // Create a hard-coded linked list:
        // 1 -> 9 -> 9 -> 9
        Node head = new Node(1);
        head.next = new Node(9);
        head.next.next = new Node(9);
        head.next.next.next = new Node(9);

        head = addOne(head);

        printList(head);
    }
}
JavaScript 
// Javascript program to add one to a linked list

class Node {
    constructor(data) {
        this.data = data;
        this.next = null;
    }
}

// Recursively add 1 from end to beginning and return
// carry after all nodes are processed.
function addWithCarry(head) {
    
    // If linked list is empty, return carry
    if (head === null) {
        return 1;
    }

    // Add carry returned by the next node call
    const res = head.data + addWithCarry(head.next);

    // Update data and return new carry
    head.data = res % 10;
    return Math.floor(res / 10);
}

function addOne(head) {
    
    // Add 1 to linked list from end to beginning
    const carry = addWithCarry(head);

    // If there is carry after updating all nodes,
    // then we need to add a new node to the linked list
    if (carry > 0) {
        const newNode = new Node(carry);
        newNode.next = head;

        // New node becomes head now
        return newNode;
    }

    return head;
}

function printList(head) {
    let curr = head;
    while (curr !== null) {
        console.log(curr.data + " ");
        curr = curr.next;
    }
    console.log();
}

// Create a hard-coded linked list:
// 1 -> 9 -> 9 -> 9
let head = new Node(1);
head.next = new Node(9);
head.next.next = new Node(9);
head.next.next.next = new Node(9);

head = addOne(head);

printList(head);

Output
2 0 0 0

Time Complexity: O(n), where n is the number of nodes in the linked list.
Auxiliary Space: O(n)

[Expected Approach - 2] Using Iterative Method - O(n) Time and O(1) Space:

The idea is to start by reversing the entire linked list. This allows you to begin processing the digits from the least significant to the most significant. Add 1 to the value of this head node. If this addition results in a carry (i.e., the new value exceeds 9), update the current node to store only the last digit (value modulo 10) and pass the carry to the next node. Continue traversing subsequent nodes until we reach the end of the list or the carry becomes 0. Reverse the linked list again to restore it to its original order.

Below is the implementation of above approach. 

C++ 
// C++ program to add one to a linked list

#include <bits/stdc++.h>
using namespace std;

class Node {
  public:
    int data;
    Node *next;

    Node(int x) {
        data = x;
        next = nullptr;
    }
};

// Function to reverse the linked list
Node* reverse(Node* head) {
    Node *curr = head, *prev = nullptr, *next;
    while (curr != nullptr) {
        next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
    }
    return prev;
}

// Function to add one to a linked list and 
// return the head node of the resultant list
Node *addOneUtil(Node *head) {
    Node *res = head;
  	Node *curr = head;
    Node *last = nullptr;

    // Initialize carry with 1 (to add one)
    int carry = 1;
    int sum;

    while (curr != nullptr) {
        // Calculate sum of carry and current node's data
        sum = carry + curr->data;

        // Update carry for next digit
        carry = (sum >= 10) ? 1 : 0;

        // Update current node's data to sum modulo 10
        curr->data = sum % 10;

        // Move to the next node
        last = curr;
        curr = curr->next;
    }

    // If there's a carry left, add a 
  	// new node with carry value
    if (carry > 0) {
        last->next = new Node(carry);
    }

    return res;
}

// Main function to add one to the linked list
Node *addOne(Node *head) {
  
    // Reverse the linked list
    head = reverse(head);

    // Add one to the reversed list
    head = addOneUtil(head);

    // Reverse the list again to restore
  	// the original order
    return reverse(head);
}

void printList(Node *head) {
    Node *curr = head;
    while (curr != nullptr) {
        cout << curr->data << " ";
        curr = curr->next;
    }
    cout << endl;
}

int main() {
  
  	// Create a hard-coded linked list:
    // 1 -> 9 -> 9 -> 9
    Node *head = new Node(1);
    head->next = new Node(9);
    head->next->next = new Node(9);
    head->next->next->next = new Node(9);

    head = addOne(head);
  
    printList(head);

    return 0;
}
C 
// C program to add 1 to a linked list

#include <stdio.h>
#include <stdlib.h>

struct Node {
    int data;
    struct Node* next;
};

struct Node* createNode(int data);

// Function to reverse the linked list
struct Node* reverse(struct Node* head) {
    struct Node* curr = head;
    struct Node* prev = NULL;
    struct Node* next;
    while (curr != NULL) {
        next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
    }
    return prev;
}

