使用自定义排序对字符串数组进行排序 [英] Sorting an array of String with custom ordering

问题描述

我有一个 String 数组:

 String[] str = {"ab" , "fog", "dog", "car", "bed"};
 Arrays.sort(str);
 System.out.println(Arrays.toString(str));

如果我使用Arrays.sort，输出是:

 [ab, bed, car, dog, fog]

但我需要实现以下顺序:

But I need to implement the following ordering:

FCBWHJLOAQUXMPVINTKGZERDYS

我想我需要实现 Comparator 并覆盖 compare 方法:

I think I need to implement Comparator and override compare method:

 Arrays.sort(str, new Comparator<String>() {

        @Override
        public int compare(String o1, String o2) {
            // TODO Auto-generated method stub
            return 0;
        }
    });

我应该如何解决这个问题?

How should I go about solving this?

推荐答案

final String ORDER= "FCBWHJLOAQUXMPVINTKGZERDYS";

Arrays.sort(str, new Comparator<String>() {

    @Override
    public int compare(String o1, String o2) {
       return ORDER.indexOf(o1) -  ORDER.indexOf(o2) ;
    }
});

您还可以添加:

o1.toUpperCase()

如果您的数组不区分大小写.

If your array is case in-sensitive.

显然 OP 不仅要比较字母，还要比较字母串，所以有点复杂:

Apparently the OP wants to compare not only letters but strings of letters, so it's a bit more complicated:

    public int compare(String o1, String o2) {
       int pos1 = 0;
       int pos2 = 0;
       for (int i = 0; i < Math.min(o1.length(), o2.length()) && pos1 == pos2; i++) {
          pos1 = ORDER.indexOf(o1.charAt(i));
          pos2 = ORDER.indexOf(o2.charAt(i));
       }

       if (pos1 == pos2 && o1.length() != o2.length()) {
           return o1.length() - o2.length();
       }

       return pos1  - pos2  ;
    }

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