如何使用Java对文本文件中的数字进行排序？

Question

16° C
15° C
18° C
13° C
17° C
19° C
21° C
20° C
16° C

import java.io.*;
import java.util.*;

public class temperatur {

    public static void main(String[] args) throws IOException {
        BufferedReader reader = null;
        PrintWriter outputStream = null;
        ArrayList<String> rows = new ArrayList<String>();

        try {
            reader  = new BufferedReader(new FileReader("temp.txt"));
            outputStream = new PrintWriter(new FileWriter("tempout.txt"));

            String file;
            while ((file = reader .readLine()) != null) {
                rows.add(file);
            }
            Collections.sort(rows);
            String[] strArr= rows.toArray(new String[0]);
            for (String cur : strArr)
                outputStream.println(cur);
        } finally {
            if (reader  != null) {
                inputStream.close();
            }
            if (outputStream != null) {
                outputStream.close();
            }
        }
    }
    
}

Answer 1

下面是如何对行中已经有的行进行排序的示例。这是假设您希望保留原始行，而不是重建它们。

解析和排序

理想情况下，使用一个类来保存行的内容和数值，例如：

//simplified, add setters, getters etc.
class Line {
  String content;
  double temp;
}

List<Line> rows = ...
String lineContent;
while ((lineContent= reader .readLine()) != null) {
   double temp = Double.parseDouble(lineContent.split("°")[0]);
   rows.add(new Line(lineContent, temp));
}

字符串通过比较字符进行排序，因此“10”将被认为小于“2”。因此，我们需要使用一个“诀窍”：

• 先按长度排序，所以2小于10
• 然后按词法排序

Collections.sort(rows, Comparator.comparing(String::length)
                                 .thenComparing(Function.identity())
                                 .reversed());

请注意，如果数字可以为负数或具有不同的格式（即不同长度的分数、不同数量的空格、行中的附加数据等），这将中断。在这些情况下，编译器需要更加复杂，复杂性越高，使用“解析和排序”方法就越容易。

Collections.sort(rows, (String left, String right) -> {
        char c1Left = left.charAt(0);
        char c1Right = right.charAt(0);
        
        int direction = -1; //1 for ascending
        
        //negative numbers are smaller than 0 or positive numbers
        int result = Integer.compare(c1Left == '-' ? -1 : 0, c1Right == '-' ? -1 : 0);
        if( result != 0) {
            return result * direction;
        }
        
        //at this point, both numbers are either both positive or both negative
        //for negative numbers we need to reserve direction since we'll only compare "absolute values"
        if( c1Right == '-' ) {
            direction *= -1;
        }
        
        String subLeft = Character.isDigit(c1Left) ? left : left.substring(1);
        String subRight = Character.isDigit(c1Right) ? right : right.substring(1);
        
        //shorter numbers are smaller than longer numbers (takes care of cases like 2 vs 10 or 3.1 vs 21.0)
        result = Integer.compare(subLeft.length(), subRight.length());
        if( result != 0) {
            return result * direction;
        }
        
        //sign and length are equal, the rest is a simple character comparison
        result = subLeft.compareTo(subRight);
        
        return result * direction;
    });