Problem Description

Given the head of a linked list, reverse the nodes of the list k at a time, and return the 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]

Follow-up: Can you solve the problem in O(1) extra memory space?

Solution

There are not much tricks here, but we need to divide and solve sub-problems. The first is to reverse a linked-list, the second is to properly connecting nodes between the k-reversed linked-lists, especially with the constant space complexity. Of course, we have to first check whether there are k nodes before reversing it.

Java Code

class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        if( k==1 ) return head;
        ListNode prev_node = new ListNode(), test = head, curr=prev_node;
        while( true ){
            int i=0;
            test = head;
            while( test!=null && i<k){
                test = test.next;
                i++;
            }
            if(i < k){
                curr.next = head;
                break;
            }
            i = 1;
            ListNode second_node = head.next;
            test = head;
            while(i++ < k){
                ListNode tmp = second_node.next;
                second_node.next = test;
                test = second_node;
                second_node = tmp;
            }
            curr.next = test;
            curr = head;
            head = second_node;
        }
        return prev_node.next;
    }
}