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
[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 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 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 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# 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 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)
[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 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 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);
}
}
# 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# 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 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.
Add 1 to a number represented as linked list | DSA Problem
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Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
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Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
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Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
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GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
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