C++实现单链表删除倒数第k个节点的方法

typedef struct Node
{
  int data;
  struct Node* next;
}node, *pLinkedList;

//head表示头结点
Node* removeLastKthNode(pLinkedList head, int k)
{
  if (NULL == head->next || k < 1)
    return head;
  pLinkedList cur = head; //1号指针
  pLinkedList ret = head; //2号指针，指向待删除节点
  pLinkedList pre = NULL; //pdel待删除节点的前一个节点
  /*
  head  1  2  3  4  5
  cur
  ret
  */
  while (k > 0 && cur != NULL)
  {
    k--;
    cur = cur->next;
  }
  //当链表走到终点时，k仍然大于0，可知k值大于链表长度
  if (k > 0 && cur == NULL)
  {
    cout << "k值大于链表长度" << endl;
    return head;
  }
  //k == 0时，1号指针和2号指针同时走,ret即为待删除的节点
  if (k == 0)
  {
    while (cur != NULL)
    {
      pre = ret;
      cur = cur->next;
      ret = ret->next;
    }
    /*k = 2时
    head  1    2   3    4   5  NULL
               pre   ret     cur
    即要求删除ret节点
    */
    pre->next = ret->next;
    free(ret);
    ret = NULL;
    return head;
  }
}