Nth Node

Getting the Nth Node

To get the node at index k (0-indexed), advance a pointer k times from the head:

struct Node *nth_node(struct Node *head, int k) {
    struct Node *cur = head;
    while (k > 0 && cur != NULL) {
        cur = cur->next;
        k--;
    }
    return cur;
}

Returns a pointer to the node, or NULL if k is out of bounds.

Example

[10] -> [20] -> [30] -> [40] -> NULL

nth_node(head, 0) → [10]   (head)
nth_node(head, 2) → [30]
nth_node(head, 9) → NULL   (out of bounds)

O(n) Access

This is the "random access" cost of linked lists. Unlike arrays, there is no pointer arithmetic shortcut — you must walk the list each time.

This is why linked lists are poor choices when you frequently need elements by index.

Your Task

Implement struct Node *nth_node(struct Node *head, int k) that returns the node at index k, or NULL if out of bounds.