Java 使用自定义顺序对字符串数组进行排序

我有一个

字符串
数组：

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

如果使用
数组.sort
，则输出为：

 [ab, bed, car, dog, fog]

但我需要实现以下顺序：

FCBWHJLOAQUXMPVINTKGZERDYS

我想我需要实现
Comparator
并重写
compare
方法：

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

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

Arrays.sort（str，newcomparator（）{
@凌驾
公共整数比较（字符串o1、字符串o2）{
//TODO自动生成的方法存根
返回0；
}
});

我该如何着手解决这个问题

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) ;
    }
});

如果数组区分大小写


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  ;
    }

public int比较（字符串o1、字符串o2）{
int pos1=0；
int pos2=0；
对于（inti=0；i
在这里您可以找到有用的链接：


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字符的位置，并在此基础上检查字符是否为greather/equal/level
 我会这样做：

将字母放在哈希表中（我们称之为orderMap）。键是字母，值是顺序索引

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;
    }
});

Arrays.sort（str，newcomparator（）{
@凌驾
公共整数比较（字符串o1、字符串o2）{
int length=o1.length>o2.length？o1.length:o2.length
对于（int i=0；i

要使其不区分大小写，只需在开头对两个字符串都执行toUpperCase（）。
花了我的时间改进了所选的答案。这样效率更高

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());
}

publicstaticvoidcustomsort（最终字符串顺序，字符串[]数组）{
字符串[]字母={“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”}；
字符串关键字=顺序；
对于（int g=0；gcareful:upper/lower case mission比比皆是。@MajidL应该区分大小写吗？@MajidL它似乎没有给出正确的结果：
{“AA”，“FB”，“A”，“FC”}
，它给出了
[AA，FB，FC，A]
。这是不正确的。提供的答案只有在您给它一个或多个作为顺序子字符串的情况下才有效（例如-JLO）您可以返回
o1.length（）-o2.length（）<代码> >代码> POS1- POS2= = 0 如果它们匹配如<代码> A<代码>和<代码> AA/代码>，例如，首先，如果您的项始终是一个字符，则应该考虑使用字符数组。使用字符串强制进行更多的错误检查。@邓肯琼斯：不，它们不是，这只是一个例子。我改变了。他们。Thanks@Sam：因此字符串中的每个字符都需要按顺序排序，然后应用“正常”字符串规则？如何将“AA”相对于“A”排序？或将“FC”相对于“FB”排序？@JoachimSauer我不知道我是否正确理解了您的观点。但顺序应该是：“FC”、“FB”、“A”、“AA”
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());
}