如何制作使用输入扫描器的线性搜索 OOP 方法

Question

所以我重构了一个只使用 main 方法的线性搜索代码。我的目标是将其转换为 OOP 方法。但我无法保存输入的整数集。

// LinearSearchDriver.java
import java.util.Scanner;

public class LinearSearchDriver {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);

        LinearSearch linearSearch = new LinearSearch();
        System.out.println("Enter number of elements");
        int numElements = in.nextInt();
        linearSearch.setNumberOfElements(numElements);

        System.out.println("Enter " + numElements + " integers");

        for (int count = 0; count < numElements; count++){
            int setIntegers = in.nextInt();
            linearSearch.setNumberOfIntegers(setIntegers);
        }
        
        System.out.println("Enter value to find");
        int search = in.nextInt();
        linearSearch.findValue(search);

    }
}

//LinearSearch.java
public class LinearSearch {
    private int c;
    private int n;
    private int array[];

    public void setNumberOfElements(int n) {
        this.n = n;
        this.array = new int[n];
    }

    public void setNumberOfIntegers(int y) {
        for (c=0; c < n; c++)
            array[c] = y;
    }

    public void findValue(int search) {
        for (c = 0; c < n; c++) {
            if (array[c] == search) {    /* Searching element is present */
                System.out.println(search + " is present at location " + (c + 1) + ".");
                break;
            }
        }

        if (c == n) { /* Searching element is absent */
            System.out.println(search + " is not present in array.");
        }
    }
}

示例输出：

但是当我输入数字 1 时，这是输出：

该程序只读取我认为的 2 号输出，最后一个数字只是保存到数组中的那个。

Answer 1

for (c = 0; c < n; c++) {
    array[c] = y;
}

是出问题的地方。您正在为数组中的每个索引设置传递给该函数的最后一个值。

您可以通过多种方式解决此问题：

1. 将数组而不是单个参数传递给函数。
2. 您可以确定数组中的当前元素数量，然后“手动”附加最新值。见this post。
3. 或者您可以简单地使用动态结构，例如 List 并将元素附加到该结构。

   以下是第 3 个选项的粗略大纲：

   public class LinearSearch {
       private List<Integer> intList;
   
       public LinearSearch() {
       }
   
       public void setNumberOfElements(int n) {
           intList = new ArrayList<>(n); //Set the capacity here like before.
       }
   
       public void setNumberOfIntegers(int y) {
           //If you want your List to always only contain the initially allowed number of elements, you could implement
           // this logic here, by adding the new value and removing the "oldest" one.
           intList.add(y);
       }
   
       public void findValue(int search) {
           if (!intList.contains(search)) { //You can put this up here and potentially skip the looping.
               System.out.println(search + " is not present in array.");
               return;
           }
           for (int n : intList) {
               if (n == search) {
                   System.out.println(search + " is present at location " + intList.indexOf(search) + ".");
                   return; //Use return to exit the method, break only exits the loop in your example, and you could print both lines.
               }
           }
       }
   }