如何从java中的以下无限循环中退出，但仍然使用唯一值将向量填充到100？

Question

i want to create vector of big length from unique values without repetition but the following code enter in infinite loop .

 public static void main(String[] args) {
        // TODO code application logic here
        Vector v1=new Vector(100);
        Vector v2=new Vector(100);
        Random r=new Random();
        int i=0;
        while(i<100){
            double val=Math.abs(r.nextGaussian());
            double h=(double)((double)Math.round(val*10)/10.0);
            if (!v1.contains(h)){
                v1.add(h);
                i++;
            }
        }
        for(int j=0;j<v1.size();j++)
            v2.set(j, v1.get(j));

    }

}

Answer 1

First, don't use raw types ( Collection<Double> , not just raw Collection ). Second, you want a Set (not a Vector ) and assuming you need to preserve the original order you can use a LinkedHashSet . You can test the Set size until you have one hundred values. And, you don't need any of those casts in assigning h (and you need to remember the count you want - if you use 10 you are limiting your value range to 1 to 10 ). Like,

Set<Double> set = new LinkedHashSet<>();
Random r = new Random();
int count = 100;
while (set.size() < count) {
    double val = Math.abs(r.nextGaussian());
    double h = Math.round(val * count) / 10.0;
    set.add(h);
}
System.out.println(set);

You could also write the while loop like

while (set.size() < count) {
    set.add(Math.round(Math.abs(r.nextGaussian()) * count) / 10.0);
}

Or, if using Java 8+, you could use a DoubleStream and distinct() with limit(int) with a lambda. Like,

Random r = new Random();
int count = 100;
Set<Double> set = DoubleStream.generate(() -> Math.round(
                    Math.abs(r.nextGaussian()) * count) / 10.0).boxed()
        .distinct().limit(count).collect(Collectors.toSet());
System.out.println(set);