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"数据结构与算法分析（Java语言描述）习题答案第三版"

Java

数据结构

算法分析

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The "数据结构与算法分析（Java语言描述）习题答案（第三版）.docx" and "数据结构与算法分析Java语言描述（第三版）课后习题答案" are resources that provide answers to exercises related to data structures and algorithm analysis using Java language. These resources are useful for students, professionals, and anyone interested in learning about data structures and algorithms in the context of Java programming. The exercises cover various topics such as processing files, algorithms, and data structures. One example given is the general way to process files, which involves writing a procedure with a specific heading to open a file, perform processing, and then close it. This indicates the practical application of data structures and algorithms in handling files and data manipulation. In addition, the exercises demonstrate mathematical concepts such as series and sequences. For example, the expression 4S = 1 + 2^2 + 3^2 + ... + 442^2 is analyzed, showing how mathematical principles and formulas are integrated into algorithm analysis. Overall, these resources provide comprehensive answers to exercises and demonstrate the integration of data structures, algorithms, and Java programming. They serve as valuable references for understanding and applying these concepts in practical situations. Whether for educational purposes or professional development, these resources offer insights into the complexities of data structures and algorithm analysis in the context of Java programming.

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}

else

{

itemL2 = iterL2.hasNext()?iterL2.next():null;

}

3.5 public static <AnyType extends Comparable<? super AnyType>>

void union(List<AnyType> L1, List<AnyType> L2,

List<AnyType> Result)

{

ListIterator<AnyType> iterL1 = L1.listIterator();

ListIterator<AnyType> iterL2 = L2.listIterator();

AnyType itemL1=null, itemL2=null;

// get first item in each list

if ( iterL1.hasNext() && iterL2.hasNext() )

{

itemL1 = iterL1.next();

itemL2 = iterL2.next();

}

while ( itemL1 != null && itemL2 != null )

{

int compareResult = itemL1.compareTo(itemL2);

if ( compareResult == 0 )

{

Result.add(itemL1);

itemL1 = iterL1.hasNext()?iterL1.next():null;

itemL2 = iterL2.hasNext()?iterL2.next():null;

}

else if ( compareResult < 0 )

{

Result.add(itemL1);

itemL1 = iterL1.hasNext()?iterL1.next():null;

}

else

{

Result.add(itemL2);

itemL2 = iterL2.hasNext()?iterL2.next():null;

}

3.6 A basic algorithm is to iterate through the list, removing every Mth item. This can be improved by two

observations. The first is that an item M distance away is the same as an item that is only M mod N away.

This is useful when M is large enough to cause iteration through the list multiple times. The second is that

an item M distance away in the forward direction is the same as an item (M – N) away in the backwards

direction. This improvement is useful when M is more than N/2 (half the list). The solution shown below uses

these two observations. Note that the list size changes as items are removed. The worst case running time is

O(Nmin(M, N)), though with the improvements given the algorithm might be significantly faster. If M = 1,

the algorithm is linear.

public static void pass(int m, int n)

{

int i, j, mPrime, numLeft;

ArrayList<Integer> L = new ArrayList<Integer>();

for (i=1; i<=n; i++)

L.add(i);

ListIterator<Integer> iter = L.listIterator();

Integer item=0;

numLeft = n;

mPrime = m % n;

for (i=0; i<n; i++)

{

mPrime = m % numleft;

if (mPrime <= numLeft/2)

{

if (iter.hasNext())

item = iter.next();

for (j=0; j<mPrime; j++)

{

if (!iter.hasNext())

iter = L.listIterator();

item = iter.next();

}

else

{

for (j=0; j<numLeft-mPrime; j++)

{

if (!iter.hasPrevious())

iter = L.listIterator(L.size());

item = iter.previous();

}

System.out.print("Removed " + item + " ");

iter.remove();

if (!iter.hasNext())

iter = L.listIterator();

System.out.println();

for (Integer x:L)

System.out.print(x + " ");

System.out.println();

numLeft--;

}

System.out.println();

}

3.7 O(N

). The trim method reduces the size of the array, requiring each add to resize it. The resize takes O(N)

time, and there are O(N) calls.

3.8 (a) Because the remove call changes the size, which would affect the loop.

(b) O(N

). Each remove from the beginning requires moving all elements forward, which takes O(N) time.

(d) No. Since it is always the first element being removed, finding the position does not take time, thus

an iterator would not help.

3.9 public void addAll( Iterable<? extends AnyType> items )

{

Iterator<? extends AnyType> iter = items.iterator();

while ( iter.hasNext() )

{

add(iter.next());

}

This runs in O(N) time, where N is the size of the items collection.

