16

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]

但我需要实现以下排序：

FCBWHJLOAQUXMPVINTKGZERDYS

我想我需要实现Comparator和覆盖compare方法：

 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？

4

4 回答 4

41 
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。

显然，OP不仅要比较字母，还要比较字母串，所以它有点复杂：

    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  ;
    }
于 2013-05-14T10:57:27.567 回答
5

我会做这样的事情：

将字母放在一个 HashTable 中（我们称之为 orderMap）。key 是字母，value 是 ORDER 中的索引。

接着：

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

    @Override
    public int compare(String o1, String o2) {
        int length = o1.length > o2.length ? o1.length: o2.length
        for(int i = 0; i < length; ++i) {
           int firstLetterIndex = orderMap.get(o1.charAt(i));
           int secondLetterIndex = orderMap.get(o2.charAt(i));

           if(firstLetterIndex == secondLetterIndex) continue;

           // First string has lower index letter (for example F) and the second has higher index letter (for example B) - that means that the first string comes before
           if(firstLetterIndex < secondLetterIndex) return 1;
           else return -1;
        }

        return 0;
    }
});

为了使其不区分大小写，只需在开头对两个字符串执行 toUpperCase() 即可。

于 2013-05-14T11:38:20.803 回答
0

在这里您可以找到有用的链接：

使用比较器进行自定义排序

在您的示例中，您需要比较类的特定属性来检查基准字符串中 char 的位置，并基于此检查它是否更大/相等/更小。

于 2013-05-14T11:03:44.830 回答
0

我花了一些时间来改进所选答案。这个效率更高

public static void customSort(final String order,String[] array){
String[] alphabets={"A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z","0","1","2","3","4","5","6","7","8","9"};
    String keyword=order;
    for(int g=0; g<alphabets.length; g++){
    String one=alphabets[g];
    if(!keyword.toUpperCase().contains(one)){keyword=keyword+one;}
    }

final String finalKeyword=keyword;
Arrays.sort(array, new Comparator<String>() {

    @Override
   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 = finalKeyword.toUpperCase().indexOf(o1.toUpperCase().charAt(i));
          pos2 = finalKeyword.toUpperCase().indexOf(o2.toUpperCase().charAt(i));
       }

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

       return pos1  - pos2  ;
    }
});
//Arrays.sort(array, Collections.reverseOrder());
}
于 2017-09-01T07:13:55.777 回答