我正在尝试将一些集合分组为不相交的集合。例如,如果我有这 5 套:
[[1, 3], [2], [1, 5], [6, 8], [1, 7]]
我想要这个结果:
[[2], [6, 8], [1, 3, 5, 7]]
这是代码:
import java.util.*;
public class SetTest {
public static void main(String[] args) {
// ---- Initialisation
Set<Set<Integer>> groups = new LinkedHashSet<Set<Integer>>();
for (int[] set : new int[][] { {1, 3 }, {2 }, {1, 5 }, {6, 8 }, {1, 7} }) {
Set<Integer> group = new TreeSet<Integer>();
for (int i : set) {
group.add(i);
}
groups.add(group);
}
System.out.println(groups);
// ---- Grouping values in disjoint sets
for (Iterator<Set<Integer>> iterator = groups.iterator(); iterator.hasNext();) {
Set<Integer> group = iterator.next();
System.out.println(String.format(" + Checking %20s in \t %s", group, groups));
for (Set<Integer> other : groups) {
if (!group.equals(other) && !Collections.disjoint(group, other)) {
other.addAll(group);
iterator.remove();
System.out.println(String.format(" - Removed %20s -> \t %s", group, groups));
break;
}
}
}
System.out.println(groups);
}
}
我在集合上使用迭代器,我想将 2 个集合组合为一个,删除其中一个。但是,我的Iterator.remove()
方法有问题。
这个程序打印的是:
[[1, 3], [2], [1, 5], [6, 8], [1, 7]]
+ Checking [1, 3] in [[1, 3], [2], [1, 5], [6, 8], [1, 7]]
- Removed [1, 3] -> [[2], [1, 3, 5], [6, 8], [1, 7]]
+ Checking [2] in [[2], [1, 3, 5], [6, 8], [1, 7]]
+ Checking [1, 3, 5] in [[2], [1, 3, 5], [6, 8], [1, 7]]
- Removed [1, 3, 5] -> [[2], [1, 3, 5], [6, 8], [1, 3, 5, 7]]
+ Checking [6, 8] in [[2], [1, 3, 5], [6, 8], [1, 3, 5, 7]]
+ Checking [1, 3, 5, 7] in [[2], [1, 3, 5], [6, 8], [1, 3, 5, 7]]
- Removed [1, 3, 5, 7] -> [[2], [1, 3, 5, 7], [6, 8], [1, 3, 5, 7]]
[[2], [1, 3, 5, 7], [6, 8], [1, 3, 5, 7]]
第一次,删除[1, 3]
按预期工作,但其余时间,它不会删除该项目。我认为这是因为我使用addAll()
,但这是为什么呢?因为我没有在groups
; 我只改变了它里面的一个元素(other
)——而且引用是一样的,对吧?