Java中带定时器的IndexOutOfBoundsException

Question

我有一个问题，希望需要一个简单明了的解决方案，我似乎想不出来。好的，所以我在玩斐波那契序列：我想提示用户输入一个整数。然后它将运行该序列，将每个元素添加到ArrayList中。然后，我想使用计时器将数组列表中的每个元素显示到控制台--每秒显示序列中的每个元素--我用一个简短的方法来完成。

public class Sequence {

static int seconds = 0;
static Timer timer;
static ArrayList<Integer> fibList = new ArrayList<Integer>();

public static void main(String[] args) {

    boolean noError = true; 
    Scanner sc = new Scanner(System.in);

    //will store sum of evens
    int tmp = 0;

    //timer variables
    int delay = 1000;
    int period = 1000;
    timer = new Timer();

    //enter user input, try catch to eliminate invalid input
    do{
        try{
            System.out.println("Please enter a number: ");
            int input = Integer.parseInt(sc.nextLine());

            //add to arraylist from user input
            for (int i=0; i<=input; i++){
                   fibList.add(fib(i));

                   //sum even numbers
            if(fib(i)%2 == 0){
                tmp += fib(i);
            }

            }

        }catch(Exception e){
            System.out.println("Not a valid Number!");

        }

        noError = false;
    }while(noError);

     System.out.println(fibList);
     System.out.println("Sum of Even Numbers is: "+tmp);


     timer.scheduleAtFixedRate(new TimerTask() {

            public void run() {      

                System.out.print(fibList.get(setInterval())+ ",");
            }
        }, delay, period);

}

    //This is where I count up through my ArrayList and probably cause the Error
//counts up for every element through the array
public static final int setInterval() {

    if (seconds >= fibList.size())
        timer.cancel();
    return seconds++;


}

//does fibonacci sequence
public static int fib(int n) {
    if (n < 2) {
         return n;
      }
      else {
   return fib(n-1)+fib(n-2);
      }

Answer 1

顺便说一句，Iluxa的方法更好。逻辑清晰，避免混淆。

但是，如果您只想修复测试条件，请将if更改为

if (seconds == fibList.size() - 1)