# C++ Program For Finding Length Of A Linked List

Write a function to count the number of nodes in a given singly linked list.

For example, the function should return 5 for linked list 1->3->1->2->1.

Iterative Solution:

```1) Initialize count as 0
2) Initialize a node pointer, current = head.
3) Do following while current is not NULL
a) current = current -> next
b) count++;
4) Return count ```

Following is the Iterative implementation of the above algorithm to find the count of nodes in a given singly linked list.

## C++

 `// Iterative C++ program to find length ` `// or count of nodes in a linked list ` `#include ` `using` `namespace` `std;`   `// Link list node` `class` `Node ` `{ ` `    ``public``:` `    ``int` `data; ` `    ``Node* next; ` `}; `   `/* Given a reference (pointer to pointer)` `   ``to the head of a list and an int, push ` `   ``a new node on the front of the list. */` `void` `push(Node** head_ref, ``int` `new_data) ` `{ ` `    ``// Allocate node` `    ``Node* new_node =``new` `Node();`   `    ``// Put in the data` `    ``new_node->data = new_data; `   `    ``// Link the old list of the ` `    ``// new node` `    ``new_node->next = (*head_ref); `   `    ``// Move the head to point to ` `    ``// the new node` `    ``(*head_ref) = new_node; ` `} `   `// Counts no. of nodes in linked list` `int` `getCount(Node* head) ` `{ ` `    ``// Initialize count ` `    ``int` `count = 0; `   `    ``// Initialize current ` `    ``Node* current = head;` `    ``while` `(current != NULL) ` `    ``{ ` `        ``count++; ` `        ``current = current->next; ` `    ``} ` `    ``return` `count; ` `} `   `// Driver code` `int` `main() ` `{ ` `    ``// Start with the empty list` `    ``Node* head = NULL; `   `    ``// Use push() to construct list ` `    ``// 1->2->1->3->1` `    ``push(&head, 1); ` `    ``push(&head, 3); ` `    ``push(&head, 1); ` `    ``push(&head, 2); ` `    ``push(&head, 1); `   `    ``// Check the count function` `    ``cout << ``"count of nodes is "` `<< ` `             ``getCount(head); ` `    ``return` `0; ` `} `   `// This is code is contributed by rathbhupendra`

Output:

`count of nodes is 5`

Time Complexity: O(n), where n represents the length of the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Recursive Solution:

```int getCount(head)
1) If head is NULL, return 0.
2) Else return 1 + getCount(head->next) ```

Following is the Recursive implementation of the above algorithm to find the count of nodes in a given singly linked list.

## C++

 `// Recursive C++ program to find length` `// or count of nodes in a linked list` `#include ` `using` `namespace` `std;`   `// Link list node` `class` `Node ` `{` `    ``public``:` `    ``int` `data;` `    ``Node* next;` `};`   `/* Given a reference (pointer to pointer) ` `   ``to the head of a list and an int, ` `   ``push a new node on the front of the list. */` `void` `push(Node** head_ref, ``int` `new_data)` `{` `    ``// Allocate node ` `    ``Node* new_node = ``new` `Node();`   `    ``// Put in the data` `    ``new_node->data = new_data;`   `    ``// Link the old list of the ` `    ``// new node ` `    ``new_node->next = (*head_ref);`   `    ``// Move the head to point to ` `    ``// the new node` `    ``(*head_ref) = new_node;` `}`   `// Recursively count number of ` `// nodes in linked list ` `int` `getCount(Node* head)` `{` `    ``// Base Case` `    ``if` `(head == NULL) ` `    ``{` `        ``return` `0;` `    ``}` `    ``// Count this node plus the rest ` `    ``// of the list` `    ``else` `    ``{` `        ``return` `1 + getCount(head->next);` `    ``}` `}`   `// Driver code` `int` `main()` `{` `    ``// Start with the empty list` `    ``Node* head = NULL;`   `    ``// Use push() to construct list` `    ``// 1->2->1->3->1` `    ``push(&head, 1);` `    ``push(&head, 3);` `    ``push(&head, 1);` `    ``push(&head, 2);` `    ``push(&head, 1);`   `    ``// Check the count function` `    ``cout << ``"Count of nodes is "` `<< ` `             ``getCount(head);` `    ``return` `0;` `}` `// This is code is contributed by rajsanghavi9`

Output:

`Count of nodes is 5`

Time Complexity: O(n), where n represents the length of the given linked list.
Auxiliary Space: O(n), for recursive stack where n represents the length of the given linked list.

Please refer complete article on Find Length of a Linked List (Iterative and Recursive) for more details!

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