我正在編寫一個計算斐波納契數字的代碼。通過這段代碼,我可以生成前n個斐波那契數列。 Stream.generate(new Supplier<Long>() {
private long n1 = 1;
private long n2 = 2;
@Override
public Long get() {
long fibonacci = n1;
考慮到我有下面的類: public class Problem2 extends Problem<Integer> {
@Override
public void run() {
result = toList(new FibSupplier(i -> (i <= 4_000_000)))
.stream()
.filter(i ->