Given a linked list with n nodes, reverse it in the following way :

- If n is even, reverse it in group of n/2 nodes.
- If n is odd, keep the middle node as it is, reverse first n/2 elements and reverse last n/2 elements.

Examples:

Input : 1 2 3 4 5 6 (n is even)

Output : 3 2 1 6 5 4Input : 1 2 3 4 5 6 7 (n is odd)

Output : 3 2 1 4 7 6 5

**Approach:** The idea is similar to Reversing a linked list in groups of size k where k is n/2. Just need to check for mid node.

- If n is even, divide the linked list into two parts i.e. first n/2 elements and last n/2 elements and reverse both the parts.
- If n is odd, divide the linked list into three parts i.e. first n/2 elements, (n/2 + 1) th element and last n/2 elements and reverse both the parts except (n/2 + 1) th element .

// C++ program to reverse given // linked list according to its size #include <bits/stdc++.h> using namespace std; struct Node { int data; Node* next; }; // Function to create a new Node Node* newNode(int data) { Node *temp = new Node; temp->data = data; temp->next = NULL; return temp; } // Prints a list. void printList(Node* head) { Node *temp = head; while (temp) { cout << temp->data << " "; temp = temp->next; } cout << endl; } /* Function to push a Node */ void push(Node** head_ref, int new_data) { Node* new_Node = new Node; new_Node->data = new_data; new_Node->next = (*head_ref); (*head_ref) = new_Node; } // Returns size of list. int getSize(Node* head) { Node* curr = head; int count = 0; while (curr) { curr = curr->next; count++; } return count; } // Function to reverse the linked // list according to its size Node* reverseSizeBy2Util(Node* head, int k, bool skipMiddle) { if (!head) return NULL; int count = 0; Node* curr = head; Node* prev = NULL; Node* next; // Reverse current block of list. while (curr && count < k) { next = curr->next; curr->next = prev; prev = curr; curr = next; count++; } // If size is even, reverse next block too. if (!skipMiddle) head->next = reverseSizeBy2Util(next, k, false); else { // if size is odd, skip next element // and reverse the block after that. head->next = next; if (next) next->next = reverseSizeBy2Util(next->next, k, true); } return prev; } Node* reverseBySizeBy2(Node* head) { // Get the size of list. int n = getSize(head); // If the size is even, no need // to skip middle Node. if (n % 2 == 0) return reverseSizeBy2Util(head, n/2, false); // If size is odd, middle Node has // to be skipped. else return reverseSizeBy2Util(head, n/2, true); } // Drivers code int main() { /* Start with the empty list */ Node* head = NULL; /* Created Linked list is 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 << "Original List : "; printList(head); cout << "Reversed List : "; Node* reversedHead = reverseBySizeBy2(head); printList(reversedHead); return 0; }

**Output:**

Original List : 1 2 3 4 5 6 7 8 9 Reversed List : 4 3 2 1 5 9 8 7 6

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