Difficulty

Hard

Description

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

Solution

Time Complexity: O(n)
Space Complexity: O(1)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        if (head == null || k == 1) return head;
		
		ListNode dummy = new ListNode(0);
		dummy.next = head;
		ListNode pre = dummy;
		ListNode cur = head;
		int round = k;

		while(cur != null)
		{
			while (round != 1 && cur.next != null)
			{
				ListNode tmp = cur.next;
				cur.next = tmp.next;
				tmp.next = pre.next;
				pre.next = tmp;
				round--;
			}
			
			if (round == 1)
			{
				pre = cur;
				cur = cur.next;
				round = k;
			} 
			else 
			{
				cur = pre.next;
				round = k;
				while (round != 1 && cur.next != null)
				{
					ListNode tmp = cur.next;
					cur.next = tmp.next;
					tmp.next = pre.next;
					pre.next = tmp;
					round--;
				}

				return dummy.next;
			}
		}

		return dummy.next;
    }
}