java - 线性搜索在应该返回下标数字时返回未找到

Question

我正在做一个 Java 作业，其中一个程序应该从用户那里读取 10 个数字，然后要求一个数字进行搜索。它对数字进行排序(升序)并对数组执行线性搜索，然后应该返回找到的下标或一条表示未找到的消息。

即使应该返回下标，我的代码也会给出“未找到”响应。你能看一下我的代码吗？我尝试改变我的 if 语句，但没有帮助。我已尽可能详细地使用了书中搜索和排序方法的示例(请随意指出错误)。

这是代码。

package program11;
import java.util.Scanner;

public class ArraySearch {

    public static void main(String[] args) {
        double[] arrayBuild = new double[10];
        Scanner input = new Scanner(System.in);

        for (int i = 0; i < arrayBuild.length; i++) {
            System.out.print("Enter a number.");
            arrayBuild[i] = input.nextDouble();
        }

        System.out.print("Enter a number to search for ");
        int objective = input.nextInt();

        linearCheck(arrayBuild, objective);

        if ((objective >= 0) && (objective < arrayBuild.length)) {
            System.out.println("Found at index: " + objective);
        } else {
            System.out.println("Not Found");
        }
    }

    public static void arraySort(double[] arrayBuild) {
        for (int i = 1; i < arrayBuild.length; i++) {
            double currentPoint = arrayBuild[i];
            int r;
            for (r = i - 1; r >= 0 && arrayBuild[r] > currentPoint; r--) {
                arrayBuild[r + 1] = arrayBuild[r];
            }
            arrayBuild[r + 1] = currentPoint;
        }
    }

    public static double linearCheck(double[] arrayBuild, int objective) {
        for (int i = 0; i < arrayBuild.length; i++) {
            if (objective == arrayBuild[i])
                return i;
        }
        return -1;
    }

}

<小时/>

编辑-新代码。现在一切都完成了，除了当我输入一个不在数组中的数字时，我得到了一个负结果。例如，输入 10 8 7 6 5 3 5 3 5 3 5 6 并搜索 11，线性搜索的结果为 -1，二分搜索的结果为 -11。我尽力听取了你的建议。我现在缺少什么？

package program11;
import java.util.Scanner;
import javax.swing.JOptionPane;
public class ArraySearch {

    public static void main(String[] args) {
        double[] arrayBuild = new double[10];
        Scanner input = new Scanner(System.in);
        int reply = 2;

for (int i = 0; i < arrayBuild.length; i++) {
    System.out.print("Enter a number.");
    arrayBuild[i] = input.nextDouble();
        }
while (reply != 1) {
        System.out.print("Enter a number to search for ");
        double objective = input.nextDouble();

arraySort(arrayBuild);
        double linearResult = linearCheck(arrayBuild, objective);

if (objective >= 0) {
        System.out.println("Linear search found at index: " + linearResult);
                }
else {
        System.out.println("Not Found (linear)");
        }
        double binaryResult = binaryCheck(arrayBuild, objective);

if (objective >= 0) {
System.out.println("Binary search found at index: " + binaryResult);
                }
else {
     System.out.println("Not Found (binary)");
        }
reply = JOptionPane.showConfirmDialog(null, "Continue?"); 
    }
}

public static void arraySort(double[] arrayBuild) {
    for (int i = 1; i < arrayBuild.length; i++) {
        double currentPoint = arrayBuild[i];
        int r;
        for (r = i - 1; r >= 0 && arrayBuild[r] > currentPoint; r--) {
            arrayBuild[r + 1] = arrayBuild[r];
        }
        arrayBuild[r + 1] = currentPoint;
    }
    }

public static double linearCheck(double[] arrayBuild, double objective) {
    for (int i = 0; i < arrayBuild.length; i++) {
        if (objective == arrayBuild[i])
            return i;
    }
return -1;
    }
public static double binaryCheck(double[] arrayBuild, double objective) {
    int low = 0;
    int high = arrayBuild.length - 1;

    while (high >= low) {
        int mid = (low + high) / 2;
        if (objective < arrayBuild[mid])
            high = mid - 1;
        else if (objective == arrayBuild[mid])
            return mid;
        else
        low = mid + 1;
    }
return -low - 1;
    }
}

Answer 1

我不确定是什么导致了您的问题，但我发现您的代码存在一些问题

1. 如果您想搜索 Int，那么我也会将 arrayBuild 设为 int 类型。或者，您可以通过将objective 的类型更改为 double 并让用户输入 double 来搜索 double

2. 当您调用linearCheck时，您不会将结果存储在任何地方，您应该将其分配给一个变量

   double foundLocation = linearCheck(arrayBuild, objective);
   

3. 您的 if 语句没有多大意义，您应该使用 linearCheck 的返回值，如下所示

   if (foundLocation >= 0) {
       System.out.println("Found at index: " + foundLocation);
   } else {
       System.out.println("Not Found");
   }
   

4. 您没有在主程序中的任何位置调用 arraySort