在Java中，我如何使列表中的数组被洗牌以相互匹配？

Question

我是一个新的编码，我想知道我如何编码这个程序，使数组是匹配的。（苹果和1号匹配，香蕉和2号匹配，芒果和3号匹配）我使用了assert fruits.size()==numbers.size()；，但它仍然不起作用，有没有办法可以修复它？

package example;

import java.util.Arrays;
import java.util.Collections;
import java.util.List;
import javax.swing.JOptionPane;

public class Example {
    
    public static void main(String[] args) {
final StringBuilder generator = new StringBuilder();
final List<String> fruits = Arrays.asList("apples", "bananas", "mangos");
final List<Integer> numbers = Arrays.asList(1, 2, 3);

Collections.shuffle(fruits);
Collections.shuffle(numbers);

assert fruits.size() == numbers.size();

for (int i = 0; i < fruits.size(); i++) {
int a = i+1;
    final String name = JOptionPane.showInputDialog(null, "Enter person " + a + "'s name");

    generator.append(name)
        .append(" likes ")
        .append(fruits.get(i))
        .append(" and the number ")
        .append(numbers.get(i))
        .append("\n");
}

JOptionPane.showMessageDialog(null, generator);
    }
}

我希望输出为：

Alex喜欢芒果和3号

约翰喜欢苹果和数字1

简喜欢苹果和2号

Answer 1

import java.util.Arrays;
import java.util.Collections;
import java.util.List;
import java.util.AbstractMap;
import javax.swing.JOptionPane;

public class Example {

    public static void main(String[] args) {
        final StringBuilder generator = new StringBuilder();

        final List<AbstractMap.SimpleEntry<String, Integer>> fruits = Arrays.asList(
                new AbstractMap.SimpleEntry("apples", 1),
                new AbstractMap.SimpleEntry("bananas", 2),
                new AbstractMap.SimpleEntry("mangos", 3)
        );

        Collections.shuffle(fruits);

        for (int i = 0; i < fruits.size(); i++) {
            int a = i + 1;
            final String name = JOptionPane.showInputDialog(null, "Enter person " + a + "'s name");

            generator.append(name)
                    .append(" likes ")
                    .append(fruits.get(i).getKey())
                    .append(" and the number ")
                    .append(fruits.get(i).getValue())
                    .append("\n");
        }

        JOptionPane.showMessageDialog(null, generator);
    }
}

输出示例：