The master of data structure solving is in a hurry

In general, we divide the data generated by the system into three types according to the structural model: structured data, semi-structured data, and unstructured data. Structured data, i.e., row data, is data that is stored in a database and can be logically expressed in a two-dimensional table structure.

The most common are numeric and textual data, which can exist in a fixed field of a file or record in some standard format. Correspondingly, the data that is not convenient to be represented by the two-dimensional logical table of the database without a fixed structure is called unstructured data, including office documents, text, XML, HTML, various reports, images and audio information, etc. We all know that structured data is easy to collect and store, and it is easy to analyze and display, and there are many mature BIs on the market....

This is a condition that scares people. You just point to BI and point to AI +1. As long as one table ends, it is terminated.

Ask for the answer to the question of Kai Ying, the calculation process and the reason, see the figure.

The array a[n][m], a[0][0] is the smallest element of the address in the array. If a[0][0] is stored at the address n, then the address of a[i][j] is n+i*m*sizeof(t)+j*sizeof(t), and sizeof(t) is the rental group or god storage unit occupied by each element.

1) Xixun pants sorting (the increment of the first sorting is 4) mu chongjian: 49 13 27 49 76 38 65 97

2) Bubbling sort (large number sinking): 38 49 65 76 13 27 49 97

3) Quick sort (pivot to 49) 27 38 13 49 76 97 65 49

4) 2-Way Merge Judgment Sort 38 49 65 97 13 76 27 49

Select B, the stack is a first-in-last-out structure, advanced 1, 2, 3, out 3, in 4, out 4, 2, 1, in 5, 6, out 6, 5

Select A, traverse the binary tree according to the hierarchy, you can use a queue Q, first put the root node of the binary tree into the queue, and then unqueue it, and output the node; If the brigade regrets that it has a left subtree, the root node of the left subtree will be queued; If it has a right subtree, the root node of the right subtree will be queued until the queue is empty.

(1) (a) of the pre-root sequence: a b c d e f posterior root sequence: b d e f c a

2) Forest sequence: a k c d e f g h i j k meso-sequence: b d e f c a i j k h g (3) This forest is converted into the corresponding binary tree as follows:

a/ \ b g\ /c h

d ie j \f k

If j starts from 1, it is c, a[9][5] has 41 data in front of it, 2000 + 41 * 4 = 2164; I take the liberty of asking, which school is my classmate.

In the tree, each node must have added one degree to the parent node except for the root node, so the total number of degrees is equal to the number of nodes -1.

The total degree in the problem is 25, so there are 26 nodes, and there are 11 nodes with a degree of 0 minus the number of nodes with degrees of 2 and 1.

The correct option is B.

"Data Structure and Problem Solving" is a book published by Tsinghua University Press in 2010, and the author is Weiss. This book is designed for a two-semester course for computer science majors, starting with an introduction to what data structures are, followed by an analysis of advanced data structures and algorithms. [1]

This is because, according to the definition of mid-order traversal, if a node has a left child, then the line point must be visited before it ends traversing its left subtree, and the last node traversed by his left subtree must be the child in the bottom right corner of his left subtree.

As shown in the figure below, node B has a left subtree, so the precursor of node b is the child g in the bottom right corner of her left subtree.

1) It's wrong.

Because n(logn) 2 cannot be represented as a*(n 2) (a is a constant).

O(N2) grows faster than O(N(logn) 2).

2) It's wrong.

Because nlog(n 2) = 2nlogn, the growth rate of n 2logn is higher than that of nlog(n 2).