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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:
![Reverse Nodes in k-Group LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct] 2](https://assets.leetcode.com/uploads/2020/10/03/reverse_ex1.jpg)
Input: head = [1,2,3,4,5], k = 2 Output: [2,1,4,3,5]
Example 2:
![Reverse Nodes in k-Group LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct] 3](https://assets.leetcode.com/uploads/2020/10/03/reverse_ex2.jpg)
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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