Leetcode#25 Reverse Nodes in k-Group

原题地址

分两步：

1. 试探。看是否还有k个节点可以翻转

2. 翻转。

[]->[]->[]->[1]->[2]->[3]->[4]->[]->[]->[]->NULL
           |------------------|
               假设要翻转这一段

翻转前：

            head           tail
             |              |
[]->[]->[]->[1]->[2]->[3]->[4]->[]->[]->[]->NULL
        |                        |
     beforeHead               afterTail


翻转后：

            tail           head
             |              |
[]->[]->[]->[4]->[3]->[2]->[1]->[]->[]->[]->NULL
        |                        |
     beforeHead               afterTail

为了方便处理，在链表首部增加了一个虚拟节点vhead（代码第21行）

代码：

void reverseList(ListNode *head, ListNode *tail) {
  ListNode *curr = head;
  ListNode *next = head->next;
  ListNode *nextNext = NULL;

  while (next != tail) {
    nextNext = next->next;
    next->next = curr;
    curr = next;
    next = nextNext;
  }

  next->next = curr;
  head->next = NULL;
}

ListNode *reverseKGroup(ListNode *head, int k) {
  if (k < 2)
    return head;

  ListNode *vhead = new ListNode(0);
  vhead->next = head;
  ListNode *tail = NULL;
  ListNode *beforeHead = vhead;
  ListNode *afterTail = NULL;
  int count = 0;

  while (head) {
    tail = head;
    count = k;
    while (tail && --count)
      tail = tail->next;
    if (count)
      break;
    afterTail = tail->next;

    reverseList(head, tail);

    beforeHead->next = tail;
    head->next = afterTail;

    beforeHead = head;
    head = afterTail;
  }

  return vhead->next;
}