每k个一组翻转链表。题意是请将input中每K个node进行翻转。这题我暂时只会递归解法，之后想通了迭代怎么写，写的清楚了我再补上。例子
Example:
Given this linked list:
1->2->3->4->5For k = 2, you should return:
2->1->4->3->5For k = 3, you should return:
3->2->1->4->5
思路是用cur指针遍历input，然后用count记录遍历的node个数。因为cur是从head开始的，所以当count === K的时候，cur已经到了第K + 1个node的地方了。此时用递归调用，反转K个node。若不足K个的时候，不会进行翻转。
时间O(n)
空间O(n)
1 /** 2 * @param {ListNode} head 3 * @param {number} k 4 * @return {ListNode} 5 */ 6 var reverseKGroup = function(head, k) { 7 // corner case 8 if (head === null || head.next === null) { 9 return head; 10 } 11 12 // normal case 13 let count = 0; 14 let cur = head; 15 while (cur !== null && count != k) { 16 cur = cur.next; 17 count++; 18 } 19 if (count === k) { 20 cur = reverseKGroup(cur, k); 21 while (count-- > 0) { 22 let next = head.next; 23 head.next = cur; 24 cur = head; 25 head = next; 26 } 27 head = cur; 28 } 29 return head; 30 };