在这个项目中（使用BlueJ，因为我是初学者）我希望将Climbers添加到具有名称，性别和年龄的ArrayList中。登山者可以添加他们爬过的山峰。以山名和身高。
我需要在Climber类中添加一个方法来返回特定登山者爬过的最高峰。
要定义方法，我是否使用
定义它public ArrayList<Mountain> getHighestMountain(Mountain mountainHeight)
我不确定如何检查山地类中的对象进行比较。仅供参考我尚未学习比较关键词，所以我希望不要使用它。
俱乐部课程：
import java.util.ArrayList;
import java.util.Scanner;
/**
* Write a description of class Club here.
*
* @author (your name)
* @version (a version number or a date)
*/
public class Club
{
// An ArrayList for storing climber details.
private ArrayList<Climber> climbers;
/**
* Constructor for objects of class Club
*/
public Club()
{
// Initialise instance variables.
climbers = new ArrayList<Climber>();
}
public void addClimber(Climber newName)
{
climbers.add(newName);
}
public Climber getClimber(String name)
{
Climber foundClimber = null;
int index = 0;
boolean searching = true;
while(searching && index < climbers.size()) {
Climber climber = climbers.get(index);
if(climber.getName().equals(name)) {
searching = false;
foundClimber = climber;
}
else {
System.out.println(name + " not found");
index++;
}
}
return foundClimber;
}
public void displayClimberList()
{
for (int item = 0; item<climbers.size();
item++) {
Climber climber = climbers.get(item);
System.out.println(climber.getName() + (" ") + climber.getAge() + (" ")
+ climber.getGender());
}
}
}
登山者课程：
import java.util.ArrayList;
/**
* Write a description of class Climber here.
*
* @author (your name)
* @version (a version number or a date)
*/
public class Climber
{
// Instance variables.
// The climber name.
private String name;
// The climber age
private int age;
// The climber gender.
private String gender;
private ArrayList<Mountain> mountains;
/**
* Constructor for objects of class Climber
*/
public Climber (String newName, int newAge, String newGender)
{
// Initialise instance variables.
name = newName;
age = newAge;
gender = newGender;
mountains = new ArrayList<Mountain>();
}
/**
* Accessor method for climber's name.
*/
public String getName()
{
return name;
}
/**
* Set the climber's name.
*/
public void setName(String newName)
{
name = newName;
}
/**
* Accessor method for climber's age.
*/
public int getAge()
{
return age;
}
/**
* Set the climber's age.
*/
public void setAge(int newAge)
{
age = newAge;
}
/**
* Set the climer's gender.
*/
public String getGender()
{
return gender;
}
/**
* Accessor method for climber's gender.
*/
public void getGender(String newGender)
{
gender = newGender;
}
public Mountain addMountain(Mountain mountain)
{
return mountain;
}
public ArrayList<Mountain> getHighestMountain(Mountain mountainHeight)
{
double maxHeight = 1;
int index = 1;
for(int i = 0; i < mountainHeight.length; i++) {
if(mountainHeight[i].getHeight()>maxHeight) {
index = i;
}
}
}
}
山地类：
/**
* Write a description of class Mountain here.
*
* @author (your name)
* @version (a version number or a date)
*/
public class Mountain
{
// Instance variables.
private double height;
private String name;
/**
* Constructor for objects of class Mountain
*/
public Mountain(String mountainName, double mountainHeight)
{
// Initialise instance variables
name = mountainName;
height = mountainHeight;
}
/**
* Accessor method for mountain name.
*/
public String getName()
{
return name;
}
/**
* Set the mountain name.
*/
public void setName(String mountainName)
{
name = mountainName;
}
/**
* Accessor method for mountain height.
*/
public double getHeight()
{
// put your code here
return height;
}
/**
* Set the mountain height.
*/
public void setHeight(double newHeight)
{
height = mountainHeight;
}
}
我试图返回此值，但您可以看到它不完整。
提前致谢！
答案 0 :(得分：0)
<强>评论
删除你的评论，他们没有添加任何价值。不要在代码中放置或保留默认注释，这会降低可读性。还应该避免澄清，尝试创建一个具有有意义名称的单独变量或方法。例如：
// check if the user is on the last page
if (!((count - i) < pageSize)) {
addNextButton();
}
可能成为：
boolean userIsOnTheLastPage = (count - i) < pageSize;
if (!userIsOnTheLastPage) {
addNextButton();
}
<强>接口
Java集合（如ArrayList）实现接口（例如List）。通常，您需要声明并传递接口，而不是您使用的实现。代码的其他部分应该只关心使用List功能，而不是如何实现它们。
所以替换;
private ArrayList<Climber> climbers;
带
private List<Climber> climbers;
也可以在方法中使用接口作为参数或返回类型。这种方法的结果是，如果您希望稍后使用不同的List实现，则只需在代码中的一个位置更改它。这是一种在代码中解耦的形式，这是一件好事。 （＆＃39;低耦合，高凝聚力＆＃39;）
<强> getClimber
您在getClimber中的搜索实现有点复杂。我这样做：
for (Climber climber : climbers) {
if(climber.getName().equals(name)) {
return climber;
}
}
System.out.println(name + " not found");
我知道你可能会试图避免从迭代中返回，但在我看来，代码的可读性会大大降低。 顺便说一句，你的代码会被打印出来，而不是找不到＆＃39;对于每个不匹配的登山者。我认为这不是你想要的，所以如果找不到匹配的登山者，我的代码只打印一次。
<强> displayClimberList
public void displayClimbers() {
for (Climber climber : climbers) {
System.out.println(climber);
}
}
System.out.println在其参数上调用toString()，因此我们可以将显示字符串格式化为Climber类：
public class Climber {
private String name;
private int age;
private String gender;
// ...
public String toString() {
return String.format("%s %s %s", name, age, gender);
}
<强>性别
我会将性别字段从String更改为枚举：
public class Climber {
private Gender gender;
枚举可能如下所示：
public enum Gender {
MALE, FEMALE;
}
（这是一个单独的顶级类，所以在一个名为Gender.java的文件中）。另请参阅the Oracle tutorial about enums。
<强> addMountain
修复addMountain方法，使其实际添加一个山：
public void addMountain(Mountain mountain) {
mountains.add(mountain);
}
<强> getHighestMountain
getHighestMountain的逻辑是有缺陷的（这是你的问题），试试这个：
public List<Mountain> getHighestMountain() {
Mountain highest = null;
for(Mountain mountain : mountains) {
if(highest == null || mountain.getHeight() > highest.getHeight()) {
highest = mountain;
}
}
return highest;
}
答案 1 :(得分：0)
变化：
//This does absolutely nothing. It takes a parameter and returns it.
public Mountain addMountain(Mountain mountain) {
return mountain;
}
public ArrayList<Mountain> getHighestMountain(Mountain mountainHeight) {
double maxHeight = 1;
int index = 1;
for(int i = 0; i < mountainHeight.length; i++) {
if(mountainHeight[i].getHeight()>maxHeight) {
index = i;
}
}
}
要：
//This adds the parameter mountain to the list of mountains for this climber
public void addMountain(Mountain mountain) {
mountains.add(mountain);
}
//This loops over the mountains for this climber and returns the highest one.
public Mountain getHighestMountain() {
Mountain highestMountain = null;
for(int i = 0; i < mountains.size(); i++) {
if (highestMountain == null) {
highestMountain = mountains.get(i);
}
else if(mountains.get(i).getHeight() > highestMountain.getHeight()) {
highestMountain = mounts.get(i);
}
}
return highestMountain;
}
您的代码示例显示缺乏对基本概念的理解，例如字段和变量，参数和返回类型等。我建议阅读语言基础知识here。