Reverse Nodes in k-Group

Given a linked list, reverse the nodes of a linked list k at a time and return its modified 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 nodes, only nodes itself may be changed. Only constant memory is allowed.

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

Think

• Consider the list like following:

  
    dummy(pre) -> l1 -> l2 -> l3 -> l4 -> l5 -> null
  

• Get the reversed segment:

  when cnt % k == 0:
  
        pre.next      node  node.next
           |            |     |
    pre -> l1 -> l2 -> l3 -> l4 -> l5 -> null
  
          last      ---->         end
            |                  |
    pre -> l1 -> l2 -> l3 -> l4 -> l5 -> null
            |     |        
        pre.next pre.next(final status)    
               \     \
                  |  -> |     |
                 cur   cur     cur.next (final status)
  

Solution

public class Solution {
    /**
     * @param head a ListNode
     * @param k an integer
     * @return a ListNode
     */
    public ListNode reverseKGroup(ListNode head, int k) {
        if(head == null || head.next == null)
            return head;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy;
        int cnt = 1;
        ListNode front = head;
        while(head != null) {
            if(cnt % k == 0) {
                pre = reverse(pre, head.next);
                head = pre.next;
            }else
                head = head.next;
            cnt++;
        }
        return dummy.next;
    }

    private ListNode reverse(ListNode pre, ListNode end) {
        ListNode last = pre.next, cur = last.next;
        while(cur != end) {
            last.next = cur.next;
            cur.next = pre.next;
            pre.next = cur;
            cur = last.next;
        }
        return last;
    }
}