Collections.sort(names, (a, b) -> a.compareTo(b));
时间: 2024-05-21 17:14:00 浏览: 10
您好!感谢您对 ChitGPT 的提问。关于您的问题,这是一个使用 lambda 表达式进行排序的示例代码。其中,sort() 方法对传入的 List 对象 names 进行排序,lambda 表达式 (a, b) -> a.compareTo(b) 则按照字母顺序比较字符串 a 和 b 的大小关系。其中 a 和 b 分别代表两个要比较的字符串。这个表达式的返回值为一个整数,如果 a 大于 b,则返回正整数;如果 a 等于 b,则返回 0;如果 a 小于 b,则返回负整数。这便是 Collections.sort() 方法内部所使用的比较器。希望这个回答能够帮到您。如果您还有其他问题,可以继续提出哦!
相关问题
import java.util.*; public class Task { public static void main(String[] args) { List<String> names = new ArrayList<>(Arrays.asList(args)); Collections.addAll(names, "John", "Kate", "Tom", "Bob", "Alice"); // 排序---------------------------------------------- /* TODO 对names集合进行排序 */ System.out.println(names); // 打乱----------------------------------------------- Random random = new Random(names.size()); /* TODO 使用random打乱names集合 */ System.out.println(names); } }集合工具类:Collections 任务 请使用集合工具类Collections,添加如下元素到集合 "John", "Kate", "Tom", "Bob", "Alice" 对集合进行排序,并输出排序后的结果,排序规则为 1.按照字符串长度从小到大排序 2.如果字符串长度相同,则按照字典序从小到大排序 打乱并输出集合 创建一个随机对象new Random(names.size()) 使用随机对象作为打乱方法的参数,打乱集合 将输出随机打乱后的结果
可以使用Collections.sort()方法对集合进行排序,排序规则可以通过传入一个Comparator对象来实现。以下是代码:
import java.util.*;
public class Task {
public static void main(String[] args) {
List<String> names = new ArrayList<>(Arrays.asList(args));
Collections.addAll(names, "John", "Kate", "Tom", "Bob", "Alice");
// 排序----------------------------------------------
Collections.sort(names, new Comparator<String>() {
@Override
public int compare(String s1, String s2) {
if (s1.length() != s2.length()) {
return s1.length() - s2.length();
} else {
return s1.compareTo(s2);
}
}
});
System.out.println(names);
// 打乱-----------------------------------------------
Random random = new Random(names.size());
Collections.shuffle(names, random);
System.out.println(names);
}
}
输出结果为:
[Bob, Tom, John, Kate, Alice]
[Bob, Alice, Tom, John, Kate]
用JavaSort the Student list by ascending order of student namesThe following is an example illustrating formatting output. You will need it for task 2 and 3. public class Main { public static void main(String[] args) { String string1 = "Hello"; String string2 = "World"; String string3 = "Java"; System.out.println(String.format("%-10s %-10s %-10s", string1, string2, string3)); } }
Sure, here's an example Java code that sorts a list of students by ascending order of their names using the `Collections.sort()` method and formats the output using `String.format()`:
```java
import java.util.ArrayList;
import java.util.Collections;
public class Student implements Comparable<Student> {
private String name;
private int age;
public Student(String name, int age) {
this.name = name;
this.age = age;
}
public String getName() {
return name;
}
public int getAge() {
return age;
}
public int compareTo(Student other) {
return this.getName().compareTo(other.getName());
}
public String toString() {
return String.format("%-10s %-10d", this.getName(), this.getAge());
}
}
public class Main {
public static void main(String[] args) {
ArrayList<Student> students = new ArrayList<>();
students.add(new Student("John", 20));
students.add(new Student("Alice", 19));
students.add(new Student("Bob", 21));
students.add(new Student("Charlie", 18));
System.out.println("Before sorting:");
for (Student student : students) {
System.out.println(student);
}
Collections.sort(students);
System.out.println("After sorting:");
for (Student student : students) {
System.out.println(student);
}
}
}
```
In this example, the `Student` class implements the `Comparable` interface and overrides the `compareTo()` method to compare students based on their names. The `toString()` method formats the output using `String.format()` with the `%10s` and `%10d` format specifiers for the name and age fields, respectively. In the `Main` class, we create a list of students, print the list before sorting, sort the list using `Collections.sort()`, and print the list after sorting. The output will be:
```
Before sorting:
John 20
Alice 19
Bob 21
Charlie 18
After sorting:
Alice 19
Bob 21
Charlie 18
John 20
```
I hope this helps! Let me know if you have any other questions.
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