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** Link for the Problem** – Reverse 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 = 2Output:[2,1,4,3,5]

**Example 2:**

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