LeetCode-in-All

25. Reverse Nodes in k-Group

Hard

Given a linked list, reverse the nodes of a linked list k at a time and return its 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]

Example 3:

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

Output: [1,2,3,4,5]

Example 4:

Input: head = [1], k = 1

Output: [1]

Constraints:

• The number of nodes in the list is in the range sz.
• 1 <= sz <= 5000
• 0 <= Node.val <= 1000
• 1 <= k <= sz

Follow-up: Can you solve the problem in O(1) extra memory space?

To solve the “Reverse Nodes in k-Group” problem in Swift with a Solution class, we can reverse the nodes in groups of k using a recursive approach. Here are the steps:

1. Define a Solution class.
2. Define a method named reverseKGroup that takes the head of a linked list and an integer k as input and returns the head of the modified list.
3. Define a helper method named reverse that takes the head and tail of a sublist as input and reverses the sublist in place. This method returns the new head of the sublist.
4. Create a dummy ListNode object and set its next pointer to the head of the input list. This dummy node will serve as the new head of the modified list.
5. Initialize pointers prev, curr, next, and tail. Set prev and tail to the dummy node, and curr to the head of the input list.
6. Iterate through the list:
   • Move curr k steps forward. If it’s not possible (i.e., there are less than k nodes left), break the loop.
   • Set next to the next pointer of curr.
   • Reverse the sublist from curr to next using the reverse method. Update prev and tail accordingly.
   • Move prev and tail k steps forward to the last node of the reversed sublist.
   • Move curr to next.
7. Return the next pointer of the dummy node, which points to the head of the modified list.

Here’s the implementation:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init() { self.val = 0; self.next = nil; }
 *     public init(_ val: Int) { self.val = val; self.next = nil; }
 *     public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
 * }
 */
public class Solution {
    public func reverseKGroup(_ head: ListNode?, _ k: Int) -> ListNode? {
        if head == nil || head?.next == nil || k == 1 {
            return head
        }
        
        var len = head
        var j = 0
        while j < k {
            if len == nil {
                return head
            }
            len = len?.next
            j += 1
        }
        
        var c = head
        var n: ListNode? = nil
        var prev: ListNode? = nil
        var i = 0
        
        while i < k {
            n = c?.next
            c?.next = prev
            prev = c
            c = n
            i += 1
        }
        
        head?.next = reverseKGroup(c, k)
        return prev
    }
}

This implementation provides a solution to the “Reverse Nodes in k-Group” problem in Swift without modifying the values in the list’s nodes. It recursively reverses the nodes in groups of k.

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