在B树中找到第k个密钥的算法？

本文介绍了在B树中找到第k个密钥的算法？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 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.