该解决方案将创建为一个类FunctionalSequence，用于表示由具有整数参数的 lambda 函数定义的惰性无限对象序列。该函数可以是迭代的，也可以不是。对于迭代情况，FunctionalSequence 类将有一个方法 initialize 用于设置起始值。
此类对象的声明如下所示：
FunctionalSequence<BigInteger> fiboSequence = new FunctionalSequence<>();
fiboSequence.
initialize(Stream.of(BigInteger.ONE,BigInteger.ONE)).
setSequenceFunction(
(i) ->
fiboSequence.get(i-2).add(fiboSequence.get(i-1))
);
请注意，就像问题中的递归 lambda 示例一样，我们不能声明对象并在一个运算符中递归地定义它。一个运算符用于声明，另一个用于定义。
FunctionalSequence 类定义：
import java.util.Iterator;
import java.util.LinkedList;
import java.util.stream.Stream;
public class FunctionalSequence<T> implements Iterable<T>{
LinkedList<CountedFlighweight<T>> realList = new LinkedList<>();
StackOverflowingFunction<Integer, T> calculate = null;
public FunctionalSequence<T> initialize(Stream<T> start){
start.forEachOrdered((T value) ->
{
realList.add(new CountedFlighweight<>());
realList.getLast().set(value);
});
return this;
}
public FunctionalSequence<T> setSequenceFunction(StackOverflowingFunction<Integer, T> calculate){
this.calculate = calculate;
return this;
}
@Override
public Iterator<T> iterator() {
return new SequenceIterator();
}
public T get(int currentIndex) throws StackOverflowError{
if(currentIndex < 0) return null;
while (currentIndex >= realList.size()){
realList.add(new CountedFlighweight<T>());
}
try {
return (T) realList.get(currentIndex).get(calculate, currentIndex);
} catch (Exception e) {
return null;
}
}
public class SequenceIterator implements Iterator<T>{
int currentIndex;
@Override
public boolean hasNext() {
return true;
}
@Override
public T next() {
T result = null;
if (currentIndex == realList.size()){
realList.add(new CountedFlighweight<T>());
}
// here the StackOverflowError catching is a pure formality, by next() we would never cause StackOverflow
try {
result = realList.get(currentIndex).get(calculate, currentIndex);
} catch (StackOverflowError e) {
}
currentIndex++;
return result;
}
}
/**
* if known is false, the value of reference is irrelevant
* if known is true, and reference is not null, reference contains the data
* if known is true, and reference is null, that means, that the appropriate data are corrupted in any way
* calculation on corrupted data should result in corrupted data.
* @author Pet
*
* @param <U>
*/
public class CountedFlighweight<U>{
private boolean known = false;
private U reference;
/**
* used for initial values setting
*/
private void set(U value){
reference = value;
known = true;
}
/**
* used for data retrieval or function counting and data saving if necessary
* @param calculate
* @param index
* @return
* @throws Exception
*/
public U get(StackOverflowingFunction<Integer, U> calculate, int index) throws StackOverflowError{
if (! known){
if(calculate == null) {
reference = null;
} else {
try {
reference = calculate.apply(index);
} catch (Exception e) {
reference = null;
}
}
}
known = true;
return reference;
}
}
@FunctionalInterface
public interface StackOverflowingFunction <K, U> {
public U apply(K index) throws StackOverflowError;
}
}
由于递归函数很容易遇到 StackOverflowError，我们应该组织递归，这样在这种情况下，整个递归序列将在没有真正遇到任何更改的情况下回滚并抛出异常。
FunctionalSequence 的使用可能是这样的：
// by iterator:
int index=0;
Iterator<BigInteger> iterator = fiboSequence.iterator();
while(index++<10){
System.out.println(iterator.next());
}
左右：
static private void tryFibo(FunctionalSequence<BigInteger> fiboSequence, int i){
long startTime = System.nanoTime();
long endTime;
try {
fiboSequence.get(i);
endTime = System.nanoTime();
System.out.println("repeated timing for f("+i+")=" + (endTime-startTime)/1000000.+" ns");
} catch (StackOverflowError e) {
endTime = System.nanoTime();
//e.printStackTrace();
System.out.println("failed counting f("+i+"), time=" + (endTime-startTime)/1000000.+" ns");
}
}
最后一个函数可以按如下方式使用：
tryFibo(fiboSequence, 1100);
tryFibo(fiboSequence, 100);
tryFibo(fiboSequence, 100);
tryFibo(fiboSequence, 200);
tryFibo(fiboSequence, 1100);
tryFibo(fiboSequence, 2100);
tryFibo(fiboSequence, 2100);
tryFibo(fiboSequence, 1100);
tryFibo(fiboSequence, 100);
tryFibo(fiboSequence, 100);
tryFibo(fiboSequence, 200);
tryFibo(fiboSequence, 1100);
以下是结果（堆栈限制为 256K 以供测试）：
1
1
2
3
5
8
13
21
34
55
failed counting f(1100), time=3.555689 ns
repeated timing for f(100)=0.213156 ns
repeated timing for f(100)=0.002444 ns
repeated timing for f(200)=0.266933 ns
repeated timing for f(1100)=5.457956 ns
repeated timing for f(2100)=3.016445 ns
repeated timing for f(2100)=0.001467 ns
repeated timing for f(1100)=0.005378 ns
repeated timing for f(100)=0.002934 ns
repeated timing for f(100)=0.002445 ns
repeated timing for f(200)=0.002445 ns
repeated timing for f(1100)=0.003911 ns
看，f(i) 对同一索引的可重复调用几乎不需要任何时间 - 没有进行迭代。由于 StackOverflowError，我们无法立即达到 f(1100)。但是在我们达到一次 f(200) 之后，f(1100) 变得可达。我们成功了！