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?

Solution

import com_github_leetcode.ListNode

/*
 * Definition for singly-linked list.
 * class ListNode(_x: Int = 0, _next: ListNode = null) {
 *   var next: ListNode = _next
 *   var x: Int = _x
 * }
 */
object Solution {
    def reverseKGroup(head: ListNode, k: Int): ListNode = {
        if (head == null || head.next == null || k == 1) {
            return head
        }
        var j = 0
        var len = head
        // Loop for checking the length of the linked list; if the linked list is less than k, then return as it is.
        while (j < k) {
            if (len == null) {
                return head
            }
            len = len.next
            j += 1
        }
        // Reverse linked list logic applied here.
        var c = head
        var n: ListNode = null
        var prev: ListNode = null
        var i = 0
        // Traverse the while loop for K times to reverse the nodes in K groups.
        while (i != k) {
            n = c.next
            c.next = prev
            prev = c
            c = n
            i += 1
        }
        // head.x is pointing to the next K group linked list; recursion for further remaining linked list.
        head.next = reverseKGroup(n, k)
        prev
    }
}

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