Given a linked list and two integers M and N. Traverse the linked list such that you retain M nodes then delete next N nodes, continue the same till end of the linked list.
Difficulty Level: Rookie
Examples:
Input: M = 2, N = 2 Linked List: 1->2->3->4->5->6->7->8 Output: Linked List: 1->2->5->6 Input: M = 3, N = 2 Linked List: 1->2->3->4->5->6->7->8->9->10 Output: Linked List: 1->2->3->6->7->8 Input: M = 1, N = 1 Linked List: 1->2->3->4->5->6->7->8->9->10 Output: Linked List: 1->3->5->7->9
The main part of the problem is to maintain proper links between nodes, make sure that all corner cases are handled. Following is C implementation of function skipMdeleteN() that skips M nodes and delete N nodes till end of list. It is assumed that M cannot be 0.
// C++ program to delete N nodes // after M nodes of a linked list #include <bits/stdc++.h> using namespace std;
// A linked list node class Node
{ public :
int data;
Node *next;
}; /* Function to insert a node at the beginning */
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 off the
// new node
new_node->next = (*head_ref);
// Move the head to point to
// the new node
(*head_ref) = new_node;
} // Function to print linked list void printList(Node *head)
{ Node *temp = head;
while (temp != NULL)
{
cout<<temp->data<< " " ;
temp = temp->next;
}
cout<<endl;
} // Function to skip M nodes and then // delete N nodes of the linked list. void skipMdeleteN(Node *head,
int M, int N)
{ Node *curr = head, *t;
int count;
// The main loop that traverses
// through the whole list
while (curr)
{
// Skip M nodes
for (count = 1; count < M &&
curr!= NULL; count++)
curr = curr->next;
// If we reached end of list,
// then return
if (curr == NULL)
return ;
// Start from next node and delete
// N nodes
t = curr->next;
for (count = 1; count<=N &&
t!= NULL; count++)
{
Node *temp = t;
t = t->next;
free (temp);
}
// Link the previous list with
// remaining nodes
curr->next = t;
// Set current pointer for next
// iteration
curr = t;
}
} // Driver code int main()
{ /* Create following linked list
1->2->3->4->5->6->7->8->9->10 */
Node* head = NULL;
int M=2, N=3;
push(&head, 10);
push(&head, 9);
push(&head, 8);
push(&head, 7);
push(&head, 6);
push(&head, 5);
push(&head, 4);
push(&head, 3);
push(&head, 2);
push(&head, 1);
cout << "M = " << M<< " N = " <<
N << "Given Linked list is :" ;
printList(head);
skipMdeleteN(head, M, N);
cout<< "Linked list after deletion is :" ;
printList(head);
return 0;
} // This code is contributed by rathbhupendra |
Output:
M = 2, N = 3 Given Linked list is : 1 2 3 4 5 6 7 8 9 10 Linked list after deletion is : 1 2 6 7
Time Complexity:
O(n) where n is number of nodes in linked list.
Auxiliary Space: O(1)
Please refer complete article on Delete N nodes after M nodes of a linked list for more details!