相关疑难解决方法(0)

我只是写了一个小方法来计算手机短信的页数.我没有选择使用Math.ceil,老实说它似乎非常难看.

这是我的代码:

public class Main {

/**
 * @param args the command line arguments
 */
public static void main(String[] args) {
   String message = "today we stumbled upon a huge performance leak while optimizing a raycasting algorithm. Much to our surprise, the Math.floor() method took almost half of the calculation time: 3 floor operations took the same amount of time as one trilinear interpolation. Since we could not belive that the floor-method could produce such a enourmous overhead, …

我想要一些类似于scala分组函数的东西.基本上,一次挑选2个元素并处理它们.以下是相同的参考:

将列表拆分为具有固定数量元素的多个列表

Lambdas确实提供了诸如groupingBy和partitioningBy之类的东西,但它们似乎都没有像Scala中的分组函数那样做.任何指针将不胜感激.

假设我有以下存储库:

public interface UserRepository extends JpaRepository<User, Long> {

    @Query("select u from User u")
    Stream<User> streamAllPaged(Pageable pageable);
}

我想进行搜索:

public Page<User> findAllUsers(Pageable page) {

    Page<User> page = null;
    try (Stream<User> users = userRepository.streamAllPaged(page)) {
            Set<User> users = users.filter(u -> user.getName().equals("foo"))
                    .collect(Collectors.toSet());
            //create page from set?
    }

}

显然我可以使用子列表并手动插入页面大小等但我想应该有更"标准"的方法来做到这一点？