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)

We strongly recommend you to minimize your browser and try this yourself first.

Below are the steps :

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.

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? head->data: 0);          // 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; }`

Output:

`List is 1999  Resultant list is 2000`