Пример #1
0
User_Node user_successor(User_Node node) {
	RBTNode rbt = rbt_successor(node->rbtnode);
	if (rbt == NULL)
		return NULL;
	else
		return (User_Node)rbt->data;
}
Пример #2
0
void rbt_inorder(rbtree_t * T,void (*visit)(rbnode_t *))
{
    rbnode_t *x;
    x = T->root;
    if (x != &T->nil) {
        x = rbt_min(T,x);
        while (x != 0) {
            visit(x);
            x = rbt_successor(T,x);
        }
    }
    return ;
}
// 删除一个结点
// 考虑的情况
// 空树:该情况不可能出现,因为y已经指向树中的某一结点
// 只有一个根节点的树
// 有多个结点的树
prbtree rbt_delete_node(prbtree T, prbt_node z)
{
     // y指向要删除的结点
     prbt_node y;
     prbt_node x;
     if (z->left == nil || z->right == nil)
     {
	  y = z;
     }
     else
     {
	  y = rbt_successor(z);
     }
     // x 指向要连接的子树
     if (y->left != nil)
     {
	  x = y->left;
     }
     else
     {
	  x = y->right;
     }
     x->parent = y->parent;
     if (y->parent == nil)
     {
	  T = x;
     }
     else if (y->parent->left == y)
     {
	  y->parent->left = x;
     }
     else
     {
	  y->parent->right = x;
     }
// 如果y != z 则y为z的后继结点,将z的key赋值为y的key
     if (y != z)
     {
	  z->key = y->key;
	  // 将z的卫星数据赋值为y的卫星数据
	  // ...
     }
     if (y->color == BLACK)
     {
	  T = rbt_delete_fixup(T, x);
     }
     return T;
}
Пример #4
0
void tweet_print_top_5_most_users() {
	if (user_database == NULL || user_database == Nil) {
		printf("트윗 데이터베이스가 비어있습니다.\n");
		return;
	}

	char user[6][1000] = { "" };
	int maxNum[6] = { 0 };

	User_Node now = (User_Node)rbt_min_node(user_database)->data;
	User_Node max = (User_Node)rbt_max_node(user_database)->data;

	do {
		maxNum[5] = get_tweet_number_of_user(now);
		strcpy(user[5], now->name);

		for (int i = 0; i <5; i++)
			for (int j = i + 1; j < 6; j++)
				if (maxNum[i] < maxNum[j]) {
					int tmp = maxNum[i];
					maxNum[i] = maxNum[j];
					maxNum[j] = tmp;

					char tmpStr[100];
					strcpy(tmpStr, user[i]);
					strcpy(user[i], user[j]);
					strcpy(user[j], tmpStr);
				}

		maxNum[5]++;

		now = (User_Node)rbt_successor(now->rbtnode)->data;
	} while (now != max);

	for (int i = 0; i < 5; i++) {
		printf("%d위 : %-20s - %2d번\n", i + 1, user[i], maxNum[i]);
	}
}
Пример #5
0
//free node, return val
void *
delete_node(struct rbtree *rbt, struct rbnode *nd)
{
    void *val = NULL;
    struct rbnode *ret = nd;
    struct rbnode *tmp, *itor;
    if (nd == NULL || rbt == NULL)
        return NULL;
    val = nd->key;
    //pthread_mutex_lock(&(rbt->lock));
    if (nd->left == &rbt->nil || nd->right == &rbt->nil)
        tmp = nd;
    else
        tmp = rbt_successor(rbt, nd);
    if (tmp->left != &rbt->nil)
        itor = tmp->left;
    else
        itor = tmp->right;
    itor->parent = tmp->parent;
    if (tmp->parent == &rbt->nil)
        rbt->root = itor;
    else {
        if (tmp == tmp->parent->left)
            tmp->parent->left = itor;
        else
            tmp->parent->right = itor;
    }
    if (tmp != itor)
        nd->key = tmp->key;
    if (tmp->color == BLACK)
        delete_fixup(rbt, itor);
    if (ret == NULL)
        printf("ret is null\n");
    free(tmp);
    rbt->size--;
    //pthread_mutex_unlock(&(rbt->lock));
    return val;
}
Пример #6
0
//free node, return val
void *
delete_node(struct rbtree *rbt, void *key)
{
    void *val = NULL;
    struct rbnode *nd = NULL;
    struct rbnode *tmp, *itor;
    nd = find_node(rbt, key);
    if (nd == NULL || rbt == NULL) {
        printf("find node error\n");
        return NULL;
    }
    val = nd->key;
    if (nd->left == &rbt->nil || nd->right == &rbt->nil)
        tmp = nd;
    else
        tmp = rbt_successor(rbt, nd);
    if (tmp->left != &rbt->nil)
        itor = tmp->left;
    else
        itor = tmp->right;
    itor->parent = tmp->parent;
    if (tmp->parent == &rbt->nil)
        rbt->root = itor;
    else {
        if (tmp == tmp->parent->left)
            tmp->parent->left = itor;
        else
            tmp->parent->right = itor;
    }
    if (tmp != itor)
        nd->key = tmp->key;
    if (tmp->color == BLACK)
        delete_fixup(rbt, itor);
    //free(tmp);
    rbt->size--;
    return tmp;
}
Пример #7
0
RBT_NODE *rbt_remove( RBT_ROOT *root , RBT_NODE *z )
{
  RBT_NODE *x ;
  RBT_NODE *y ;

  if ( z == &( root->nil ) ) {
    return NULL ;
  }

  HQTRACE( rbt_debug ,
           ( "rbt_remove: [%x] %x" , root , z->key )) ;

  if ( z->left == &( root->nil ) || z->right == &( root->nil ) ) {
    y = z ;
  }
  else {
    y = rbt_successor( root , z ) ;
  }

  HQASSERT( y != NULL , "Should rbt_successor return NULL?" ) ;

  if ( y->left != &( root->nil ) ) {
    x = y->left ;
  }
  else {
    x = y->right ;
  }

  x->p = y->p ;

  if ( y->p == &( root->nil ) ) {
    root->node = x ;
  }
  else {
    if ( y == y->p->left ) {
      y->p->left = x ;
    }
    else {
      y->p->right = x ;
    }
  }

  if ( y != z ) {
    uintptr_t tk ;
    void *td ;

    /* The example pseudo-code just copies data and "other fields"
       from y to z. That's not good enough when any one of them is a
       pointer: we end up with one data struct being doubly referenced
       and the other one orphaned. Solution: swap the pointers
       instead. */

    tk = z->key ;
    z->key = y->key ;
    y->key = tk ;

    td = z->data ;
    z->data = y->data ;
    y->data = td ;
  }

  if ( y->red == FALSE ) {
    rbt_remove_fixup( root , x ) ;
  }

  root->count-- ;
  rbt_validate( root , root->node , NULL ) ;

  return y ;
}