Write a C++ Program to GetNth() function that takes a linked list and an integer index and returns the data value stored in the node at that index position.
Example:
Input: 1->10->30->14, index = 2
Output: 30
The node at index 2 is 30
Algorithm:
1. Initialize count = 0
2. Loop through the linkList
a. If count is equal to the passed index then return current
node
b. Increment count
c. change current to point to next of the current.
Implementation:
C++
// C++ program to find n'th // node in linked list #include <assert.h> #include <bits/stdc++.h> 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;
} // Takes head pointer of // the linked list and index // as arguments and return // data at index int GetNth(Node* head, int index)
{ Node* current = head;
// The index of the
// node we're currently
// looking at
int count = 0;
while (current != NULL) {
if (count == index)
return (current->data);
count++;
current = current->next;
}
/* If we get to this line,
the caller was asking
for a non-existent element
so we assert fail */
assert (0);
} // Driver Code int main()
{ // Start with the empty list
Node* head = NULL;
// Use push() to construct list
// 1->12->1->4->1
push(&head, 1);
push(&head, 4);
push(&head, 1);
push(&head, 12);
push(&head, 1);
// Check the count function
cout << "Element at index 3 is " << GetNth(head, 3);
return 0;
} // This code is contributed by rathbhupendra |
Output
Element at index 3 is 4
Time Complexity: O(n)
Space complexity: O(1) using only constant variables
C++ Program For Writing A Function To Get Nth Node In A Linked List using Recursion:
Algorithm:
getnth(node,n)
1. Initialize count = 0
2. if count==n
return node->data
3. else
return getnth(node->next,n-1)
Implementation:
C++
// C++ program to find n'th node in // linked list using recursion #include <bits/stdc++.h> using namespace std;
// Link list node struct Node {
int data;
struct 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( struct Node** head_ref, int new_data)
{ // Allocate node
struct Node* new_node
= ( struct Node*) malloc ( sizeof ( struct 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;
} /* Takes head pointer of the linked list and index as arguments and return data at index.
(Don't use another variable)*/
int GetNth( struct Node* head, int n)
{ // If length of the list is less
// than the given index, return -1
if (head == NULL)
return -1;
// If n equal to 0 return node->data
if (n == 0)
return head->data;
// Increase head to next pointer
// n - 1: decrease the number of
// recursions until n = 0
return GetNth(head->next, n - 1);
} // Driver code int main()
{ // Start with the empty list
struct Node* head = NULL;
// Use push() to construct list
// 1->12->1->4->1
push(&head, 1);
push(&head, 4);
push(&head, 1);
push(&head, 12);
push(&head, 1);
// Check the count function
printf ( "Element at index 3 is %d" , GetNth(head, 3));
getchar ();
} |
Output
Element at index 3 is 4
Time Complexity: O(n)
Space Complexity: O(n) since using constant space to create nodes.
Please refer complete article on Write a function to get Nth node in a Linked List for more details!