binary-search-tree 相关问题

我有一个示例 XML 输入文件，其中包含以下内容：我有一个示例 XML 输入文件，其中包含以下内容： <?xml version="1.0" encoding="utf-8"?> <infobases> <infobase author="Chartered Professional Accountants of Canada" levelDefOrder="Level 1,Level 2,Level 3,Level 4,Level 5,Level 6,Level 7,Level 8,Level 9,Level 10,Level 11,Level 12,Level 13,Normal Level" levels="Level 1,Level 2,Level 3,Level 4,Level 5,Level 6,Level 7,Level 8,Level 9,Level 10,Level 11,Level 12,Level 13" name="info_a" title="CPA Canada Standards and Guidance Collection"> <file level="Level 1" heading="XYZ1-L1">XYZ1-L1 <file level="Level 2" heading="XYZ1-L12">XYZ1-L12 <file level="Level 3" heading="XYZ1-L123">XYZ1-L123</file> <file level="Level 3" heading="XYZ1-L123">XYZ1-L123</file> <file level="Level 3" heading="XYZ1-L123">XYZ1-L123</file> </file> </file> <file level="Level 1" heading="XYZ2-L1">XYZ2-L1</file> <file level="Level 1" heading="XYZ2-L1">XYZ2-L1 <file level="Level 2" heading="XYZ2-L12">XYZ2-L12</file> <file level="Level 2" heading="XYZ2-L123">XYZ2-L123 <file level="Level 3" heading="XYZ1-L123">XYZ1-L123 <file level="Level 4" heading="XYZ1-L123">XYZ1-L123</file> </file> </file> </file> </infobase> </infobases> 我想编写一个脚本来识别文件元素级别属性，并在打印文件元素的标题时给出适当的缩进。文件元素标题将获得 .ditamap 扩展名，该扩展名将其他文件元素作为子元素并忽略其他子元素（不在示例 xml 文件中）。如果文件元素没有子文件元素，则会添加 .dita 扩展名。我用 javascript 编写了一个脚本，它倾向于进行正确的缩进，但分配的扩展名不正确。我得到所有文件元素标题的 .dita。这是代码： const fs = require('fs'); const XmlStream = require('xml-stream'); // Create a readable stream from the XML file const stream = fs.createReadStream('input1.xml'); // Create a writable stream to the output text file const outputStream = fs.createWriteStream('output.txt'); // Create a new XML stream parser const xmlParser = new XmlStream(stream); // Function to print headings with proper indentation function printHeadingsToFile(file, indentation = '') { // Calculate the indentation based on the level of the file const levelIndentation = ' '.repeat(parseInt(file.$.level.substr(6)) - 1); // Determine file extension based on child elements const fileExtension = file.file ? 'ditamap' : 'dita'; // Write the heading with indentation to the output file outputStream.write(`${indentation}${levelIndentation}${file.$.heading}.${fileExtension}\n`); // Check if there are nested files if (file.file) { // If nested files exist, recursively print their headings with increased indentation file.file.forEach(nestedFile => { printHeadingsToFile(nestedFile, `${indentation}${levelIndentation}`); }); } } // Event listener for when a new XML element is encountered xmlParser.on('startElement: file', function(element) { // Print headings for each file printHeadingsToFile(element); }); // Event listener for when the parsing ends xmlParser.on('end', function() { console.log('Parsing finished.'); // Close the output stream after finishing writing outputStream.end(); }); // Event listener for any errors during parsing xmlParser.on('error', function(err) { console.error('Error during parsing:', err); // Close the output stream if there is an error outputStream.end(); }); 我在这里得到的输出如下： XYZ1-L1.dita XYZ1-L12.dita XYZ1-L123.dita XYZ1-L123.dita XYZ1-L123.dita XYZ2-L1.dita XYZ2-L1.dita XYZ2-L12.dita XYZ2-L123.dita XYZ1-L123.dita XYZ1-L123.dita 预期输出： XYZ1-L1.ditamap XYZ1-L12.ditamap XYZ1-L123.dita XYZ1-L123.dita XYZ1-L123.dita XYZ2-L1.dita XYZ2-L1.ditamap XYZ2-L12.dita XYZ2-L123.ditamap XYZ1-L123.ditamap XYZ1-L123.dita 找到了使用不同 xml 解析器包的解决方案。 const fs = require('fs'); const sax = require('sax'); // Create a SAX parser const parser = sax.parser(true); // Create a write stream to a text file const outputStream = fs.createWriteStream('output.txt'); // Array to store file attributes let fileAttributes = []; // Event handlers for the SAX parser parser.onopentag = function (node) { // Check if the current node is a <file> element if (node.name === 'file') { // Extract attributes and push them to the array fileAttributes.push(node.attributes); } }; parser.onend = function () { // Parsing ends, log the final output //console.log('File attributes:', fileAttributes); // Function to determine the file extension function getFileExtension(currentLevel, nextLevel) { if (currentLevel < nextLevel) { return '.ditamap'; } else { return '.dita'; } } // Iterate through the fileAttributes array for (let i = 0; i < fileAttributes.length; i++) { const currentFile = fileAttributes[i]; const nextFile = fileAttributes[i + 1]; // Get the level numbers const currentLevel = parseInt(currentFile.level.match(/\d+/)[0]); const nextLevel = nextFile ? parseInt(nextFile.level.match(/\d+/)[0]) : 0; // Get the file extension const extension = getFileExtension(currentLevel, nextLevel); // Indentation based on the level const indentation = ' '.repeat((currentLevel - 1) * 4); // Prepare the text to write const textToWrite = indentation + currentFile.heading + extension + '\n'; // Write the text to the output stream outputStream.write(textToWrite); } // Close the output stream when done outputStream.end(); }; // Read XML file const xmlData = fs.readFileSync('input.xml', 'utf8'); // Parse XML data parser.write(xmlData).close(); 结果输出： L1.ditamap L2.ditamap L3.dita L3.dita L3.dita L1.dita L1.ditamap L2.dita L2.ditamap L3.ditamap L4.ditamap Level5Text.dita

javascript recursion linked-list scripting binary-search-tree

回答 1 投票 0

查找存储在二叉搜索树中的单词的所有前缀

我有一个存储字符串的二叉搜索树。字符串按字典顺序进行比较（使用 c 中的 strcmp）。给定一些输入字符串，我想检索该字符串的所有前缀的列表......

string algorithm binary-search-tree

回答 1 投票 0

查找存储在 BST 中的单词的所有前缀

我有一个存储字符串的二叉搜索树。字符串按字典顺序进行比较（使用 c 中的 strcmp）。给定一些输入字符串，我想检索该字符串的所有前缀的列表...

string algorithm binary-search-tree

回答 1 投票 0

在二叉搜索树中插入重复值

我可以在二叉搜索树中插入相同的值吗？如果可以的话，这个插入应该发生在具有该值的现有节点的左侧还是右侧？我可以向这棵树插入 23 吗？