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 :
- Reverse given linked list. For example, 1-> 9-> 9 -> 9 is converted to 9-> 9 -> 9 ->1.
- 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.
- Reverse modified linked list and return head.
Below is the implementation of above steps.
C++
// 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 |
chevron_right
filter_none
C
// 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; } |
chevron_right
filter_none
Java
// 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 |
chevron_right
filter_none
Python3
# 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 |
chevron_right
filter_none
C#
// 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 |
chevron_right
filter_none
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++
// 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; } |
chevron_right
filter_none
Java
// 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 |
chevron_right
filter_none
Python
# Recursive Python program to add 1 to a linked list # Node class class Node: # Constructor to initialize the node object def __init__(self, data): self.data = data self.next = None # Function to create a new node with given data def newNode(data): new_node = Node(0) new_node.data = data new_node.next = None return new_node # Recursively add 1 from end to beginning and returns # carry after all nodes are processed. def addWithCarry(head): # If linked list is empty, then # return carry if (head == None): return 1 # Add carry returned be next node call res = head.data + addWithCarry(head.next) # Update data and return new carry head.data = int((res) % 10) return int((res) / 10) # This function mainly uses addWithCarry(). def addOne(head): # Add 1 to linked list from end to beginning 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): newNode = Node(0) newNode.data = carry newNode.next = head return newNode # New node becomes head now return head # A utility function to print a linked list def printList(node): while (node != None): print( node.data,end = "") node = node.next print("\n") # Driver program to test above function head = newNode(1) head.next = newNode(9) head.next.next = newNode(9) head.next.next.next = newNode(9) print("List is ") printList(head) head = addOne(head) print("\nResultant list is ") printList(head) # This code is contributed by Arnab Kundu |
chevron_right
filter_none
C#
// 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 */ |
chevron_right
filter_none
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
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
Recommended Posts:
- Add one to a number represented as linked list | Set 2
- Multiply two numbers represented as linked lists into a third list
- Reverse all the word in a String represented as a Linked List
- Add two numbers represented by linked lists | Set 1
- Add two numbers represented by linked lists | Set 2
- Add Two Numbers Represented by Linked Lists | Set 3
- Create new linked list from two given linked list with greater element at each node
- Subtract Two Numbers represented as Linked Lists
- Compare two strings represented as linked lists
- Multiply two numbers represented by Linked Lists
- Compare numbers represented by Linked Lists
- Add the given digit to a number stored in a linked list
- Difference between Singly linked list and Doubly linked list
- Merge a linked list into another linked list at alternate positions
- XOR Linked List - A Memory Efficient Doubly Linked List | Set 1
- XOR Linked List – A Memory Efficient Doubly Linked List | Set 2
- Convert singly linked list into circular linked list
- Number of connected components in a doubly linked list
- Sum of all nodes in a doubly linked list divisible by a given number K
- Add the given digit to a number stored in a linked list using recursion
