Program to find average of all nodes in a Linked List
Given a singly linked list. The task is to find the average of all nodes of the given singly linked list.
Examples:
Input: 7->6->8->4->1 Output: 26 Average of nodes: (7 + 6 + 8 + 4 + 1 ) / 5 = 5.2 Input: 1->7->3->9->11->5 Output: 6
Iterative Solution:
- Initialise a pointer ptr with the head of the linked list and a sum variable with 0.
- Start traversing the linked list using a loop until all the nodes get traversed.
- Add the value of current node to the sum i.e. sum += ptr -> data .
- Increment the pointer to the next node of linked list i.e. ptr = ptr ->next .
- Divide sum by total number of node and Return the average.
Below is the implementation of the above approach:
// C++ implementation to find the average of // nodes of the Linked List #include <bits/stdc++.h> using namespace std; /* A Linked list node */ struct Node { int data; struct Node* next; }; // function to insert a node at the // beginning of the linked list void push( struct Node** head_ref, int new_data) { /* allocate node */ struct Node* new_node = new Node; /* put in the data */ new_node->data = new_data; /* link the old list to the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } // Function to iteratively find the avg of // nodes of the given linked list float avgOfNodes( struct Node* head) { // if head = NULL if (!head) return -1; int count = 0; // Initialize count int sum = 0; float avg = 0.0; struct Node* current = head; // Initialize current while (current != NULL) { count++; sum += current->data; current = current->next; } // calculate average avg = ( double )sum / count; return avg; } // Driver Code int main() { struct Node* head = NULL; // create linked list 7->6->8->4->1 push(&head, 7); push(&head, 6); push(&head, 8); push(&head, 4); push(&head, 1); cout << "Average of nodes = " << avgOfNodes(head); return 0; } |
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Output:
Average of nodes = 5.2
Time complexity : O(n)
Where n is equal to number of nodes.
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Improved By : VishalBachchas