Question

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，则输出为：

 [ab, bed, car, dog, fog]

但我需要实现以下排序：

FCBWHJLOAQUXMPVINTKGZERDYS

我认为我需要实施Comparator并覆盖compare方法：

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

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

我该如何解决这个问题？

Answer 1

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()

如果您的阵列是区分大小写的。

显然，OP不仅要比较字母而且要比较字母串，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  ;
    }

Answer 2

我会做这样的事情：

将字母放在HashTable中（让我们称之为orderMap）。键是字母，值是ORDER中的索引。

然后：

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;
    }
});

为了使它不区分大小写，只需在开头为两个字符串添加toUpperCase（）。

Answer 3

在这里你可以找到有用的链接：

Using comparator to make custom sort

在你的例子中，比较你需要的类的特定属性，以检查基准字符串中的char的可能性，并基于此检查它是否更好/相等/更小。

Answer 4

花时间改进所选答案。这更有效率

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());
}

使用自定义排序对String数组进行排序

4 个答案: