Find the product of last N nodes of the given Linked List in C++

C++Server Side ProgrammingProgramming

Consider we have few elements in a linked list. We have to find the multiplication result of last n number of elements. The value of n is also given. So if the list is like [5, 7, 3, 5, 6, 9], and n = 3, then result will be 5 * 6 * 9 = 270.

The process is straight forward. We simply read the current element starting from left side, then add the elements into stack. After filling up the stack, remove n elements and multiply them with the prod. (initially prod is 1), when n number of elements are traversed, then stop.

Example

#include<iostream>
#include<stack>
using namespace std;
   class Node{
   public:
      int data;
      Node *next;
   };
   Node* getNode(int data){
      Node *newNode = new Node;
      newNode->data = data;
      newNode->next = NULL;
      return newNode;
   }
   void append(struct Node** start, int key) {
      Node* new_node = getNode(key);
      Node *p = (*start);
      if(p == NULL){
         (*start) = new_node;
         return;
      }
      while(p->next != NULL){
         p = p->next;
      }
      p->next = new_node;
   }
   long long prodLastNElements(Node *start, int n) {
      if(n <= 0)
         return 0;
      stack<int> stk;
      long long res = 1;
      Node* temp = start;
      while (temp != NULL) {
         stk.push(temp->data);
         temp = temp->next;
      }
      while(n--){
         res *= stk.top();
         stk.pop();
      }
   return res;
}
int main() {
   Node *start = NULL;
   int arr[] = {5, 7, 3, 5, 6, 9};
   int size = sizeof(arr)/sizeof(arr[0]);
   int n = 3;
   for(int i = 0; i<size; i++){
      append(&start, arr[i]);
   }
   cout << "Product of last n elements: " << prodLastNElements(start, n);
}

Output

Product of last n elements: 270
raja
Published on 04-Nov-2019 14:29:11
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