Wil*_*lls 5 java memory sorting memory-management file
我有一个文件列表,我想对它进行排序并提取上次修改的前 3 个文件。
约束:由于下游应用程序的兼容性问题,我无法使用 Java 7
File[] files = directory.listFiles();
Arrays.sort(files, new Comparator<File>(){
public int compare(File f1, File f2)
{
return Long.valueOf(f1.lastModified()).compareTo(f2.lastModified());
} });
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public static void sortFilesDesc(File[] files) {
Arrays.sort(files, new Comparator() {
public int compare(Object o1, Object o2) {
if ((File)o1).lastModified().compareTo((File)o2).lastModified()) {
return -1;
} else if (((File) o1).lastModified() < ((File) o2).lastModified()) {
return +1;
} else {
return 0;
}
}
});
}
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以上两种解决方案需要更多的时间来执行和内存。我的文件列表包含大约 300 个 tar 文件,每个文件大小为 200MB。所以它消耗更多的时间和内存。
有没有办法有效地处理这个问题?
每个比较操作都使用一个内存较高的文件对象,有没有办法释放内存并有效地处理它?
You can do it much faster.
Arrays.sort(...) uses "quick sort", which takes ~ n * ln(n) operations.
This example takes only one iteration trough the whole array, which is ~ n operations.
public static void sortFilesDesc(File[] files) {
File firstMostRecent = null;
File secondMostRecent = null;
File thirdMostRecent = null;
for (File file : files) {
if ((firstMostRecent == null)
|| (firstMostRecent.lastModified() < file.lastModified())) {
thirdMostRecent = secondMostRecent;
secondMostRecent = firstMostRecent;
firstMostRecent = file;
} else if ((secondMostRecent == null)
|| (secondMostRecent.lastModified() < file.lastModified())) {
thirdMostRecent = secondMostRecent;
secondMostRecent = file;
} else if ((thirdMostRecent == null)
|| (thirdMostRecent.lastModified() < file.lastModified())) {
thirdMostRecent = file;
}
}
}
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On small numbers of files you won't see much difference, but even for tens of files the difference will be significant, for bigger numbers - dramatic.
The code to check the algorithm (please put in a correct files structure):
package com.hk.basicjava.clasload.tests2;
import java.io.File;
import java.util.Date;
class MyFile extends File {
private long time = 0;
public MyFile(String name, long timeMills) {
super(name);
time = timeMills;
}
@Override
public long lastModified() {
return time;
}
}
public class Files {
/**
* @param args
*/
public static void main(String[] args) {
File[] files = new File[5];
files[0] = new MyFile("File1", new Date(2013,1,15, 7,0).getTime());
files[1] = new MyFile("File2", new Date(2013,1,15, 7,40).getTime());
files[2] = new MyFile("File3", new Date(2013,1,15, 5,0).getTime());
files[3] = new MyFile("File4", new Date(2013,1,15, 10,0).getTime());
files[4] = new MyFile("File5", new Date(2013,1,15, 4,0).getTime());
sortFilesDesc(files);
}
public static void sortFilesDesc(File[] files) {
File firstMostRecent = null;
File secondMostRecent = null;
File thirdMostRecent = null;
for (File file : files) {
if ((firstMostRecent == null)
|| (firstMostRecent.lastModified() < file.lastModified())) {
thirdMostRecent = secondMostRecent;
secondMostRecent = firstMostRecent;
firstMostRecent = file;
} else if ((secondMostRecent == null)
|| (secondMostRecent.lastModified() < file.lastModified())) {
thirdMostRecent = secondMostRecent;
secondMostRecent = file;
} else if ((thirdMostRecent == null)
|| (thirdMostRecent.lastModified() < file.lastModified())) {
thirdMostRecent = file;
}
}
System.out.println("firstMostRecent : " + firstMostRecent.getName());
System.out.println("secondMostRecent : " + secondMostRecent.getName());
System.out.println("thirdMostRecent : " + thirdMostRecent.getName());
}
}
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