如何让两个随机Feat生成器不输出重复结果 附现有Java实现代码

现有代码的可优化点 #

• 两个存储Feat的数组内容完全一致，没有重复定义的必要
• 数组实际长度为17，你当前使用Math.random() * 16只能生成0-15的随机数，最后一个元素Shield Master永远不会被选中

实现方案1：简单循环去重（改动最小） #

逻辑非常直接，生成Feat2的下标后判断是否和Feat1的下标相同，如果相同就重新生成，直到两个值不同即可，修改后代码如下：

System.out.println("Feat 1:");
// 复用同一个数组即可
String[] feats = {"Actor", "Athlete", "Dragon Hide", "Dual Wielder", "Durable", "Elven Accuracy", "Firearm Specialist", "Healer", "Heavily Armored", "Martial Adept", "Mobile", "Moderately Armored", "Mounted Combatant", "Resilient", "Ritual Caster", "Ambidextrous", "Shield Master"};
// 修正随机数范围，直接用数组长度适配，后续修改数组内容不用改随机数参数
int feat1 = (int) (Math.random() * feats.length);
System.out.println(feats[feat1]);

System.out.println("\nFeat 2:");
int feat2;
// 循环直到生成和feat1不同的下标
do {
    feat2 = (int) (Math.random() * feats.length);
} while (feat2 == feat1);
System.out.println(feats[feat2]);
System.out.println("\n");

实现方案2：洗牌法（支持扩展取N个不重复元素） #

如果后续有取更多不重复Feat的需求，用集合洗牌的方式更简洁，不需要额外写多次判断逻辑：

import java.util.Arrays;
import java.util.Collections;
import java.util.List;

// ... 你的其他业务代码
System.out.println("Feat 1:");
String[] feats = {"Actor", "Athlete", "Dragon Hide", "Dual Wielder", "Durable", "Elven Accuracy", "Firearm Specialist", "Healer", "Heavily Armored", "Martial Adept", "Mobile", "Moderately Armored", "Mounted Combatant", "Resilient", "Ritual Caster", "Ambidextrous", "Shield Master"};
List<String> featList = Arrays.asList(feats);
// 打乱列表整体顺序
Collections.shuffle(featList);
// 直接取前两个元素，天然不重复
System.out.println(featList.get(0));
System.out.println("\nFeat 2:");
System.out.println(featList.get(1));
System.out.println("\n");

两种方案都能稳定实现两个Feat不重复的需求，方案1适合仅取2个元素的轻量化场景，方案2适合后续可能需要扩展取更多不重复元素的场景。

内容的提问来源于stack exchange，提问作者Jihyorno