Given a singly circular linked list. The task is to find the sum and product of nodes which are divisible by K of the given linked list.
Input : List = 5->6->7->8->9->10->11->11 K = 11 Output : Sum = 22, Product = 121 Input : List = 15->7->3->9->11->5 K = 5 Output : Product = 75, Sum = 20
- Initialize a pointer current with the head of the circular linked list and a sum variable sum with 0 and a product variable product with 1.
- Start traversing the linked list using a do while loop until all the nodes get traversed.
- If current node data is divisible by given key.
- Add the value of current node to the sum i.e. sum = sum + current -> data.
- Multiply the value of current node to the product i.e. product = product * current -> data.
- Increment the pointer to the next node of linked list i.e. temp = temp -> next.
- Print the sum and product.
Below is the implementation of the above approach:
Initial List: 5 6 7 8 9 10 11 11 Sum = 22, Product = 121
- Sum and Product of the nodes of a Singly Linked List which are divisible by K
- Delete all Prime Nodes from a Circular Singly Linked List
- Product of the nodes of a Singly Linked List
- Sum and Product of all Prime Nodes of a Singly Linked List
- Product of all nodes in a doubly linked list divisible by a given number K
- Circular Singly Linked List | Insertion
- Convert singly linked list into circular linked list
- Sum of the nodes of a Singly Linked List
- Find minimum and maximum elements in singly Circular Linked List
- Alternate Odd and Even Nodes in a Singly Linked List
- Reverse alternate K nodes in a Singly Linked List
- Delete all Prime Nodes from a Singly Linked List
- Find the common nodes in two singly linked list
- Delete all Non-Prime Nodes from a Singly Linked List
- Count of Prime Nodes of a Singly Linked List
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Improved By : andrew1234