// Function to add one to a linked list and 
// return the head node of the resultant list
struct Node* addOneUtil(struct Node* head) {
    struct Node* res = head;
    struct Node* curr = head;
    struct Node* last = NULL;

    // Initialize carry with 1 (to add one)
    int carry = 1;
    int sum;

    while (curr != NULL) {
        // Calculate sum of carry and current node's data
        sum = carry + curr->data;

        // Update carry for next digit
        carry = (sum >= 10) ? 1 : 0;

        // Update current node's data to sum modulo 10
        curr->data = sum % 10;

        // Move to the next node
        last = curr;
        curr = curr->next;
    }

    // If there's a carry left, add a new
  	// node with carry value
    if (carry > 0) {
        last->next = createNode(carry);
    }

    return res;
}

// Main function to add one to the linked list
struct Node* addOne(struct Node* head) {
  
    // Reverse the linked list
    head = reverse(head);

    // Add one to the reversed list
    head = addOneUtil(head);

    // Reverse the list again to restore 
  	// the original order
    return reverse(head);
}

void printList(struct Node* head) {
    struct Node* curr = head;
    while (curr != NULL) {
        printf("%d ", curr->data);
        curr = curr->next;
    }
    printf("\n");
}

struct Node* createNode(int data) {
    struct Node* newNode = 
      (struct Node*)malloc(sizeof(struct Node));
    newNode->data = data;
    newNode->next = NULL;
    return newNode;
}

int main() {

    // Create a hard-coded linked list:
    // 1 -> 9 -> 9 -> 9
    struct Node* head = createNode(1);
    head->next = createNode(9);
    head->next->next = createNode(9);
    head->next->next->next = createNode(9);

    head = addOne(head);

    printList(head);

    return 0;
}
Java 
// Java program to add 1 to a linked list

class Node {
    int data;
    Node next;

    Node(int x) {
        this.data = x;
        this.next = null;
    }
}

// Function to reverse the linked list
class GfG {
    static Node reverse(Node head) {
        Node curr = head, prev = null, next;
        while (curr != null) {
            next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        return prev;
    }

    // Function to add one to a linked list and 
    // return the head node of the resultant list
    static Node addOneUtil(Node head) {
        Node res = head;
        Node curr = head;
        Node last = null;

        // Initialize carry with 1 (to add one)
        int carry = 1;
        int sum;

        while (curr != null) {
          
            // Calculate sum of carry 
          	// and current node's data
            sum = carry + curr.data;

            // Update carry for next digit
            carry = (sum >= 10) ? 1 : 0;

            // Update current node's data to sum modulo 10
            curr.data = sum % 10;

            // Move to the next node
            last = curr;
            curr = curr.next;
        }

        // If there's a carry left, add a new
      	// node with carry value
        if (carry > 0) {
            last.next = new Node(carry);
        }

        return res;
    }

    // Main function to add one to the linked list
    static Node addOne(Node head) {
      
        // Reverse the linked list
        head = reverse(head);

        // Add one to the reversed list
        head = addOneUtil(head);

        // Reverse the list again to restore
      	//the original order
        return reverse(head);
    }

    static void printList(Node head) {
        Node curr = head;
        while (curr != null) {
            System.out.print(curr.data + " ");
            curr = curr.next;
        }
        System.out.println();
    }

    public static void main(String[] args) {

        // Create a hard-coded linked list:
        // 1 -> 9 -> 9 -> 9
        Node head = new Node(1);
        head.next = new Node(9);
        head.next.next = new Node(9);
        head.next.next.next = new Node(9);

        head = addOne(head);

        printList(head);
    }
}
Python 
# Python3 program to add 1 to a linked list

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

# Function to reverse the linked list
def reverse(head):
    curr = head
    prev = None
    while curr:
        next = curr.next
        curr.next = prev
        prev = curr
        curr = next
    return prev

# Function to add one to a linked list and
# return the head node of the resultant list
def addOneUtil(head):
    res = head
    curr = head
    last = None

    # Initialize carry with 1 (to add one)
    carry = 1

    while curr:
      
