Given a linked list, reverse the nodes of a linked listkat a time and return its modified list.

kis 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 ofkthen 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.

For example,
Given this linked list:1->2->3->4->5

Fork= 2, you should return:2->1->4->3->5

Fork= 3, you should return:3->2->1->4->5

Solution: 先找出第1~k个node（少于k个直接返回head），翻转，再把head接到helper(kth.next)返回的node。返回的是kth node

    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode kth = head;
        int num = 1;
        while (kth != null && num < k) {
            num++;
            kth = kth.next;
        }
        if (kth == null) {
            return head;
        }
        ListNode nextK = kth.next;
        ListNode prev = null, cur = head;
        while (cur != nextK) {
            ListNode next = cur.next;
            cur.next = prev;
            prev = cur;
            cur = next;
        }
        head.next = reverseKGroup(nextK, k);
        return kth;
    }