我正在尝试使用带有自定义TreeSet的{{1}}来清除列表中的重复对象。
对于此代码：
Comparator控制台上的实际输出：[p1-> 1，p2-> 2，p3-> 3，p4-> 4，p1-> 1，p3-> 3]
控制台上的预期输出：[p1-＆gt; 1，p2-> 2，p3-> 3，p4-> 4]
为什么我在集合中再次获得class ASDF {
int i
Pass ref
new(Pass p, int i) {
this.ref = p
this.i=i
}
public static def void main(String[] args) {
val list = new TreeSet(
new Comparator<ASDF> {
override compare(ASDF obj1, ASDF obj2) {
if (obj1.ref == obj2.ref && obj1.i == obj2.i) {
return 0
}
return 1
}
}
)
val a1 = new ASDF(new Pass("p1"), 1)
val a2 = new ASDF(new Pass("p2"), 2)
val a3 = new ASDF(new Pass("p3"), 3)
val a4 = new ASDF(new Pass("p4"), 4)
list.addAll(
a1, a2, a3, a4
,
a1, a2, a3, a4
,
a1, a2, a3, a4
,
a1, a2, a3, a4
)
println(list.map['''«ref.s»->«i»'''])
}
}
class Pass {
@Accessors
String s
new (String s) {
this.s=s
}
}
和p1？ p3在哪里出错？
注意：这只是一个示例代码段。在我的“真实”代码中，我无法覆盖Comparator或equals
答案 0 :(得分：4)
If you break contract of the Comparator, don't expect TreeSet to behave correctly.
Namely, here you failed to make the comparator symmetric:
Pass p1 = new Pass("p1");
Pass p2 = new Pass("p2");
compare(p1, p2); // returns 1
compare(p2, p1); // also returns 1 - not good
If you don't care (or cannot even define) about total ordering, you'd do much better with a properly defined hashCode() and equals() and a HashSet.
If you cannot override those (as you write), create a Key class that will hold the relevant attributes, define Key's hashCode() and equals() and use a HashMap<Key, Pass>. Alternatively, Key might be just a wrapper that holds a reference to Pass and derives it's hash code and equals info from Pass' fields.
答案 1 :(得分：1)
Your problem is that you violate the contract of Comparator which says that if compare(a, b) returns a positive integer, compare(b, a) has to return a negative integer. Besides other requirements like transitive comparings. In your implementation you either return 0 or 1 which does not follow this contract, so the TreeSet cannot work properly of course.