25. Reverse Nodes in k-Group

Problem

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.

Example:

Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

Note:

• Only constant extra memory is allowed.
• You may not alter the values in the list's nodes, only nodes itself may be changed.

Solution (Java)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        if (head == null || head.next == null || k == 1) {
            return head;
        }
        int j = 0;
        ListNode 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.
        ListNode c = head;
        ListNode n = null;
        ListNode prev = null;
        int 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;
    }
}

Solution (Javascript)

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} k
 * @return {ListNode}
 */
var reverseKGroup = function(head, k) {
  if (!head || k < 2) return head;

  var count = 0;
  var now = head;
  var last = head;
  var tmp = null;

  while (now && count < k) {
    now = now.next;
    count++;
  }

  if (count === k) {
    now = reverseKGroup(now, k);
    while (count-- > 0) {
      tmp = last.next;
      last.next = now;
      now = last;
      last = tmp;
    }
    last = now;
  }

  return last;
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(1).