Given a linked list, write a function to reverse every k nodes (where k is an input to the function).
Example:
Input: 1->2->3->4->5->6->7->8->NULL, K = 3
Output: 3->2->1->6->5->4->8->7->NULL
Input: 1->2->3->4->5->6->7->8->NULL, K = 5
Output: 5->4->3->2->1->8->7->6->NULL
Algorithm: reverse(head, k)
- Reverse the first sub-list of size k. While reversing keep track of the next node and previous node. Let the pointer to the next node be next and pointer to the previous node be prev. See this post for reversing a linked list.
- head->next = reverse(next, k) ( Recursively call for rest of the list and link the two sub-lists )
- Return prev ( prev becomes the new head of the list (see the diagrams of an iterative method of this post )
Below is image shows how the reverse function works:
Below is the implementation of the above approach:
C++
// C++ program to reverse a linked list // in groups of given size #include <bits/stdc++.h> using namespace std; // Link list node class Node { public: int data; Node* next; }; /* Reverses the linked list in groups of size k and returns the pointer to the new head node. */Node* reverse(Node* head, int k) { // Base case if (!head) return NULL; Node* current = head; Node* next = NULL; Node* prev = NULL; int count = 0; // Reverse first k nodes of the // linked list while (current != NULL && count < k) { next = current->next; current->next = prev; prev = current; current = next; count++; } /* next is now a pointer to (k+1)th node Recursively call for the list starting from current. And make rest of the list as next of first node */ if (next != NULL) head->next = reverse(next, k); // prev is new head of the input list return prev; } // UTILITY FUNCTIONS // Function to push a node 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; } // Function to print linked list void printList(Node* node) { while (node != NULL) { cout << node->data << " "; node = node->next; } } // Driver code int main() { // Start with the empty list Node* head = NULL; /* Create Linked list 1->2->3->4->5->6->7->8->9 */ 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 << "Given linked list "; printList(head); head = reverse(head, 3); cout << "Reversed Linked list "; printList(head); return (0); } // This code is contributed by rathbhupendra |
Output:
Given Linked List 1 2 3 4 5 6 7 8 9 Reversed list 3 2 1 6 5 4 9 8 7
Complexity Analysis:
- Time Complexity: O(n).
Traversal of list is done only once and it has ‘n’ elements. - Auxiliary Space: O(n/k).
For each Linked List of size n, n/k or (n/k)+1 calls will be made during the recursion.
Please refer complete article on Reverse a Linked List in groups of given size | Set 1 for more details!
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