在B树中找到第k个密钥的算法？ [英] Algorithm to find k-th key in a B-tree?

问题描述

我试图了解如何考虑如何在B树中获取第k个键/元素。即使使用步骤而不是代码，它也将大有帮助。谢谢

I'm trying to understand how I should think about getting the k-th key/element in a B-tree. Even if it's steps instead of code, it will still help a lot. Thanks

编辑：要清除，我要的是B树中第k个最小的键。

To clear up, I'm asking for the k-th smallest key in the B-tree.

推荐答案

好吧，经过几个不眠之夜，我设法做到了，对于任何想知道如何做的人，这里都是伪代码（第一个元素k = 0）：

Ok so, after a few sleepless hours I managed to do it, and for anyone who will wonder how, here it goes in pseudocode (k=0 for first element):

get_k-th(current, k):

for i = 0 to current.number_of_children_nodes
    int size = size_of_B-tree(current.child[i])
    if(k <= size-1)
        return get_k-th(current.child[i], k)
    else if(k == size && i < current.number_of_children_nodes)
        return current.key[i]
    else if (is_leaf_node(current) && k < current.number_of_children_nodes)
        return node.key[k]
    k = k - size - 1;

return null

我知道这可能看起来很奇怪，但这就是事实我想出了，幸运的是它起作用了。也许有一种方法可以使这段代码更清晰，更高效，但是我希望它能很好地帮助其他可能像我一样遇到同样障碍的人。

I know this might look kinda weird, but it's what I came up with and thankfully it works. There might be a way to make this code clearer, and probably more efficient, but I hope it's good enough to help anyone else who might get stuck on the same obstacle as I did.

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