Difficulty: Hard

这道题给 Hard 其实没必要，简单而言就是一个大的链表翻转套一个小的链表翻转，但在处理的时候要小心节点操作顺序。

这段代码还可以处理得更简洁一点，比如加一个过头元，通过计数来判断是不是一组，等等

class Solution {
public:
  ListNode *reverseKGroup(ListNode *head, int k) {
    if (k == 1)
      return head;
    ListNode *p = head;
    ListNode *newhead = head;
    ListNode *lastend = NULL;
    while (p) {
      ListNode *begin = p;
      for (int i = 0; i < k - 1; i++) {
        p = p->next;
        if (!p)
          return newhead;
      }
      ListNode *end = p;
      if (newhead == head)
        newhead = end;
      if (lastend)
        lastend->next = end;
      p = p->next;
      rev(begin, k);
      begin->next = p;
      lastend = begin;
    }
    return newhead;
  }
  ListNode *rev(ListNode *begin, int k) {
    ListNode *p = begin;
    ListNode *pp = p->next;
    ListNode *ppp;
    for (int i = 1; i <= k - 1; i++) {
      ppp = pp->next;
      pp->next = p;
      p = pp;
      pp = ppp;
    }
    return pp;
  }
};