我的班级有问题,我无法解决。我想为我的Interval1
类实现一个迭代器。这是我的代码:
public class Interval1 {
private double first;
private double last;
private double step;
private IntervalIterator e;
public Interval1(double first, double last, double step, IntervalIterator e) {
//chequear los datos
this.first = first;
this.last = last;
this.step = step;
this.e = new IntervalIterator();
}
public double at(int index) {
checkIndex(index);
return first + index*step;
}
private void checkIndex(int index) {
if(index < 0 || index >= size())
throw new IndexOutOfBoundsException("Invalid Index: " + index);
}
public int size() {
return (int) ((last - first) / step);
}
public IntervalIterator e() {
for (Double i : this)
System.out.println(i);
return e;
}
}
这是我正在使用的课程IntervalIterator
,这会在size()
和at
中出现错误:
public class IntervalIterator implements Iterator<Double> {
private int index = 0;
private Double at;
public boolean hasNext() {
return index < size();
}
public Double next() {
if (!hasNext()) {
throw new NoSuchElementException();
}
return at(index++);
}
public void remove() {
if (Interval1 <= 0) {
return throw new UnsupportedOperationException("remove");
} else {
if (Interval1 >= 0) {
//chekear el array e eliminarlo
}
}
}
}
答案 0 :(得分:0)
我使用java8流来返回Iterator<Double>
:
public class Interval
implements Iterable<Double> {
private final double first;
private final double last;
private final double step;
private final long size;
public Interval(double first, double last, double step) {
this.first = first;
this.last = last;
this.step = step;
this.size = (long) ((last - first) / step) + 1;
}
@Override
public Iterator<Double> iterator() {
return DoubleStream.iterate(this.first, n -> n + this.step)
.limit(this.size)
.iterator();
}
// TODO getters
}
我已将final
修饰符添加到字段中,并使Interval
类实现Iterable<Double>
,以便可以迭代它。我还修复了size
的计算(需要加1)。
我使用了DoubleStream.iterate()
method,它接收种子和接收流的当前元素作为输入的函数,并返回以下元素(为此我添加了{{ 1}}到当前元素)。我还必须使用DoubleStream.limit()
method,因为step
生成的流是无限的。
用法:
DoubleStream.iterate()
以上代码生成以下输出:
Interval interval = new Interval(3.0, 9.0, 1.5);
for (double n : interval) {
System.out.println(n);
}