Add 1 to a number represented as linked list (original) (raw)

Last Updated : 3 Apr, 2026

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

Try It Yourself

Table of Content

• Recursive - O(n) Time and O(n) Space:
• Iterative - O(n) Time and O(1) Space:

Recursive - O(n) Time and O(n) Space:

• **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 andinsert it at head. C++ `

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

#include 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);

`

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

Iterative - O(n) Time and O(1) Space:

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. This gives us 0->0->0->2
3. After traversing, if the **carry is not equals to 0, create new node with the data as carry andinsert it at head. In this case, the carry is 0.
4. Reverse modified linked list and return head. This gives us 2->0->0->0 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);

`

**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.