如何获得无重复的随机字符串值？

I want to fetch only a single company name and I want to fetch it only once. So if it already was fetched, it should not be fetched again.

我想只获取一个公司名称，我只想获取一次。因此，如果已经获取它，则不应再次获取它。

Here is the code:

这是代码：

private static String[] billercompanies = { 
    "1st",      
    "TELUS Communications",
    "Rogers Cablesystems",
    "Shaw Cable",
    "TELUS Mobility Inc",
    "Nanaimo Regional District of",
    "Credit Union MasterCard",
    }; 



public static String GetBillerCompany(){
    String randomBillerComp = "";
    randomBillerComp = (billercompanies[new Random().nextInt(billercompanies.length)]);
    return randomBillerComp;
}

2 个解决方案

#1

Make a copy in a List and remove the element when it was already fetched.

在List中创建一个副本，并在已经获取该元素时删除该元素。

Arrays.asList(array) is not modifiable but you can wrap it in a full featured List.

Arrays.asList（array）不可修改，但您可以将其包装在功能齐全的List中。

List<String> billercompaniesList = new ArrayList<>(Arrays.asList(billercompanies));

String randomBillerComp = "";

Random random = new Random();
// first retrieval
int index = random.nextInt(billercompaniesList.size()); 
randomBillerComp = billercompaniesList.get(index);
billercompaniesList.remove(index);  


// second retrieval
index = random.nextInt(billercompaniesList.size()); 
randomBillerComp = billercompaniesList.get(index);
billercompaniesList.remove(index);  

// and so for    

#2

Just shuffle the array you want using Collections

只需使用Collections将您想要的数组随机播放

Collections.shuffle(List);

So simply create a list from your array

所以只需从数组中创建一个列表

List<E> list = Arrays.asList(array);

Then shuffle it using the method above

然后使用上面的方法将其洗牌

Collections.shuffle(list);

Your list can be read from left to right as it was random. So simply save the index

您的列表可以从左到右阅读，因为它是随机的。所以只需保存索引即可

int currentIndex = 0;

public E getRandom(){
    //If at the end, start over
    if(++currentIndex == list.size()) { 
         currentIndex = 0;
         shuffle(list);
    }

    return list.get(currentIndex);
}

Each time you want to forget the duplicate list you already used, simply shuffle the array again

每次要忘记已经使用过的重复列表时，只需再次对阵列进行洗牌即可

Collections.shuffle(list);

---

Without index

You could simply remove the first value each time, once the list is empty, recreate it with the original array. As Ole V.V. pointer out, a List generated by Arrays.asList(E[]) doesn't support the remove methods so it is necessary to generate a new instance from it.

您可以每次删除第一个值，一旦列表为空，使用原始数组重新创建它。作为Ole V.V.指针输出，由Arrays.asList（E []）生成的List不支持remove方法，因此需要从中生成一个新实例。

Here is a quick and simple class using this solution :

以下是使用此解决方案的快速简单的类：

public class RandomList<E>{
    E[] array;
    List<E> list;

    public RandomList(E[] array){
        this.array = array;
        buildList(array);
    }

    public E getRandom(){
        if(list.isEmpty()) buildList(array);
        return list.remove(0);
    }

    public void buildList(E[] array){
        list = new ArrayList<E>(Arrays.asList(array));
        Collections.shuffle(list);
    }
}

And the test was done with this small code :

测试是用这个小代码完成的：

Integer[] array = {1,2,3,4,5};
RandomList<Integer> rl = new RandomList(array);
int i = 0;
while(i++ < 10)
        System.out.println(rl.getRandom());