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In this post, you will find the solution for the Reverse Nodes in k-Group in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemReverse Nodes in k-Group– LeetCode Problem

Reverse Nodes in k-Group– LeetCode Problem

Problem:

Given the head of a linked list, reverse the nodes of the list k at a time, and return the modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is.

You may not alter the values in the list’s nodes, only nodes themselves may be changed.

Example 1:

Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]

Example 2:

Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]

Constraints:

  • The number of nodes in the list is n.
  • 1 <= k <= n <= 5000
  • 0 <= Node.val <= 1000
Reverse Nodes in k-Group– LeetCode Solutions
class Solution {
 public:
  ListNode* reverseKGroup(ListNode* head, int k) {
    if (!head || k == 1)
      return head;

    const int length = getLength(head);
    ListNode dummy(0, head);
    ListNode* prev = &dummy;
    ListNode* curr = head;

    for (int i = 0; i < length / k; ++i) {
      for (int j = 0; j < k - 1; ++j) {
        ListNode* next = curr->next;
        curr->next = next->next;
        next->next = prev->next;
        prev->next = next;
      }
      prev = curr;
      curr = curr->next;
    }

    return dummy.next;
  }

 private:
  int getLength(ListNode* head) {
    int length = 0;
    for (ListNode* curr = head; curr; curr = curr->next)
      ++length;
    return length;
  }
};
class Solution {
  public ListNode reverseKGroup(ListNode head, int k) {
    if (head == null || k == 1)
      return head;

    final int length = getLength(head);
    ListNode dummy = new ListNode(0, head);
    ListNode prev = dummy;
    ListNode curr = head;

    for (int i = 0; i < length / k; ++i) {
      for (int j = 0; j < k - 1; ++j) {
        ListNode next = curr.next;
        curr.next = next.next;
        next.next = prev.next;
        prev.next = next;
      }
      prev = curr;
      curr = curr.next;
    }

    return dummy.next;
  }

  private int getLength(ListNode head) {
    int length = 0;
    for (ListNode curr = head; curr != null; curr = curr.next)
      ++length;
    return length;
  }
}
class Solution:
  def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
    if not head or k == 1:
      return head

    def getLength(head: ListNode) -> int:
      length = 0
      while head:
        length += 1
        head = head.next
      return length

    length = getLength(head)
    dummy = ListNode(0, head)
    prev = dummy
    curr = head

    for _ in range(length // k):
      for _ in range(k - 1):
        next = curr.next
        curr.next = next.next
        next.next = prev.next
        prev.next = next
      prev = curr
      curr = curr.next

    return dummy.next

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