        # Calculate sum of carry and current node's data
        sum = carry + curr.data

        # Update carry for next digit
        carry = 1 if sum >= 10 else 0

        # Update current node's data to sum modulo 10
        curr.data = sum % 10

        # Move to the next node
        last = curr
        curr = curr.next

    # If there's a carry left, add a new
    # node with carry value
    if carry > 0:
        last.next = Node(carry)

    return res

# Main function to add one to the linked list
def addOne(head):
  
    # Reverse the linked list
    head = reverse(head)

    # Add one to the reversed list
    head = addOneUtil(head)

    # Reverse the list again to restore
    # the original order
    return reverse(head)

def printList(head):
    curr = head
    while curr:
        print(curr.data, end=" ")
        curr = curr.next
    print()

if __name__ == '__main__':
  
    # Create a hard-coded linked list:
    # 1 -> 9 -> 9 -> 9
    head = Node(1)
    head.next = Node(9)
    head.next.next = Node(9)
    head.next.next.next = Node(9)

    head = addOne(head)

    printList(head)
C# 
// C# program to add 1 to a linked list

using System;

class Node {
    public int data;
    public Node next;

    public Node(int x) {
        this.data = x;
        this.next = null;
    }
}

class GfG {

    // Function to reverse the linked list
    static Node Reverse(Node head) {
        Node curr = head, prev = null, next = null;
        while (curr != null) {
            next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        return prev;
    }

    // Function to add one to a linked list and 
    // return the head node of the resultant list
    static Node AddOneUtil(Node head) {
        Node res = head;
        Node curr = head;
        Node last = null;

        // Initialize carry with 1 (to add one)
        int carry = 1;
        int sum;

        while (curr != null) {
            // Calculate sum of carry and 
          	// current node's data
            sum = carry + curr.data;

            // Update carry for next digit
            carry = (sum >= 10) ? 1 : 0;

            // Update current node's data to sum modulo 10
            curr.data = sum % 10;

            // Move to the next node
            last = curr;
            curr = curr.next;
        }

        // If there's a carry left, add a new 
       	//	node with carry value
        if (carry > 0) {
            last.next = new Node(carry);
        }

        return res;
    }

    // Main function to add one to the linked list
    static Node AddOne(Node head) {
      
        // Reverse the linked list
        head = Reverse(head);

        // Add one to the reversed list
        head = AddOneUtil(head);

        // Reverse the list again to restore
      	// the original order
        return Reverse(head);
    }

    static void PrintList(Node head) {
        Node curr = head;
        while (curr != null) {
            Console.Write(curr.data + " ");
            curr = curr.next;
        }
        Console.WriteLine();
    }

    static void Main() {

        // Create a hard-coded linked list:
        // 1 -> 9 -> 9 -> 9
        Node head = new Node(1);
        head.next = new Node(9);
        head.next.next = new Node(9);
        head.next.next.next = new Node(9);

        head = AddOne(head);

        PrintList(head);
    }
}
JavaScript 
// Javascript program to add 1 to a linked list 

class Node {
    constructor(data) {
        this.data = data;
        this.next = null;
    }
}

// Function to reverse the linked list
function reverse(head) {
    let curr = head, prev = null, next;
    while (curr != null) {
        next = curr.next;
        curr.next = prev;
        prev = curr;
        curr = next;
    }
    return prev;
}

// Function to add one to a linked list and 
// return the head node of the resultant list
function addOneUtil(head) {
    let res = head;
    let curr = head;
    let last = null;

    // Initialize carry with 1 (to add one)
    let carry = 1;
    let sum;

    while (curr != null) {
    
        // Calculate sum of carry and current node's data
        sum = carry + curr.data;

        // Update carry for next digit
        carry = (sum >= 10) ? 1 : 0;

        // Update current node's data to sum modulo 10
        curr.data = sum % 10;

        // Move to the next node
        last = curr;
        curr = curr.next;
    }

    // If there's a carry left, add a new
    // node with carry value
    if (carry > 0) {
        last.next = new Node(carry);
    }

    return res;
}

// Main function to add one to the linked list
function addOne(head) {

    // Reverse the linked list
    head = reverse(head);

    // Add one to the reversed list
    head = addOneUtil(head);

    // Reverse the list again to restore
    // the original order
    return reverse(head);
}

function printList(head) {
    let curr = head;
    while (curr != null) {
        console.log(curr.data + " ");
        curr = curr.next;
    }
    console.log();
}

// Create a hard-coded linked list:
// 1 -> 9 -> 9 -> 9
let head = new Node(1);
head.next = new Node(9);
head.next.next = new Node(9);
head.next.next.next = new Node(9);

head = addOne(head);

printList(head);

Output
2 0 0 0

Time Complexity: O(n), where n is the number of nodes in the linked list.
Auxiliary Space: O(1), As constant extra space is used.


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