java - 根据条件(Java/Guava)从列表中提取第一个匹配项？

Question

我有以下对象的列表:

class ResourcePermissionDTO {
  PermissionType permissionType;
  ... 
}

其中 PermissionType 是以下枚举:

public enum PermissionType {
   DENY, READ_ONLY, READ_WRITE;
}

所以，列表看起来像:

List<ResourcePermissionDTO> myResourcePermissions = ...

我想要的是返回在 myResourcePermissions 中找到的第一个具有最严格权限的 ResourcePermissionDTO。目前我有以下内容，但它有点困惑，我想可能有更好的方法使用 Google Guava/function idioms 来做到这一点？

private ResourcePermissionDTO returnTheFirstMostRestrictivePermissionFoundIn(final List<ResourcePermissionDTO> resourcePermissionDTOs) {
    if (resourcePermissionDTOs.isEmpty()) {
      return null;
    }

    final List<ResourcePermissionDTO> resourcePermissionDTOsWithReadWritePermissionOfDeny = Lists.newArrayList();
    final List<ResourcePermissionDTO> resourcePermissionDTOsWithReadWritePermissionOfReadOnly = Lists.newArrayList();
    final List<ResourcePermissionDTO> resourcePermissionDTOsWithReadWritePermissionOfReadWrite = Lists.newArrayList();

    for (final ResourcePermissionDTO resourcePermissionDTO : resourcePermissionDTOs) {
      switch (resourcePermissionDTO.getPermissionType()) {
      case DENY:
        resourcePermissionDTOsWithReadWritePermissionOfDeny.add(resourcePermissionDTO);
        break;
      case READ_ONLY:
        resourcePermissionDTOsWithReadWritePermissionOfReadOnly.add(resourcePermissionDTO);
        break;
      case READ_WRITE:
        resourcePermissionDTOsWithReadWritePermissionOfReadWrite.add(resourcePermissionDTO);
        break;
      default:
        break;
      }
    }

    if (!resourcePermissionDTOsWithReadWritePermissionOfDeny.isEmpty()) {
      return resourcePermissionDTOsWithReadWritePermissionOfDeny.get(0);
    } else if (!resourcePermissionDTOsWithReadWritePermissionOfReadOnly.isEmpty()) {
      return resourcePermissionDTOsWithReadWritePermissionOfReadOnly.get(0);
    } else if (!resourcePermissionDTOsWithReadWritePermissionOfReadWrite.isEmpty()) {
      return resourcePermissionDTOsWithReadWritePermissionOfReadWrite.get(0);
    } else {
      return null;
    }
  }

Answer 1

即使采用命令式风格，也可以简化如下:

List<ResourcePermissionDTO> permissions = ...;
ResourcePermissionDTO result = null;

for (ResourcePermissionDTO p: permissions) {
    if (result == null || isStronger(p.getPermissionType(), result.getPermissionType())) {
        result = p;
        if (result.getPermissionType() == PermissionType.DENY) break; // (1)
    }
}

return result;

如果您更喜欢函数式风格，则可以使用 reduce() 重现完全相同的结果(尽管在 (1) 处没有短路优化)。 Guava 不支持 reduce()，因此以下示例使用 Java 8:

return permissions.stream().reduce((result, p) -> {
    return isStronger(p.getPermissionType(), result.getPermissionType()) ? p : result;
}).orElse(null);