25. Reverse Nodes in k-Group 还未解决！！！

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 ofk 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.

For 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

public ListNode reverseKGroup(ListNode head, int k) {    ListNode curr = head;    int count = 0;    while (curr != null && count != k) { // find the k+1 node        curr = curr.next;        count++;    }    if (count == k) { // if k+1 node is found        curr = reverseKGroup(curr, k); // reverse list with k+1 node as head        // head - head-pointer to direct part,         // curr - head-pointer to reversed part;        while (count-- > 0) { // reverse current k-group:             ListNode tmp = head.next; // tmp - next head in direct part            head.next = curr; // preappending "direct" head to the reversed list             curr = head; // move head of reversed part to a new node            head = tmp; // move "direct" head to the next node in direct part        }        head = curr;    }    return head;}