# Reverse alternate K nodes in a Singly Linked List – Iterative Solution

Given a linked list and an integer **K**, the task is to reverse every alternate **K** nodes.

**Examples:**

Input:1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> NULL, K = 3

Output:3 2 1 4 5 6 9 8 7

Input:1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> NULL, K = 5

Output:5 4 3 2 1 6 7 8 9

**Approach:** We have already discussed a recursive solution here. In this post, we will discuss an iterative solution to the above problem. While traversing we process 2k nodes in one iteration and keep track of the first and last node of the group of k-nodes in the given linked list using the join and tail pointer. After reversing the k nodes of the linked list, we join the last node of the reversed list, pointed by the tail pointer, with the first node of the original list, pointed by the join pointer. We then move the current pointer until we skip the next k nodes.

The tail now becomes the last node of the normal list (which is pointed by the updated tail pointer) and join points to the first of the reversed list and they are then joined. We repeat this process until all the nodes are processed in the same way.

Below is the implementation of the above approach:

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Link list node ` `class` `Node { ` `public` `: ` ` ` `int` `data; ` ` ` `Node* next; ` `}; ` ` ` `/* Function to reverse alternate k nodes and ` `return the pointer to the new head node */` `Node* kAltReverse(` `struct` `Node* head, ` `int` `k) ` `{ ` ` ` `Node* prev = NULL; ` ` ` `Node* curr = head; ` ` ` `Node* temp = NULL; ` ` ` `Node* tail = NULL; ` ` ` `Node* newHead = NULL; ` ` ` `Node* join = NULL; ` ` ` `int` `t = 0; ` ` ` ` ` `// Traverse till the end of the linked list ` ` ` `while` `(curr) { ` ` ` `t = k; ` ` ` `join = curr; ` ` ` `prev = NULL; ` ` ` ` ` `/* Reverse alternative group of k nodes ` ` ` `// of the given linked list */` ` ` `while` `(curr && t--) { ` ` ` `temp = curr->next; ` ` ` `curr->next = prev; ` ` ` `prev = curr; ` ` ` `curr = temp; ` ` ` `} ` ` ` ` ` `// Sets the new head of the input list ` ` ` `if` `(!newHead) ` ` ` `newHead = prev; ` ` ` ` ` `/* Tail pointer keeps track of the last node ` ` ` `of the k-reversed linked list. The tail pointer ` ` ` `is then joined with the first node of the ` ` ` `next k-nodes of the linked list */` ` ` `if` `(tail) ` ` ` `tail->next = prev; ` ` ` ` ` `tail = join; ` ` ` `tail->next = curr; ` ` ` ` ` `t = k; ` ` ` ` ` `/* Traverse through the next k nodes ` ` ` `which will not be reversed */` ` ` `while` `(curr && t--) { ` ` ` `prev = curr; ` ` ` `curr = curr->next; ` ` ` `} ` ` ` ` ` `/* Tail pointer keeps track of the last ` ` ` `node of the k nodes traversed above */` ` ` `tail = prev; ` ` ` `} ` ` ` ` ` `// newHead is new head of the modified list ` ` ` `return` `newHead; ` `} ` ` ` `// Function to insert a node at ` `// the head of the linked 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 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 the linked list ` `void` `printList(Node* node) ` `{ ` ` ` `int` `count = 0; ` ` ` `while` `(node != NULL) { ` ` ` `cout << node->data << ` `" "` `; ` ` ` `node = node->next; ` ` ` `count++; ` ` ` `} ` `} ` ` ` `// Driver code ` `int` `main(` `void` `) ` `{ ` ` ` ` ` `// Start with the empty list ` ` ` `Node* head = NULL; ` ` ` `int` `i; ` ` ` ` ` `// Create a list 1->2->3->4->...->10 ` ` ` `for` `(i = 10; i > 0; i--) ` ` ` `push(&head, i); ` ` ` ` ` `int` `k = 3; ` ` ` ` ` `cout << ` `"Given linked list \n"` `; ` ` ` `printList(head); ` ` ` `head = kAltReverse(head, k); ` ` ` ` ` `cout << ` `"\n Modified Linked list \n"` `; ` ` ` `printList(head); ` ` ` ` ` `return` `(0); ` `} ` |

*chevron_right*

*filter_none*

**Output:**

Given linked list 1 2 3 4 5 6 7 8 9 10 Modified Linked list 3 2 1 4 5 6 9 8 7 10

**Time Complexity:** O(n)

**Space Complexity:** O(1)

## Recommended Posts:

- Reverse alternate K nodes in a Singly Linked List
- Alternate Odd and Even Nodes in a Singly Linked List
- Print the alternate nodes of linked list (Iterative Method)
- Given a linked list, reverse alternate nodes and append at the end
- Print the last k nodes of the linked list in reverse order | Iterative Approaches
- C Program to reverse each node value in Singly Linked List
- Sum of the nodes of a Singly Linked List
- Product of the nodes of a Singly Linked List
- Sum of the alternate nodes of linked list
- Count of Prime Nodes of a Singly Linked List
- Delete all Prime Nodes from a Singly Linked List
- Delete all Non-Prime Nodes from a Singly Linked List
- Sum and Product of the nodes of a Singly Linked List which are divisible by K
- Find the common nodes in two singly linked list
- Sum and Product of all Prime Nodes of a Singly Linked List

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