Access level modifiers的区别

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ArrayList刚用到的几个method

1. ArrayList  合并2个arraylist's, 即把第二个合并到第一个里面去,而不是产生一个新的

Method Signiture:

addAll(Collection<? extends E> c)

Appends all of the elements in the specified collection to the end of this list, in the order that they are returned by the specified collection's Iterator.

List<String> newList = new ArrayList<String>(listOne);
newList.addAll(listTwo);

以上即为stackoverflow上的建议代码.

 

 

 

File, formater

public class WordCount {

	// EDIT DIR PATH WHEN NEEDED, DEFINED AS A CONSTANT
	private static final String FILE_DIR = "//Users//charles//Documents//workspace//assignment1//";

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// list to store file names
		List<String> fileNameList = new ArrayList<String>();
		// total size of files
		int total = 0;

		// input file names
		System.out.println("Enter the files to be calculated(\"q\" to indicate no more files to input): ");

		// read all possible file names
		try{
			BufferedReader bufferRead = new BufferedReader(new InputStreamReader(System.in));
		String s = bufferRead.readLine();
			while(!s.equals("q")){
				fileNameList.add(s);
				s = bufferRead.readLine();
			}

		}catch(IOException e){
			e.printStackTrace();
		}

		// attempt to open files 
		for(String fileName : fileNameList){
			File file = new File(FILE_DIR.concat(fileName));

			// if file exists, format print it, and recalculate total size
			// exception if file cannot be accessed
			try{
				if(file.exists()){

					System.out.println(String.format("%7d", file.length()) +" "+ fileName);
					total += file.length();
				}
			}catch(SecurityException e){
				System.out.println("Cannot access file: " + fileName);
			}
		}

		// print total size
		System.out.println(String.format("%7d", total) + " total");

	}
// this is the comment i made from GVim, good !~~~
}

 

fdgdg

How to find files with specific extension

import java.io.*;
 
public class FindCertainExtension {
 
	private static final String FILE_DIR = "c:\\folder";
	private static final String FILE_TEXT_EXT = ".jpg";
 
	public static void main(String args[]) {
		new FindCertainExtension().listFile(FILE_DIR, FILE_TEXT_EXT);
	}
 
	public void listFile(String folder, String ext) {
 
		GenericExtFilter filter = new GenericExtFilter(ext);
 
		File dir = new File(folder);
 
		if(dir.isDirectory()==false){
			System.out.println("Directory does not exists : " + FILE_DIR);
			return;
		}
 
		// list out all the file name and filter by the extension
		String[] list = dir.list(filter);
 
		if (list.length == 0) {
			System.out.println("no files end with : " + ext);
			return;
		}
 
		for (String file : list) {
			String temp = new StringBuffer(FILE_DIR).append(File.separator)
					.append(file).toString();
			System.out.println("file : " + temp);
		}
	}
 
	// inner class, generic extension filter
	public class GenericExtFilter implements FilenameFilter {
 
		private String ext;
 
		public GenericExtFilter(String ext) {
			this.ext = ext;
		}
 
		public boolean accept(File dir, String name) {
			return (name.endsWith(ext));
		}
	}
}

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先留个底,过两天过来补完

calculate actual time的经典复杂方式,记忆犹新,还是想贴出来

private void calculateTime(int month,int year){ i=true; while (i==true){/* The logic is a loop using.*/ if(month==4||month==6||month==9||month==11){ // If the month only has 30 days,but the day greater than 30, if(new_d>30){ new_d=new_d-30;month++; // then the day should minus 30 and the month should add 1. }else i=false; }else if(month==2){ if ((year%4==0&&year%100!=0)||(year%400==0&&year%3200!=0)){ /* The leap year and in Febuary.*/ if(new_d>29){ /* Febuary will has 29 days.*/ new_d=new_d-29;month++; }else i=false; }else{ /* Febuary in a normal year.*/ if(new_d>28){ /* Febuary only has 28 days.*/ new_d=new_d-28;month++; }else i=false; } }else if (month==1||month==3||month==5||month==7||month==8||month==10||month==12){ if(new_d>31) { new_d=new_d-31;month++; }else i=false; } new_m=month; if(month>12){ //If month is greater than 12, new_y=year+month/12; //both the year and the month will be changed. new_m=month%12; month=new_m; year=new_y; } } //The loop can ensure that the month and the year keep changing until the day is in a correct range. }
 
PS. 到现在还是想不明白,当时为了计算正确的叠加后的月份与日子,尽然想了很久,最后还是借鉴了别人的答案。不过就是一步一步得叠加秒与分钟,叠加到月的时候用个循环罢了,为什么当时就觉得那么复杂呢?
由此我又要检讨自己了。这是一个相当正常的思路,quote
 
if(month>12){              //If month is greater than 12,
new_y=year+month/12;   //both the year and the month will be changed.
 
可是我却想了很久,就是因为我没有把思路理清,为何没有理清呢?就是源于没有把我觉得复杂的可能性列在纸上,连笔都不愿意动的人,怎么能够独立思考一道题,更不要说得到最终的个人最佳答案,我该狠狠得反省了。很多结果的生成并不是因为实力的不足够,而是源于信心的不充足,对自己的不信任,不能够冷静得把逻辑理清楚,是浮躁了!成功是需要静下心来想办法达到的,这次冒险过了,下次就不能luck out。
 
谨以此文逼迫自己谨记假期梦中的haunting headache。