如何在不使用arraylist的情况下仅拾取一次随机字符串

Question

有没有办法从数组中选择一个随机字符串，而不再选择该字符串。所有字符串只拾取一次后，程序结束。

这是我当前的系统，publicstaticvoidmain（String[]argv）{

    //String [] names = new String[5];
    String [] names = {"Snoopy", "Donald Duck", "Mickey Mouse", "Little Timmy", "Spiderman"};
    //put a list of names in an array
    //randomly pick a name out of the array
    //how to: randomly pick a number as the index
    //range: 0 to the number of names - 1


    int n = -1;
    int previous = n;
    int storing1 = -1;
    int storing2 = -1;
    int storing3 = -1;
    int storing4 = -1;

    System.out.println(previous);
    while (true) {

        System.out.println("Hit \"enter\" to spin");
        Scanner sc = new Scanner(System.in);
        String s = sc.nextLine();
        //if (s.equalsIgnoreCase("s")) {

        storing4 = storing3;
        storing3 = storing2;
        storing2 = storing1;
        storing1 = previous;
        previous = n;
        n = (int) (Math.random() * names.length);

        if (n == previous || n == storing1 || n == storing2 || n == storing3 || n == storing4) {
            System.out.println("This person has already been picked");

            break;

        }

        System.out.print("n = " + n + " ");
        String picked = names[n];
        System.out.println("Congratulations, " + picked + " was picked!");

        }
        System.out.println(previous);
        System.out.println(storing1);
        System.out.println(storing2);
        System.out.println(storing3);
        System.out.println(storing4); //for my own use, to tell if the storing system works properly



        }

我的问题是，控制台将返回“此人已被选中”等等。并不是说节目结束系统不起作用，因为Rest会过早地起作用。

[1]: https://i.stack.imgur.com/54Eef.png [如您所见，程序结束，尽管没有选择所有五个名称，因为它之前选择了相同的名称][1]如果这看起来很简单，我几个月前才开始使用java。我认为我不允许使用ArrayList，这是迄今为止我找到的唯一解决方案。非常感谢。

Answer 1

只需随机化列表的顺序，然后逐个打印出字符串。

Answer 2

这个问题可以随机解决。

您可以随机打乱这个数组{名称}，然后选择名称[索引]（索引从0开始，以names.length-1结束）。

如何洗牌此数组：

Random r = new Random();
for (int i = 0; i < names.length; i++){
    int nextSwapIndex = r.nextInt(names.length);
    String temp = names[nextSwapIndex];
    names[nextSwapIndex] = names[i];
    names[i] = temp;
}

Answer 3

如果您不能使用带有shuffle的列表来存储随机名称，那么您可能可以使用索引列表并将其替换为shuffle：

String [] names = {"Snoopy", "Donald Duck", "Mickey Mouse", "Little Timmy", "Spiderman"};
List<Integer> indices = IntStream.of(0,1,2,3,4).boxed().collect(Collectors.toList());
Collections.shuffle(indices);

// now iterate the indices and consume random names, one time only
for (int index : indices) {
    System.out.println("Random name: " + names[index]);
}