Question

我有n个游戏对象树。用户随机选择树中的某些对象并要删除。问题在于某些对象是另一对象的子对象。事实证明，每次删除层次结构内的节点后，我都必须遍历所有选定的节点并将其删除。算法比O（n ^ 2）快吗？

更新：为了更加清楚我的需求，我编写了伪代码：

struct TreeNode
{
    vector<TreeNode*> childs;
};

void removeNodeHierarchy(list<TreeNode*>& nodes, TreeNode *n)
{
    for(TreeNode *child : n->childs())
        removeNodeHierarchy(nodes, child);

    nodes.remove(n); // complexity nodes.size()
    delete n;
}

// The function I'm trying to write 
// Problem: Total complexity = nodes.size() * nodes.size()
void removeNodes(list<TreeNode*>& nodes, TreeNode *root)
{
    while (!nodes.empty()) // complexity nodes.size()
    {
        TreeNode *n = nodes.first();
        removeNodeHierarchy(nodes, n);
    }
}

void main()
{
    TreeNode *tree = ...
    list<TreeNode*> nodes = ...

    removeNodes(nodes, tree);
}

Answer 1

In order to search that node, it will take you O(n^d), where d is the depth of the tree. So it will be faster if d < 2 which I think will almost never be the case for a big game tree. Are you sure that n from n-ary and O(n^2) are the same symbol?

有效地从n元树中删除节点列表

1 个答案: