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

/*
 * Example:
 * var li = ListNode(5)
 * var v = li.`val`
 * Definition for singly-linked list.
 * class ListNode(var `val`: Int) {
 *     var next: ListNode? = null
 * }
 */
class Solution {
    fun reverseKGroup(head: ListNode?, k: Int): ListNode? {
        if (head?.next == null || k == 1) {
            return head
        }
        var j = 0
        var len = head
        // loop for checking the length of the linklist, if the linklist is less than k, then return
        // as it is.
        while (j < k) {
            if (len == null) {
                return head
            }
            len = len.next
            j++
        }
        // 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 node in K groups.
        while (i != k) {
            n = c!!.next
            c.next = prev
            prev = c
            c = n
            i++
        }
        // C1 is pointing to 1st node of K group, which is now going to point to the next K group
        // linklist.
        // recursion, for futher remaining linked list.
        head.next = reverseKGroup(n, k)
        return prev
    }
}

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