排序的ArrayList&LT;串GT;自定义比较 [英] Sorting an ArrayList<String> with custom Comparator
问题描述
我想排序的的ArrayList&LT;串GT;
使用自定义的比较
。我的要求是, XX
字符串应该是第一个字符串,其他人应该遵循自然顺序。
I am trying to sort an ArrayList<String>
using custom Comparator
. My requirement is that XX
String should be the first String, others should follow natural ordering.
我需要:[XX,XX,1,5,9,A,D,G,Q,Z]
我所得到的[1,5,9,A,D,G,Q,Z,XX,XX]
I need : [XX, XX, 1, 5, 9, A, D, G, Q, Z]
What I am getting is [1, 5, 9, A, D, G, Q, Z, XX, XX]
以下是我的code:
public class Test {
public static void main(String[] args)
{
List<String> list = new ArrayList<String>();
list.add("Z");
list.add("5");
list.add("D");
list.add("G");
list.add("XX");
list.add("9");
list.add("Q");
list.add("XX");
list.add("1");
list.add("A");
Collections.sort(list, new CustomComparator());
System.out.println(list);
}
}
class CustomComparator implements Comparator<String>
{
@Override
public int compare(String o1, String o2) {
if(o2.equals("XX")) {
return -1;
}
return o1.compareTo(o2);
}
}
修改:如果我改变比较逻辑:
EDIT: if i change comparator logic to:
@Override
public int compare(String o1, String o2) {
if(o2.equals("XX")) {
return 1;
}
return o1.compareTo(o2);
}
我越来越:
[1,XX,9,A,Q,XX,5,D,G,Z]
[1, XX, 9, A, Q, XX, 5, D, G, Z]
请让我知道如何着手。先谢谢了。
Please let me know how to proceed. Thanks in advance.
推荐答案
使用此比较实施
@Override
public int compare(String o1, String o2) {
if(o2.equals("XX")) {
return o1.equals("XX") ? 0 : 1;
} else if (o1.equals("XX")) {
return -1;
}
return o1.compareTo(o2);
}
原因:超越的问题时使用-1或1它来保证O1和O2的所有可能的元组明确的秩序,看到的的javadoc :
Reason: Beyond the question when to use -1 or 1 it is important to guarantee a clear order for all possible tuples of o1 and o2, see javadoc:
...]
实现者必须确保的sgn(比较(X,Y))== -sgn(比较(Y,X)),对所有的x和y。 (这意味着比较(X,Y)当且仅当比较(Y,X)抛出一个异常必须抛出异常。)
[...] The implementor must ensure that sgn(compare(x, y)) == -sgn(compare(y, x)) for all x and y. (This implies that compare(x, y) must throw an exception if and only if compare(y, x) throws an exception.)
实现程序还必须确保关系是可传递的:((比较(X,Y)> 0)及及(对比(Y,Z)> 0))表示比较(X,Z)> 0。
[...]
The implementor must also ensure that the relation is transitive: ((compare(x, y)>0) && (compare(y, z)>0)) implies compare(x, z)>0. [...]
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