使用自定义排序对字符串数组进行排序
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?
答
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 ;
}