java - 在数组中搜索字符串出现的次数

Question

在我的 java 类中，我们使用 junit test 来测试我们的方法。本节介绍了如何使用接口(interface)。

我遇到问题的这个特定方法应该在每个索引处搜索数组，寻找匹配的字符串作为输入。

在junit测试中我有

    void test() 
    {
        MyClassList labTest = new MyClassList("CIS 120", "Fall", "Professor Awesome");

        MyStudent george = new MyStudent("George","Lucas", "555-555-5555","george.lucas@starwars.com");
        MyStudent gene = new MyStudent("Gene", "Roddenberry","666-666-6666", "gene.roddenberry@startrek.com");
        MyStudent jordan = new MyStudent("Jordan" ,"Robberts", "755-555-5555", "jordan.robberts@wheeloftime.com");

        labTest.insert(george);
        labTest.insert(gene);
        labTest.insert(jordan);

        System.out.println(labTest.toString());
        System.out.println(labTest.contains(george));
        System.out.println(labTest.search("George"));

这是您用于方法搜索的代码:

注意

 protected MyStudent [] log;
 protected int lastIndex = -1;

是全局变量

package Lab2;

import java.util.Arrays;
import java.util.Scanner;

import Lab2.ClassListInterFace;

public class MyClassList implements ClassListInterFace {
    protected String course;
    protected String semester;
    protected String teacherLastName;
    protected MyStudent[] log;
    protected int lastIndex = -1;

    public MyClassList(String currCourse,  String currSemester, String lastName, int maxSize) {
        log = new MyStudent[maxSize];
        this.course = currCourse;
        this.semester = currSemester;
        this.teacherLastName = lastName;
    }

    public MyClassList( String currCourse,  String currSemester, String lastName)

    {
        log = new MyStudent[100];
        this.course = currCourse;
        this.semester = currSemester;
        this.teacherLastName = lastName;
    }

    public void insert(MyStudent element) {
        lastIndex++;
        log[lastIndex] = element;
    }

    public boolean isFull() {
        if (lastIndex == (log.length - 1))
            return true;
        else
            return false;
    }

    public int size() {

        return (lastIndex + 1);
    }


    public void clear() 
    {
          for (int i = 0; i <= lastIndex; i++)
              log[i] = null;
            lastIndex = -1;

    }

    public String getName() {
         return teacherLastName;
    }


public boolean contains(MyStudent element) {

    boolean found = false;
    for( int location = 0;location <= lastIndex; location++)
    {
      if (element.equals(log[location]))  // if they match
        found = true;
}
    return found;
}

public String toString()
{
    String message = "Course " + course + "\n Semester " + semester + "\n Proffessor " + teacherLastName + "\n"; 
            for (int i = 0; i <= lastIndex; i++) {
                message += log[i].toString();
            }
    return message;
}

public int search(String x) 

{
    int answer = 0;
     for(int i = 0; i < log.length; i++) 
     {
            if(x.equalsIgnoreCase(log[i]))
                answer++;
     }

    return answer;

}

这是我根据老师给我们引用的一些代码得出的，并做了一些调整。

Answer 1

这看起来可以通过 for 循环更优雅地完成。根据我的经验，使用专为此类事情设计的更简洁的控制结构可以大大减少出错的机会。虽然我不确定您正在寻求帮助的具体问题，但我确实注意到，如果您确实找到了匹配项，您会跳过下一个元素而不检查它是否也 一场比赛。

    int location = 0;
    while (location <= lastIndex) 
    {
        if (x.equalsIgnoreCase(log[location])) 
        { // if they match
            answer ++;
            location ++;  //<-- Here's where you increment twice upon finding a match!
        }
        location++;  //To fix with the smallest number of changes, just put this in an else clause
    }

通过将其更改为 for 循环，整个 block 可以减少到大约一半的行数和一半的变量。见下文:

    for(int i = 0; i < log.length; i++) {
        if(x.equalsIgnoreCase(log[i].firstName))
            answer++;
    }

这更容易阅读，并且更不容易出错。试图跟踪过多的变量或划分常见操作(例如增加当前位置)只会引发问题。这个四行代码在功能上等同于上面的代码块(除了它在找到匹配项时不会跳过以下条目这一事实之外)，但程序员犯错误的机会要少得多。始终，当您可以摆脱它时，请使用为该任务设计的控制流结构。