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 C++ implementation of above steps.
// 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
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 C++ implementation of recursive solution.
// 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
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
Recommended Posts:
- Multiply two numbers represented as linked lists into a third list
- Add two numbers represented by linked lists | Set 2
- Add two numbers represented by linked lists | Set 1
- XOR Linked List - A Memory Efficient Doubly Linked List | Set 1
- Convert singly linked list into circular linked list
- Difference between Singly linked list and Doubly linked list
- XOR Linked List – A Memory Efficient Doubly Linked List | Set 2
- Merge a linked list into another linked list at alternate positions
- Subtract Two Numbers represented as Linked Lists
- Multiply two numbers represented by Linked Lists
- Compare two strings represented as linked lists
- Number of connected components in a doubly linked list
- Sum of all nodes in a doubly linked list divisible by a given number K
- Large number arithmetic using doubly linked list
- Maximum and Minimum element of a linked list which is divisible by a given number k