分類收集

Question

的收藏我想排序基於前兩列值的文件內容：

我的文件內容爲：

文件名：ComplexMapFileReader.txt

而且conetent是：

B 12 hgjhgjgjgjgjgjhgjgjgjgjgjgj 
B 3 jgjdSAGdjgsdKJADJgjfgAJFL 
B 6 jygtjgjgjgjgjgjhgjhbj 
C 7 uiyuiyikykk 
C 2 wrteyytyuiiiiyjhg 
A 1 hikhkhkhkjhkhkhjkhkh 
A 4 khkhkkhkhkhkhkh 
A 2 khkhkhkhkhkhkhkhkhkhkhkhh 
A 11 jhgjsGJDGjAGFjgfjgjfgj 
A 5 gjgjgsadgajgjDGkkhkhdDKjhkjd 
B 5 hdskdhkljdhKAJHDKjahsd 
A 12 kjhkdjhfknkjflkjfsdlkjf 
C 3 kjhfekJEHFKHefkh 
B 34 khkhkjhkjhkqlwwjljlllljj 
A 21 iuhEWHRIekhrlkHRKErhwkrhk;h;kfnk.nfas 
C 11 jlklkqwwlklklklllklkkkkjjj   

我對相同的代碼是：

package CollectionExamples; 

import java.io.BufferedReader; 
import java.io.FileNotFoundException; 
import java.io.FileReader; 
import java.io.IOException; 
import java.util.HashMap; 
import java.util.Map; 
import java.util.Map.Entry; 
import java.util.TreeMap; 

public class MapFileReader { 

    private static final String FILENAME = "D:\\ComplexMapFileReader.txt"; 

    public static void main(String[] args) { 

     BufferedReader br = null; 
     FileReader fr = null; 

     Map<Integer, String> innerMap = null; 
     Map<String, TreeMap> outerMap = new TreeMap(); 
     try { 
      fr = new FileReader(FILENAME); 
      br = new BufferedReader(fr); 
      String sCurrentLine; 
      br = new BufferedReader(new FileReader(FILENAME)); 

      try { 
       while ((sCurrentLine = br.readLine()) != null) { 

        String[] s = sCurrentLine.split(" "); 

        if (!outerMap.containsKey(s[0])) { 
         innerMap = new TreeMap(); 
         innerMap.put(Integer.valueOf(s[1]), s[2]); 
         outerMap.put(s[0], (TreeMap) innerMap); 
        } else { 
         innerMap.put(Integer.valueOf(s[1]), s[2]); 
         outerMap.put(s[0], (TreeMap) innerMap); 
        } 

       } 

       String s0 = null; 
       String s1 = null; 
       String s2 = null; 

       for (Entry<String, TreeMap> map1 : outerMap.entrySet()) { 

        s0 = map1.getKey(); 

        for (Entry<Integer, String> map2 : innerMap.entrySet()) { 

         s1 = " "+map2.getKey() + ""; 
         s2 =" "+ map2.getValue(); 
         System.out.println("Each line equals:" + s0 + s1 + s2); 
         System.out.println("#############"); 

        } 

       } 

      } catch (IOException e) { 
       // TODO Auto-generated catch block 
       e.printStackTrace(); 
      } 

     } catch (FileNotFoundException e) { 
      // TODO Auto-generated catch block 
      e.printStackTrace(); 
     } 

    } 

} 

上面的代碼沒有給我想要的結果。但有序當上述文件是在某種形式的像（字母表中的所有B的一起或A的一起），那麼文件內容如下所示給出正確的排序：

B 12 hgjhgjgjgjgjgjhgjgjgjgjgjgj 
B 3 jgjdSAGdjgsdKJADJgjfgAJFL 
B 6 jygtjgjgjgjgjgjhgjhbj 
C 7 uiyuiyikykk 
C 2 wrteyytyuiiiiyjhg 
A 1 hikhkhkhkjhkhkhjkhkh 
A 4 khkhkkhkhkhkhkh 
A 2 khkhkhkhkhkhkhkhkhkhkhkhh 
A 11 jhgjsGJDGjAGFjgfjgjfgj 
A 5 gjgjgsadgajgjDGkkhkhdDKjhkjd 