3.10 public void removeAll( Iterable<? extends AnyType> items )

{

AnyType item, element;

Iterator<? extends AnyType> iterItems = items.iterator();

while ( iterItems.hasNext ( ) )

{

item = iterItems.next();

Iterator<? extends AnyType> iterList = iterator();

while ( iterList.hasNext ( ) )

{

element = iterList.next();

if ( element.equals(item) )

iterList.remove();

}

This runs in O(MN), where M is the size of the items, and N is the size of the list.

3.11 import java.util.*;

public class SingleList

剩余63页未读，继续阅读

import os from openpyxl import load_workbook import docx # 读取Excel文件 workbook = load_workbook(filename=r'D:\好记薪\工商资料生成\表格.xlsx') sheet = workbook['需调用文档'] # 判断C列是否为空 if sheet['C2'].value: # 合并A、B文档 merge_docx(['A.docx', 'B.docx'], 'AB.docx') if sheet['C3'].value or sheet['C4'].value or sheet['C5'].value or sheet['C6'].value: # 合并A、B、C文档 merge_docx(['A.docx', 'B.docx', 'C.docx'], 'ABC.docx') if sheet['C7'].value: # 合并A、B、C、D文档 merge_docx(['A.docx', 'B.docx', 'C.docx', 'D.docx'], 'ABCD.docx') if sheet['C8'].value: # 合并A、B、C、E文档 merge_docx(['A.docx', 'B.docx', 'C.docx', 'E.docx'], 'ABCE.docx') # 合并docx文件 def merge_docx(file_list, merged_file_name): # 如果合并后的文件已存在，则直接返回 if os.path.exists(merged_file_name): return # 创建合并后的文档对象 merged_doc = docx.Document() # 合并文件内容 for file_name in file_list: doc = docx.Document(file_name) for para in doc.paragraphs: merged_doc.add_paragraph(para.text) for table in doc.tables: merged_doc.add_table(table.rows, table.columns) # 保存合并后的文档 merged_doc.save(merged_file_name)，报错：NameError: name 'merge_docx' is not defined

你的代码中调用了`merge_docx`函数，但是在调用之前并没有定义该函数，因此导致了`NameError: name 'merge_docx' is not defined`错误。为了解决这个问题，你应该把`merge_docx`函数定义提到调用的前面，或者把它...

修改代码：filename = row[0] + '.docx' # 文件名为第一个单元格中的内容加上.docx # 保存文档 doc.save(r'G:\工作文件\产业办\2022年\产业帮扶\产业奖补\3.经营主体奖补\县级业务\15.绩效资料\实施方案') # 将文档保存为指定的文件名 # 重命名并另存文件 os.rename(r'G:\工作文件\产业办\2022年\产业帮扶\产业奖补\3.经营主体奖补\县级业务\15.绩效资料\实施方案.docx', filename, row[0] + '.docx') # 将文件名改为第一个单元格中的内容加上.docx，并保存。

row = ['第一行内容'] # 假设这是第一行的数据 filename = row[0] + '.docx' # 文件名为第一个单元格中的内容加上.docx doc.save(r'G:\工作文件\产业办\2022年\产业帮扶\产业奖补\3.经营主体奖补\县级业务\15.绩效...

接口文档示例.docx

通过接口文档示例.docx，开发人员可以清晰地了解每个接口的作用、输入输出数据格式、调用方式等信息，从而更加方便地进行接口的开发和集成。在接口文档示例.docx中，一般会包括接口的基本信息、使用说明、示例代码...

创建一个模板docx，如果xlsx的sheet2中D2不为空则将r'D:\好记薪\工商资料生成\A公司登记申请书.docx'加之模板末尾并保留格式，如果xlsx的sheet2中E2不为空则将r'D:\好记薪\工商资料生成\B股东会决议.docx'加之模板末尾并保留格式，如果xlsx的sheet2中F2不为空则将r'D:\好记薪\工商资料生成\C章程.docx'加之模板末尾并保留格式，如果xlsx的sheet2中G2不为空则将r'D:\好记薪\工商资料生成\D股权转让协议.docx'加之模板末尾并保留格式，如果xlsx的sheet2中H2不为空则将r'D:\好记薪\工商资料生成\E有关债务清偿及担保情况说明.docx'加之模板末尾并保留格式.

以下是实现您所描述功能的代码： ```python from openpyxl import load_workbook from docxcompose.composer import Composer from docx import Document # 加载Excel文件 wb = load_workbook(r'D:\好记薪\工商...

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