Reverse Nodes in k-Group

Question

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

LeetCodeleetcode.com/problems/reverse-nodes-in-k-group

Solution

/*
@param: Head of ListNode, Integer k, reverse linkedlist every k node;
@return: Head of reversed LinkedList;
Algorithm: dummy node -> head;
before reverse, check if there is enough nodes to be reversed;
every reverse return prehead of the linkedList;
*/
public ListNode reverseKGroup(ListNode head, int k) {
    if (head == null || head.next == null) return head;        
    
    ListNode dummy = new ListNode(0);
    dummy.next = head; head = dummy;
    
    //reverse every group
    while (head != null) {
        head = reverseKNodes(head,k);
    }
    return dummy.next;
}

// return the prehead of linkedList
public ListNode reverseKNodes(ListNode head, int k) {
    ListNode nk = head;
    //check if there is enough nodes
    for (int i = 0; i < k; i++) {
        nk = nk.next;
        if (nk == null) return null;            
    }
    
    // head -> n1 -> n2 -> n3 -> nk -> nk+1..-> null
    // head(pre) -> cur
    // head -> nk -> nk-1 ->... -> n1 -> nk+1 -> null
    
    // pre -> n1 -> n2 -> n3;
    // pre    cur  temp
    // null <- pre <- cur -> n3;
    ListNode n1 = head.next;
    ListNode pre = null;
    ListNode cur = n1;
    for (int i = 0; i < k; i++) {
        ListNode temp = cur.next;
        cur.next = pre;
        pre = cur;
        cur = temp;  
    }
    n1.next = cur;
    head.next = pre;
    return n1;
}