数独提示类调用：替代switch语句的更优实现方案咨询

原代码示例 #

switch (num) {
      case NAKED_SINGLE:
        return NakedSingle.getHint(numberMap, penBoard, pencilBoard, solvedBoard, hints);
      case NAKED_PAIR:
        return NakedPair.getHint(numberMap, penBoard, pencilBoard, solvedBoard, hints);
      case NAKED_TRIPLE:
        return NakedTriple.getHint(numberMap, penBoard, pencilBoard, solvedBoard, hints);
      case NAKED_QUAD:
        return NakedQuad.getHint(numberMap, penBoard, pencilBoard, solvedBoard, hints);
      case HIDDEN_SINGLE:
        return HiddenSingle.getHint(numberMap, penBoard, pencilBoard, solvedBoard, hints);
      case HIDDEN_PAIR:
        return HiddenPair.getHint(numberMap, penBoard, pencilBoard, solvedBoard, hints);
      case HIDDEN_TRIPLE:
        return HiddenTriple.getHint(numberMap, penBoard, pencilBoard, solvedBoard, hints);
      case HIDDEN_QUAD:
        return HiddenQuad.getHint(numberMap, penBoard, pencilBoard, solvedBoard, hints);

// more cases
}

问题说明 #

上述代码中，num为int类型，所有数独提示类都包含签名一致的getHint方法（返回String类型）。当前实现的痛点是：新增提示类型时，switch语句会持续膨胀，维护成本逐步升高。

替代方案 #

方案1：枚举+函数式接口

首先定义匹配getHint方法签名的函数式接口：

@FunctionalInterface
interface HintProvider {
    String getHint(Map<Integer, List<Point>> numberMap, int[][] penBoard, int[][][] pencilBoard, int[][] solvedBoard, List<String> hints);
}

接着创建枚举，将int常量与对应类的getHint方法绑定：

enum HintType {
    NAKED_SINGLE(NAKED_SINGLE_VALUE, NakedSingle::getHint),
    NAKED_PAIR(NAKED_PAIR_VALUE, NakedPair::getHint),
    NAKED_TRIPLE(NAKED_TRIPLE_VALUE, NakedTriple::getHint),
    NAKED_QUAD(NAKED_QUAD_VALUE, NakedQuad::getHint),
    HIDDEN_SINGLE(HIDDEN_SINGLE_VALUE, HiddenSingle::getHint),
    HIDDEN_PAIR(HIDDEN_PAIR_VALUE, HiddenPair::getHint),
    HIDDEN_TRIPLE(HIDDEN_TRIPLE_VALUE, HiddenTriple::getHint),
    HIDDEN_QUAD(HIDDEN_QUAD_VALUE, HiddenQuad::getHint);

    private final int code;
    private final HintProvider provider;

    HintType(int code, HintProvider provider) {
        this.code = code;
        this.provider = provider;
    }

    public static HintProvider getProvider(int num) {
        for (HintType type : HintType.values()) {
            if (type.code == num) {
                return type.provider;
            }
        }
        throw new IllegalArgumentException("Unknown hint type: " + num);
    }
}

调用时只需一行代码，新增提示类型仅需在枚举中添加项：

return HintType.getProvider(num).getHint(numberMap, penBoard, pencilBoard, solvedBoard, hints);

方案2：Map映射初始化

提前用HashMap建立int值与方法引用的映射：

private static final Map<Integer, HintProvider> HINT_MAP = new HashMap<>();

static {
    HINT_MAP.put(NAKED_SINGLE, NakedSingle::getHint);
    HINT_MAP.put(NAKED_PAIR, NakedPair::getHint);
    HINT_MAP.put(NAKED_TRIPLE, NakedTriple::getHint);
    HINT_MAP.put(NAKED_QUAD, NakedQuad::getHint);
    HINT_MAP.put(HIDDEN_SINGLE, HiddenSingle::getHint);
    HINT_MAP.put(HIDDEN_PAIR, HiddenPair::getHint);
    HINT_MAP.put(HIDDEN_TRIPLE, HiddenTriple::getHint);
    HINT_MAP.put(HIDDEN_QUAD, HiddenQuad::getHint);
    // 新增类型直接添加put语句即可
}

// 调用逻辑
return HINT_MAP.get(num).getHint(numberMap, penBoard, pencilBoard, solvedBoard, hints);

这种方式比枚举更灵活，支持运行时动态添加映射，但固定类型场景下枚举的类型安全性更强。

方案3：反射（不推荐）

通过反射根据num匹配类名调用方法，但会带来性能损耗和可读性下降，仅适合特殊场景，一般不建议使用。

内容的提问来源于stack exchange，提问作者Anu