从List< Object>获取一些值爪哇

问题描述

限时删除！！

我认为它很简单，但是我需要一种简单的方法来解决它.

I think it quite simple but I need one ways to solve it easy.

我的问题是:我有列表对象Student包含100个对象

My problem is: I have list object Student include 100 objects

public class Student {

private String id;
private String name;
private int age;


public String getId() {
    return id;
}
public void setId(String id) {
    this.id = id;
}

public String getName() {
    return name;
}
public void setName(String name) {
    this.name = name;
}

public int getAge() {
    return age;
}
public void setAge(int age) {
    this.age = age;
}

现在，我只想让一些年龄为"ABC"，"BCD"，"DEF"的学生

Now, I only want to get age of some student has name "ABC", "BCD", "DEF"

所以我会这样做:

Map<String, String> data = new HashMap<>();
    List<String> listSpecification = new ArrayList<>(Arrays.asList("ABC", "BCD", "DEF"));
    for (Student stu : list<Student>) {
        for (int i = 0; i < listSpecification.size(); i++) {
            if (stu.getName().equalsIgnoreCase(listSpecification.get(i))) {
                data.put(pro.getName(), pro.getValue());
            }
        }
    }

但是，如果我这样做，我认为它对性能不好.那么任何人都可以帮助我另一个案例吗?

But if I do it, I think it not good about performance. So anyone can help me another case ?

推荐答案

您的算法的复杂度为O(m * n)，其中m是学生数，n是名称数.使用流API或Collection.contains的建议不会改变这一点.

Your algorithm has complexity O(m*n) where m is the number of students and n is the number of names. The suggestions to use streaming APIs or Collection.contains won't change that.

您可以使用从学生姓名到Student的映射，将其转换为O(m + n)算法:

You can convert this to an O(m + n) algorithm by using a map from student name to Student:

Map<String, Student> studentsByName = new HashMap<>();
for (Student s : studentList) {
    studentsByName.put(s.getName().toLowerCase(), s);
}

这是O(m)运算.然后，您可以遍历规范列表中的所有名称，将每个名称转换为小写，然后直接拔出每个学生.这是一个O(n)操作.还有一个更高的效率，那就是，如果您需要为同一个学生列表处理多个规范列表，则只需要执行一次第一步即可.

That's an O(m) operation. Then you can iterate through all the names in the specification list, convert each to lower case, and pull out each student directly. This is an O(n) operation. There's a further efficiency in that you need do the first step only once if you need to process multiple specification lists for the same student list.

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