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

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */
function reverseKGroup(head: ListNode | null, k: number): ListNode | null {
    if (head === null || head.next === null || k === 1) {
        return head
    }
    let j = 0
    let 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++
    }
    // Reverse linked list logic applied here.
    let c = head
    let n = null
    let prev = null
    let 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++
    }
    // 'head' now points to the last node of the reversed K-group.
    // Recursion for further remaining linked list.
    head.next = reverseKGroup(n, k)
    return prev
}

export { reverseKGroup }

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