第一题：

题目打印：

java写法：

public static void showTree(int level, File parentFolderPath) {        if (parentFolderPath.isDirectory()) {            File[] childFiles = parentFolderPath.listFiles();            for (File file : childFiles) {                showNameByLevel(level);                System.out.println(file.getName());                if (file.isDirectory()) {                    showTree(level + 1, file);                }            }        }    }    public static void showNameByLevel(int level) {        StringBuffer stringBuffer = new StringBuffer();        if (level > 0) {            for (int i = 0; i < level; i++) {                stringBuffer.append("\t");            }        }        if(stringBuffer.length() >0) System.out.print("|" + stringBuffer);        System.out.println("|");        if (stringBuffer.length()>0) System.out.print("|" + stringBuffer);        System.out.print("----");    }    @Test    public void c23() {        File file = new File("D:\\TOOL\\IDEASpace\\mavedome\\src\\test\\java");//        printlen(file, 0);        showTree(0, file);    }

python写法：

def fileCntIn(currPath):    '''''汇总当前目录下文件数'''    sum = 0    for root, dirs, files in os.walk(currPath, topdown=False):        for flie in files:            sum += len(flie)    return sum    # return sum([len(files) for root, dirs, files in os.walk(currPath,topdown=False)])def dirsTree(startPath):    '''''树形打印出目录结构'''    for root, dirs, files in os.walk(startPath):        # 获取当前目录下文件数        fileCount = fileCntIn(root)        # 获取当前目录相对输入目录的层级关系,整数类型        level = root.replace(startPath, '').count(os.sep)        # 树形结构显示关键语句        # 根据目录的层级关系，重复显示'| '间隔符，        # 第一层 '| '        # 第二层 '| | '        # 第三层 '| | | '        # 依此类推...        # 在每一层结束时，合并输出 '|____'        indent = '| ' * 1 * level + '|____'        print('%s%s -r:%s' % (indent, os.path.split(root)[1], fileCount))        for file in files:            indent = '| ' * 1 * (level + 1) + '|____'            print('%s%s' % (indent, file))if __name__ == '__main__':    dirsTree(os.getcwd())

第二题：

判断一些哪里有问题，如果有问题，怎么修改。

FileInputStream fileInputStream = null;        FileOutputStream fileOutputStream = null;        try {            fileInputStream = new FileInputStream("ttt");            fileOutputStream = new FileOutputStream("bbb");        }

修改为：

FileInputStream fileInputStream = null;        FileOutputStream fileOutputStream = null;        try {            fileInputStream = new FileInputStream("ttt");            fileOutputStream = new FileOutputStream("bbb");            int len = 0;            while ((len = fileInputStream.read()) != 0) {                fileOutputStream.write(len);            }        } catch (FileNotFoundException e) {            e.printStackTrace();        } catch (IOException e) {            e.printStackTrace();        }finally {            if (fileInputStream != null) {                try {                    fileInputStream.close();                } catch (IOException e) {                    e.printStackTrace();                }            }            if (fileOutputStream != null) {                try {                    fileOutputStream.close();                } catch (IOException e) {                    e.printStackTrace();                }            }        }

写法：

public static void main(String[] args) {        FileInputStream fis = null;        FileOutputStream fos = null;        try {            fis = new FileInputStream("aaa.txt");            fos = new FileOutputStream("bbb.txt");            int b;            while ((b = fis.read()) != -1) {                fos.write(b);            }        } catch (FileNotFoundException e) {            e.printStackTrace();        } catch (IOException e) {            e.printStackTrace();        } finally {            if (fis != null) {                try {                    fis.close();                } catch (IOException e) {                    e.printStackTrace();                }            }            if (fos != null) {                try {                    fos.close();                } catch (IOException e) {                    e.printStackTrace();                }            }        }    }}