Question

我有100条记录[1 - ＆gt; 100]，我想在此获得随机50条记录，如何在java中做？感谢。

Answer 1

Set<T> set;

List<T> list = new ArrayList<T>(set);
Collections.shuffle(list);
List<T> random50 = list.subList(0, 50);

Answer 2

您可以获得50个随机值。

Random rand = new Random();

List<Integer> ints = new ArrayList<Integer>();
for(int i = 0; i < 50; i++)
    ints.add(rand.nextInt(100)+1);

您可以使用随机播放以随机顺序获取50个唯一值。

List<Integer> ints = new ArrayList<Integer>();
for(int i = 1; i <= 100; i++)
    ints.add(i);
Collections.shuffle(ints);
ints = ints.subList(0, 50);

Answer 3

生成四位长唯一代码的唯一可靠方法。

我发现首先要声明4个整数变量，为它们分配1到9之间的随机数。

然后我将这些整数转换为字符串，将它们连接在一起，使它们形成一个四位长的字符串，然后将结果字符串转换为整数。

得到的四位随机整数存储在一个数组中。

＆＃34;请注意!!我是java＆＃34;

的新手

import javax.swing.JOptionPane;
public class Rund4gen {
/**
 * @param args the command line arguments
 */
public static void main(String[] args) {
   // I begin by creating an array to store all my numbers
   String userInput = JOptionPane.showInputDialog("How many 4 digit long numbers would you like to generate?");
   int input = Integer.parseInt(userInput);
   // Now lets convert user input to a string
   int[] passCode = new int[input];
   // We need to loop as many time as the user specified
   for(int i = 0; i < input; i++){
       // Here I declare my integer variables
       int one, two, three, four;
       // For each of the integer variable I assign a rundom number
       one = (int)Math.floor((Math.random()*9)+1);
       two = (int)Math.floor((Math.random()*9)+1);
       three = (int)Math.floor((Math.random()*9)+1);
       four = (int)Math.floor((Math.random()*9)+1);
       // I need to convert my digits into a string in order to join them
       String n1 = String.valueOf(one);
       String n2 = String.valueOf(two);
       String n3 = String.valueOf(three);
       String n4 = String.valueOf(four);
       // Once conversion is complete then I join them as follows
       String nV = n1+n2+n3+n4;
       // Once joined, I then need to convert the joined result into an integer
       int nF = Integer.parseInt(nV);
       // I then store the result in an array as follows
       passCode[i] = nF;
   }
   // Now I need to print each value in the array
   for(int c = 0; c < passCode.length; c++){
       System.out.print(passCode[c]+"\n");
   }
   // Finally I thank the user for participating or not
   //JOptionPane.showMessageDialog(null,"Thank you for participating");
   System.exit(0);
}

}

Answer 4

fun generateCode(@IntRange(from = 1, to = 9) digits: Int, uniqueDigits: Boolean = false): String {
    var number = 0
    val numbersSet = mutableSetOf<Int>()
    for (i in 1..digits) {
        var x: Int
        do {
            x = Random.nextInt(9)
        } while (uniqueDigits && numbersSet.contains(x))

        numbersSet.add(x)

        number += (10.0.pow((digits - i).toDouble()) * x).toInt()
    }
    return String.format("%0${digits}d", number)
}

这是允许生成长度为 1-9 的数字代码的 Kotlin 代码。使用 uniqueDigits = true 使数字不重复。

如何获取集合中的随机范围数

4 个答案: