作业错题集锦（pta英文数据结构）

A graph with 30 vertices and 40 edges must have at most

twenty one connected component(s).

• 要计算最大连通分量，应该尽量先把边凑完，然后后面的每一个顶点分别是一个独立的连通分量
• 所以10 * 9 / 2 > 40 第一个连通分量用10个顶点，剩下20个顶点就是20个连通分量
• 最后加上第一个连通分量

Since the speed of a printer cannot match the speed of a computer, a buffer is designed to temperarily store the data from a computer so that later the printer can retrieve data in order. Then the proper structure of the buffer shall be a:

A.stack

B.queue

C.tree

D.graph

• 概念题

To build a heap from N records, the best time complexity is:

A.O(logN)

B.O(N)

C.O(NlogN)

D.O(N^2)

Heapify

从最后一个非叶子节点一直到根结点进行堆化的调整。如果当前节点小于某个自己的孩子节点（大根堆中），那么当前节点和这个孩子交换。Heapify是一种类似下沉的操作，HeapInsert是一种类似上浮的操作。

这种建堆的时间复杂度是O(N)

怎么找到第一个非叶子节点

根节点在数组中的索引为1，所以第一个非叶子节点的计算公式为: last_non_leaf = arr.length/2。

如果根节点在数组中的索引为0，那么第一个非叶子节点的计算公式为: last_non_leav = (arr.length - 2)/2

可以设最后一个非叶子节点位置为x，那么最后一个叶子节点一定是(2x+1) 或者(2x+2)中的一个，然后可以建立方程求解。

Which one of the following relations is correct about the extra space taken by heap sort, quick sort and merge sort?

A.heap sort < merge sort < quick sort

B.heap sort > merge sort > quick sor

C.heap sort < quick sort < merge sort

D.heap sort > quick sort > merge sort

空间复杂度堆排序O(1),快速排序O(logN),归并排序O(N);

The average search time of searching a hash table with N elements is:

A.O(1)

B.O(logN)

C.O(N)

D.cannot be determined

这个题有点争议，按照道理来讲平均查找时间复杂度是O(1),如果有冲突的话可能会退化到O(N)

目前只能先记一下答案。如果说的有不对的地方大家可以在评论区指出

If a binary search tree of N nodes is complete, which one of the following statements is FALSE?

A.the average search time for all nodes is O(logN)

B.the minimum key must be at a leaf node

C.the maximum key must be at a leaf node

D.the median node must either be the root or in the left subtree

可能会出现最后一个节点没有右子树的情况

To sort { 8, 3, 9, 11, 2, 1, 4, 7, 5, 10, 6 } by Shell Sort, if we obtain ( 4, 2, 1, 8, 3, 5, 10, 6, 9, 11, 7 ) after the first run, and ( 1, 2, 3, 5, 4, 6, 7, 8, 9, 11, 10 ) after the second run, then the increments of these two runs must be __ , respectively.

A.3 and 1

B.3 and 2

C.5 and 2

D.5 and 3

哈希排序可以看上一篇博客

To delete X from a splay tree, the operations are: (1) Find X; (2)Remove X; (3) FindMin(T

R); and the last operation is:

A.Make T L the right child of the root of T R

B.Make T L the left child of the root of T R

C.Make T R the right child of the root of T L

D.Make T R the left child of the root of T L

Let T be a tree of N nodes created by union-by-size without path compression, then the minimum depth of T may be

A.1

B.logN

C.N−1

D.N/2

Given an integer sequence 25、84、21、47、15、27、68、35、20, after the Heapsort's initial heap building, this interger sequence is __.

A.47 20 21 25 84 27 68 35 15

B.15 21 20 25 84 27 68 35 47

C.15 20 21 25 84 27 68 35 47

D.25 15 20 47 35 21 68 84 27

Given an integer sequence 25、84、21、47、15、27、68、35、20, after the first run of the shell sorting by an increment 3, the integer sequence is __.

A.25 15 20 27 35 21 68 84 47

B.15 21 20 25 84 27 68 35 47

C.25 15 20 47 21 35 68 84 27

D.25 15 20 47 35 21 68 84 27

希尔排序可以看上一篇博客

One of the following algorithms: Selection sort, Shell sort, Insertion sort, is applied to sort the sequence (2, 12, 16, 88, 1, 5, 10）in ascending order. If the resulting sequences of the first two runs are (2, 12 , 16, 10, 1, 5, 88) and (2, 12, 5, 10, 1, 16, 88), then the algorithm must be ____.

A.Selection sort

B.Shell sort

C.Insertion sort

D.uncertain

For a graph, if each vertex has an even degree, we can find an Euler circuit that visits every vertex exactly once.

T F

Linear probing is equivalent to double hashing with a secondary hash function of Hash 2(k)=1 .

T F

Quadratic probing is equivalent to double hashing with a secondary hash function of Hash (k)=k.

T F

In a binary tree, if node A is a descendant of node B, A may precede B in the inorder traversal sequence.

T F

To prove the correctness of a greedy algorithm, we must prove that an optimal solution to the original problem always makes the greedy choice, so that the greedy choice is always safe.

T F

To sort N records by quick sort, the worst-case time complexity is Ω(NlogN).

T F