，結果是所期望的。（在排序字母和然後按號碼）：

輸出：

Each line equals:A 1 hikhkhkhkjhkhkhjkhkh 
############# 
Each line equals:A 2 khkhkhkhkhkhkhkhkhkhkhkhh 
############# 
Each line equals:A 4 khkhkkhkhkhkhkh 
############# 
Each line equals:A 5 gjgjgsadgajgjDGkkhkhdDKjhkjd 
############# 
Each line equals:A 11 jhgjsGJDGjAGFjgfjgjfgj 
############# 
Each line equals:B 1 hikhkhkhkjhkhkhjkhkh 
############# 
Each line equals:B 2 khkhkhkhkhkhkhkhkhkhkhkhh 
############# 
Each line equals:B 4 khkhkkhkhkhkhkh 
############# 
Each line equals:B 5 gjgjgsadgajgjDGkkhkhdDKjhkjd 
############# 
Each line equals:B 11 jhgjsGJDGjAGFjgfjgjfgj 
############# 
Each line equals:C 1 hikhkhkhkjhkhkhjkhkh 
############# 
Each line equals:C 2 khkhkhkhkhkhkhkhkhkhkhkhh 
############# 
Each line equals:C 4 khkhkkhkhkhkhkh 
############# 
Each line equals:C 5 gjgjgsadgajgjDGkkhkhdDKjhkjd 
############# 
Each line equals:C 11 jhgjsGJDGjAGFjgfjgjfgj 
############# 

但是當文件具有內容下面的一個，我沒有得到期望的結果。我知道它是地圖，它將覆蓋以下情況下的先前值。但有沒有什麼辦法可以更好地解決這個問題。

B 12 hgjhgjgjgjgjgjhgjgjgjgjgjgj 
    B 3 jgjdSAGdjgsdKJADJgjfgAJFL 
    B 6 jygtjgjgjgjgjgjhgjhbj 
    C 7 uiyuiyikykk 
    C 2 wrteyytyuiiiiyjhg 
    A 1 hikhkhkhkjhkhkhjkhkh 
    A 4 khkhkkhkhkhkhkh 
    A 2 khkhkhkhkhkhkhkhkhkhkhkhh 
    A 11 jhgjsGJDGjAGFjgfjgjfgj 
    A 5 gjgjgsadgajgjDGkkhkhdDKjhkjd 
    B 5 hdskdhkljdhKAJHDKjahsd 
    A 12 kjhkdjhfknkjflkjfsdlkjf 
    C 3 kjhfekJEHFKHefkh 
    B 34 khkhkjhkjhkqlwwjljlllljj 
    A 21 iuhEWHRIekhrlkHRKErhwkrhk;h;kfnk.nfas 
    C 11 jlklkqwwlklklklllklkkkkjjj   

Answer 1

我認爲這是一個更好的方式來表示每個記錄對象的行。實際上，我們不需要TreeMap進行排序。如果你想添加更多的字段或者改變比較邏輯，它會更加方便。對於記錄列表，我們只需要實現Comparable接口來實現您的比較邏輯。

例如：

import java.io.BufferedReader; 
import java.io.FileNotFoundException; 
import java.io.FileReader; 
import java.io.IOException; 
import java.util.ArrayList; 
import java.util.Collections; 
import java.util.List; 

public class MapFileReader { 

    private static final String FILENAME = "D:\\ComplexMapFileReader.txt"; 


    private static class Record { 
     private char ch; 
     private int num; 
     private String string; 

     public void setCh(char ch) { 
      this.ch = ch; 
     } 

     public char getCh() { 
      return ch; 
     } 

     public void setNum(int num) { 
      this.num = num; 
     } 

     public int getNum() { 
      return num; 
     } 

     public String getString() { 
      return string; 
     } 

     public void setString(String string) { 
      this.string = string; 
     } 

    } 

    public static void main(String[] args) { 

     BufferedReader br; 
     try { 
      String sCurrentLine; 
      br = new BufferedReader(new FileReader(FILENAME)); 

      List<Record> records = new ArrayList<>(); 

      try { 
       while ((sCurrentLine = br.readLine()) != null) { 

        String[] s = sCurrentLine.split(" "); 

        if (s.length != 3) { 
         // Filter invalid records 
         continue; 
        } 

        Record r = new Record(); 
        r.ch = s[0].charAt(0); 
        r.num = Integer.parseInt(s[1]); 
        r.string = s[2]; 

        records.add(r); 
       } 

       Collections.sort(records, new Comparator<Record>() { 
        @Override 
        public int compare(Record o1, Record o2) { 
         if (o1.ch == o2.ch) { 
          return o1.num < o2.num ? -1 : (o1.num == o2.num ? 0 : 1); 
         } 
         return o1.ch < o2.ch ? -1 : 1; 
        } 
       }); 

       for (Record record : records) { 
        System.out.println(String.format("Each line equals: %s %s %s", record.ch, record.num, record.string)); 
        System.out.println("#############"); 
       } 

      } catch (IOException e) { 
       // TODO Auto-generated catch block 
       e.printStackTrace(); 
      } 

     } catch (FileNotFoundException e) { 
      // TODO Auto-generated catch block 
      e.printStackTrace(); 
     } 

    } 

}