java - 当用户在Java中输入数组大小时，如何生成从0到100的随机数而不重复？

Question

Write a program to generate any number of random integers in 0 to 100range. Your program should get the size as a parameter and return thenumbers as an array.

这是我得到的问题，下面是我试过的代码，但我得到了这段代码的重复。如何生成不重复的随机数？

    import java.util.Arrays;
    import java.util.Random;
    import java.util.Scanner;
    
    public class Main {
    
        public static void main(String[] args) {
            Scanner sc = new Scanner(System.in);
            
            System.out.println("size:\t");
            int n = sc.nextInt();
    
            int[] arr = new int[n];
    
            Random rand = new Random(); //instance of a random class.
            int upperbound = 101; //generate random values from 0-100
    
            for(int i=0;i<n;i++) {
                arr[i] = rand.nextInt(upperbound);
            }
            System.out.println("Random Numbers: "+ Arrays.toString(arr));
            
        }
    }
     

Answer 1

您可以创建一个包含从 0 到 100 的所有值的列表，然后将其打乱并提取该列表的前 n 个元素，如下所示:

public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        System.out.println("size:\t");
        int n = sc.nextInt();

        int bound = 100;
        
        List<Integer> items = IntStream.range(0, bound + 1)
                .boxed()
                .collect(Collectors.toList());

        Collections.shuffle(items);

        int[] arr = new int[Math.min(n, bound + 1)];
        for(int i = 0; i < arr.length; i++){
            arr[i] = items.get(i);
        }

        System.out.println("Random Numbers: "+ Arrays.toString(arr));
    }