在不删除重复项的情况下对树集进行排序

Question

myData{
itemId,
location,
ExpDtTm,
userId
}

我试图像这样对 TreeSet 进行排序：

 TreeSet<myData> sorted = new TreeSet<>(Comparator.comparing(myData -> myData.ExpDtTm));

但是如果任何日期相同，它们将不会被添加到树集中，这就会出现问题，这被证明是一个问题。如果有人知道该怎么做，我们将不胜感激。

我也试过了

TreeSet<myData> sorted = new TreeSet<>(Comparator.comparing(myData -> myData.ExpDtTm).thencomparing(myData -> myData.itemId));

但是除了两个 lambda 中的 myData 之外，它都停止了。

Answer 1

A TreeSet 是 Set, which explicitly forbids duplicate entries by design. If you want a sorted collection that includes duplicates, use an implementation of List (e.g. ArrayList), and sort it at the end by calling Collections.sort(myList, Comparator.comparing(myData -> myData.ExpDtTm)); 的实现。

Answer 2

发生这种情况的原因是因为比较器不仅用于确定顺序，还用于确定两个对象是否相等（而不是 hashCode，HashSet 将使用):

a TreeSet instance performs all element comparisons using its compareTo (or compare) method, so two elements that are deemed equal by this method are, from the standpoint of the set, equal.

如果您有两个对象，实际上并不相同，只是碰巧具有相同的日期，您可以使用对象的 identity hash code（实际上是内存地址）作为次要排序条件:

TreeSet<myData> sorted = 
    new TreeSet<>(Comparator.comparing(myData -> myData.ExpDtTm)
                            .thenComparingInt(System::identityHashCode));

Answer 3

这就是我最后做的事情：

final SortedSet<myData> sorted = new TreeSet<>(new Comparator<myData>() {
            @Override
            public int compare(myData o1, myData o2) {
                 int comparedValue = o1.expDtTm.compareTo(o2.expDtTm);
                 if(comparedValue == 0){
                       comparedValue = 1;
                 }
            }
});

在该 compareValue if 语句中，您可以添加任何其他您可能希望用于排序的变量。

Answer 4

Employee.Java

private String name;

public Employee(String name) {
    this.name = name;
}
public void setName(String name) {
    this.name = name;
}

public String getName() {
    return this.name;
}

public String toString() {
    return "Employee[ id=" + name + "]";
}

Driver Class

TreeSetTest.Java

    TreeSet<Employee> treeSet = new TreeSet<>(new Comparator<Employee>() {
        @Override
        public int compare(Employee o1, Employee o2) {
            int index = o1.getName().compareTo(o2.getName());
            return index==0?1:index;
        }
    });
    treeSet.add(new Employee("Harvansh"));
    treeSet.add(new Employee("Harvansh1"));
    treeSet.add(new Employee("Harvansh"));
    treeSet.add(new Employee("Harvansh2"));
    treeSet.add(new Employee("Harvansh"));
    treeSet.add(new Employee("Harvansh3"));
    treeSet.add(new Employee("Harvansh"));
    treeSet.forEach(System.out::println);

输出

员工[id=Harvansh]

员工[id=Harvansh1]

员工[id=Harvansh2]

员工[id=Harvansh3]

Here Is The Complete Solution With Output

在不删除重复项的情况下对树集进行排序

Sorting tree sets without removing duplicates

java

lambda

comparator

treeset

sortedset