25. reverse nodes in k

 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class  {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        head = dummy;
        // head : reverse的前一个
        while(head != null) {
            head = reverse(head,  k);
        }
        return dummy.next;

    }
    // 如果不够k个，不reverse, return 一系列的前一个node
    private ListNode reverse(ListNode head, int k) {
        // 因为如果不够k个，不reverse， 所以要先判断有没有够k个，不能上来就reverse
        // dummy 和nk 同时找到，n1 同时找到 nk+1
        ListNode nk = head;
        for (int i = 0; i < k; i++) {
            nk = nk.next;
            if (nk == null) {
                return null;
            }
        }
        ListNode n0 = head;
        ListNode n1 = head.next;
        ListNode nkplus = nk.next;
        // now we have n0, n1,nk,nk+1;

        ListNode prev, curt;
        prev = n1;
       // 如果先调整头尾顺序的话 curt = n1.next; 完全不对！！！
        curt = n1.next;

        // prev n1--nk-1
        // curt n2 -- nk
        for (int i = 1; i < k; i++) {
            ListNode temp = curt.next;
            curt.next = prev;
            prev = curt;
            curt = temp;
        }
        n0.next = nk;
        n1.next = nkplus;
        return n1;
    }
}