Question

通过变量步骤递增然后递减两个限制之间的值的最佳方法是什么。例如：以＆＃34; 0＆＃34;开始，停在＆＃34; 10＆＃34;，逐步＆＃34; 1＆＃34;：0,1,2 ... 9,10,9， 8 ... 2,1,0,1,2 ......到目前为止（像yoyo）。

我的解决方案是：

public void something() {
    new Thread(new Runnable() {
        float counter = 0;
        int step = 5;

        @Override
        public void run() {
            while (true) {
                if (counter >= 100)
                    step = -5;
                if (counter <= 0)
                    step = 5;

                counter += step;
                // do something ...

                try {
                    Thread.sleep(50);
                } catch (InterruptedException e) {
                }
            }
        }
    }).start();
}

Answer 1

我建议将行为封装在自己的类中。类似的东西：

public class Yoyo {
  private final int from;
  private final int to;

  private int current;
  private int step;

  public Yoyo(int from, int to, int step) {
    if (step > to - from || to <= from) throw new IllegalArgumentException("invalid arguments");
    this.from = from;
    this.to = to;
    this.current = from - step;
    this.step = step;
  }

  public synchronized int next() {
    if (current + step > to || current + step < from) step = -step;
    return current += step;
  }

  public static void main(String[] args) {
    Yoyo y = new Yoyo(1, 10, 1);
    for (int i = 0; i < 25; i++) System.out.println(y.next());
  }
}

Answer 2

您的答案似乎工作正常，但您也可以使用两个for循环：

for(int i=0;i<=10;i++) {
    counter = i;
    // Sleep here
}
for(int i=9;i>=0;i--) {
    counter = i;
    // Sleep here
}

在两个限制之间反弹值的最佳方法是什么？

2 个答案: