Question

该方法返回一些空跳过一些对象。该方法的目标是根据他们的期中＆＃39;返回分类人员列表。并且＆＃39; final＆＃39; （＆＃39;一个人的中期＆＃39;紧接着是他的＆＃39;最终＆＃39;）smb请帮忙修复它？这是代码

public static Exam[] collateExams(Exam[] exams)
    {

        Exam[] r = new Exam[exams.length]; 
        int index = 0;

        for (int i = 0; (i < exams.length) && (index < exams.length/2); i++)
        {
            if (exams[i].getExamType() == 'm')
            {
                r[index*2] = new Exam(exams[i].getFirstName(), exams[i].getLastName(), exams[i].getID(), "midterm", exams[i].getScore());
                for(int j = 0; (j < exams.length) && (index < exams.length/2); j++)
                {
                    if((exams[j].getExamType() == 'f') && (exams[i].getID() == exams[j].getID()))
                    {
                        r[index*2 + 1] = new Exam(exams[i].getFirstName(), exams[i].getLastName(), exams[i].getID(), "final", exams[i].getScore());
                    }
                }
            }
            index++;
        }
        return r;

这是输出：

null
null
Bill Gates 6 midterm  90
Bill Gates 6 final  90
James Gosling 3 midterm  100
James Gosling 3 final  100
Sergey Brin 22 midterm  98
null
Dennis Ritchie 5 midterm  94
Dennis Ritchie 5 final  94
Steve Jobs 9 midterm  95
Steve Jobs 9 final  95

这里是未排序的列表，它作为参数传递：

Steve Jobs 9 final  91
Bill Gates 6 midterm  90
James Gosling 3 midterm  100
Sergey Brin 22 midterm  98
Dennis Ritchie 5 midterm  94
Steve Jobs 9 midterm  95
Dennis Ritchie 5 final  100
Jeff Dean 7 midterm  100
Bill Gates 6 final  96
Jeff Dean 7 final  100
Sergey Brin 27 final  97
James Gosling 3 final  100

Answer 1

你的整个班级都发布了同样的问题，或者你是Joshua mark II。现在，其他变体只表示一个用于循环和其他限制，所以我将忽略所有这些并解决上面混乱的双循环。这将解决上面的循环：

public static Exam[] collateExams(Exam[] exams)
{
     Exam[] r = new Exam[exams.length]; 
     int index = 0;

     for (int i = 0; i < exams.length ; i++)
     {
         if (exams[i].getExamType() == 'm')
         {
             r[index*2] = new Exam(exams[i].getFirstName(), exams[i].getLastName(), exams[i].getID(), 'm', exams[i].getScore());
             for(int j = 0; j < exams.length; j++)
             {
                 if((exams[j].getExamType() == 'f') && (exams[i].getID() == exams[j].getID()))
                 {
                     r[index*2 + 1] = new Exam(exams[j].getFirstName(), exams[j].getLastName(), exams[j].getID(), 'f', exams[j].getScore());
                 }
             }
             index++;
         }
     }
     return r;    
 }

输出：

Sorted list: 
Exam [fn=Bill, ln=Gates, id=6, ex=m, score=90]
Exam [fn=Bill, ln=Gates, id=6, ex=f, score=96]
Exam [fn=James, ln=Gosling, id=3, ex=m, score=100]
Exam [fn=James, ln=Gosling, id=3, ex=f, score=100]
Exam [fn=Dennis, ln=Ritchie, id=5, ex=m, score=94]
Exam [fn=Dennis, ln=Ritchie, id=5, ex=f, score=100]
Exam [fn=Steve, ln=Jobs, id=9, ex=m, score=95]
Exam [fn=Steve, ln=Jobs, id=9, ex=f, score=91]
Exam [fn=Jeff, ln=Dean, id=7, ex=m, score=100]
Exam [fn=Jeff, ln=Dean, id=7, ex=f, score=100]
Exam [fn=Sergey, ln=Brin, id=22, ex=m, score=99]
Exam [fn=Sergey, ln=Brin, id=22, ex=f, score=97]

Answer 2

[编辑] @Michael，你的代码给我的输出只有一个null：

    Sorted list: 
Bill Gates 6 midterm 90
Bill Gates 6 final 96
James Gosling 3 midterm 100
James Gosling 3 final 100
Sergey Brin 22 midterm 98
null
Dennis Ritchie 5 midterm 94
Dennis Ritchie 5 final 100
Steve Jobs 9 midterm 95
Steve Jobs 9 final 91
Jeff Dean 7 midterm 100
Jeff Dean 7 final 100

这里是具有toString方法的Exam类：

class Exam {

    private String firstName;
    private String lastName;
    private int ID;
    private String examType;
    private int score;


    public Exam(String firstName, String lastName, int ID, String examType, int score)
    {
        this.firstName = firstName;
        this.lastName = lastName;
        this.ID = ID;
        this.examType = examType;
        this.score = score;
    }

    public String getFirstName()
    {
        return this.firstName;
    }

    public String getLastName()
    {
        return this.lastName;
    }

    public int getID()
    {
        return this.ID;
    }

    public char getExamType()
    {
        char examTypeCasted = 0;
        examTypeCasted = examType.charAt(0);
        return Character.toUpperCase(examTypeCasted);
    }

    public int getScore()
    {
        return this.score;
    }

    public String toString()
    {
        return this.firstName + " " + this.lastName + " " + this.ID + " " + this.examType + " " + this.score;
    }

    public boolean equals(Exam e)
    {
        if(this.equals(e))
            return true;
        else
        return false;
    }

}

Answer 3

这些显示 null 的项目永远不会被初始化，这就是为什么..

public class Example2 {

String array[] = new String[15];

public Example2(){

    for(int j=0; j<array.length-1; j++)
        array[j]="j";


    for(int j=0; j<array.length; j++)
        System.out.println(array[j]);
}


public static void main(String[] args){
    new Example2();
}

}

这个例子将打印14（“J”）和一个null。为什么？原因数组[15]未初始化。所以在你的例子中，我确信这些项目显示 null 原因永远不会被初始化。

在输出中返回null

3 个答案: