这天是去公司实习的最后一天,也是原定启程的那天。但大约在前几日前得到消息因为香港愤青闹事于是我的航班(AA192)从当天的下午18时推迟到了第二天早上8时。于是我的计划也从中午出发乘坐港深大巴变成了晚上出发去口岸再在机场躺一晚。 时间突然变得充裕了起来。 虽然如此,早上的时光依然宝贵,仍需要使命般的去公司蹭最后一顿的早饭。因为某些原因没坐成班车,于是带着刚买的21寸小箱子坐了公交去深大北门,过了天桥,再走了几步路便到了腾大。此时大概不早了,腾大13楼餐厅也没有往常的那么挤。 我乘势点了平时最爱吃的汤圆丸子并随机配上了若干馒头,拍完照留存纪念后便开始吃了。 因为动作比较慢,到了工位发现自己大概是最后一个了。而那几天不知道为啥头总是有些晕(可能没睡够),于是这之后便去腾大22楼的医务室配了点药。由于第一次去那,差点迷路走进了腾讯云的工位群十分尴尬。好在最终还是找到了企鹅诊所(公司自营的),医生稍微给我检查了并问了最近的一些情况最终认为我大概还好,但我跟她说了当晚会离开深圳去美国,她还是很好心的给我配了一些要(顺便要了点创口贴)。最后当我掏出工卡准备付钱的时候突然得知公司员工在企鹅诊所看病不用钱,然后我怀着诧异+感激涕零的心情和医生道了谢并返回了工位。
AVL is a great data structure, which get rid of the disadvantage of ordinary BST. As we all know, for BST, when the data is bad enough, it's very likely become a linked list, which will make the time complexity very bad $O(n)$ The rotation of AVL can prevent us from this problem.
Height: NULL? -1 : max{height(leftchild), height(rightchild}+1 Balance: $|height(left)-height(right)|$ If balance factor is greater than 1, we say this tree is unbalanced.
Since the speed to access memory is much more faster than the speed to access disk, therefore, we need a data structure to make the height as small as possible.
The hard disk support to load a chunk of data in one time
order: the number of children of each node, or 1+max number of element of each node
Find is easy, we need only find the right interval through root to the leaf. Also one property the BTree has is that for each chunk of data if one node, it's sorted. So we can use binary search to do it.
\section{Introduction}
\subsection{Problem Background}
\subsubsection{This is Third}
$E=mc^2$
\[ E=mc^2. \]
\begin{equation}
\int ^{2}_{1}\oint^{b}_{a}dxd\theta
\end{equation}
\[ \begin{pmatrix} a&b\\c&d \end{pmatrix} \quad
\begin{bmatrix} a&b\\c&d \end{bmatrix} \quad
\begin{Bmatrix} a&b\\c&d \end{Bmatrix} \quad
\begin{vmatrix} a&b\\c&d \end{vmatrix} \quad
\begin{Vmatrix} a&b\\c&d \end{Vmatrix} \]
前天晚上参加了头条的后端面试,果然被拒了,在这里开个坑回忆一下
The bugs I have submitted patch for