25. Reverse Nodes in k-Group

使用递归
在进行K个节点翻转的时候需要设置变量来存储游标，当前节点和父节点。
采用在翻转操作结束后进行长度的检查，牺牲了最后一组不足k个的节点对两次翻转的时间，但是省去了每次递归先进行长度计算，较为效率。

/**
 * 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||head.next==null||k==1)return head;
        ListNode res = head;
        ListNode parent = head,current = null;
        int count = 1;
        head = head.next;
        while(head!=null&&count<k){
            current = head;
            head = head.next;
            current.next = parent;            
            parent = current;            
            count ++ ;
        }
        if(count<k){
            ListNode tmp = current.next;
            current.next = null;
            for(int i=0;i<count-1;i++){
                current = tmp;
                tmp = tmp.next;
                current.next = parent;
                parent = current;
            }
            return current;
        }
        res.next = reverseKGroup(head, k);
        res = current;
      
        return res;
    }
}