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我必须创建一个接受用户输入（数字）的程序，然后该程序应该具有该数字并对数组应用搜索并通过匹配索引和用户输入的数字来输出相应的标题。但是，在运行时，什么也没有发生。我在我的代码中设置了断路器，并注意到 for 循环（搜索算法）存在问题。请帮助我，让我知道我的搜索算法出了什么问题。我要做的是使用用户输入的数量来匹配索引，然后输出存储在索引中的书名。

       private void btnFindActionPerformed(java.awt.event.ActionEvent evt) {                                        
    // TODO add your handling code here:  

    // declares an array 
   String[] listOfBooks = new String [101];

   // assigns index in array to book title 
   listOfBooks[1] = "The Adventures of Tom Sawyer"; 
   listOfBooks[2] = "Huckleberry Finn"; 
   listOfBooks[4] = "The Sword in the Stone";
   listOfBooks[6] = "Stuart Little";
   listOfBooks[10] = "Treasure Island";
   listOfBooks[12] = "Test";
   listOfBooks[14] = "Alice's Adventures in Wonderland";
   listOfBooks[20] = "Twenty Thousand Leagues Under the Sea";
   listOfBooks[24] = "Peter Pan";
   listOfBooks[26] = "Charlotte's Web";
   listOfBooks[31] = "A Little Princess";
   listOfBooks[32] = "Little Women";
   listOfBooks[33] = "Black Beauty";
   listOfBooks[35] = "The Merry Adventures of Robin Hood";
   listOfBooks[40] = "Robinson Crusoe";
   listOfBooks[46] = "Anne of Green Gables";
   listOfBooks[50] = "Little House in the Big Woods";
   listOfBooks[52] = "Swiss Family Robinson";
   listOfBooks[54] = "The Lion, the Witch and the Wardrobe";
   listOfBooks[54] = "Heidi";
   listOfBooks[66] = "A Winkle in Time";
   listOfBooks[100] = "Mary Poppins";

    // gets user input 
    String numberInput = txtNumberInput.getText();
    int number = Integer.parseInt(numberInput);

    // Linear search to match index number  and user input number
        for(int i = 0; i < listOfBooks.length - 1; i++) {
        if (listOfBooks.get(i) == number) {
        txtLinearOutput.setText(listOfBooks[i]);
        break; 
        }


    }

* if 语句中的 listOfBooks.get 有问题。此外，我需要应用二进制搜索，仅使用二进制方法搜索相同的数组。需要帮助来应用这种类型的二分搜索。

我怎样才能做出检查 int 数是否等于索引的语句？

请注意，以下代码只是我必须应用的示例。变量都是出于示例目的：

public static Boolean binarySearch(String [ ] A, int left, int right, String V){
     int middle;

     if (left > right) {
         return false;
     }

     middle = (left + right)/2;
     int compare = V.compareTo(A[middle]);
     if (compare == 0) {
         return true;
     }
     if (compare < 0) {
         return binarySearch(A, left, middle-1, V);
     } else {
         return binarySearch(A, middle + 1, right, V);
     }
 }
4

5 回答 5

4

你可以通过给出这样的数字来避免for loop和检查条件：txtLinearOutput.setText(listOfBooks[number-1]);

删除您的代码

// Linear search to match index number  and user input number
for(int i = 0; i < listOfBooks.length - 1; i++) {
    if (listOfBooks.get(i) == number) {
     txtLinearOutput.setText(listOfBooks[i]);
     break; 
}

和

try{
     int number = Integer.parseInt(numberInput);
     if(number>0 && number<101){
       txtLinearOutput.setText(listOfBooks[number-1]);
     }else{
        // out of range
     }
 }catch(Exception e){
   // handle exception here
 }
于 2015-08-19T05:54:00.503 回答
2

您正在比较if (listOfBooks.get(i) == number)它是错误的，您应该比较：if (i == number)，因为您需要比较元素位置。

于 2015-08-19T05:57:28.223 回答
0

这不是二进制搜索答案。只是一个实现HashMap。看看它。

HashMap<String, Integer> books = new HashMap();
books.put("abc", 1);
books.put("xyz", 2);
books.put("pqr", 3);
books.put("lmo", 4);

System.out.println(books.getValue("abc");

使用内置的 BinarySearch。

String []arr = new String[15];
arr[0] = "abc";
arr[5] = "prq";
arr[7] = "lmo";
arr[10] = "xyz";
System.out.println(Arrays.binarySearch(arr, "lmo"));

如何Strings使用二分搜索进行比较。

String[] array = new String[4];
        array[0] = "abc";
        array[1] = "lmo";
        array[2] = "pqr";
        array[3] = "xyz";
        int first, last, middle;
        first = 0;
        last = array.length - 1;
        middle = (first + last) / 2;
        String key = "abc";
        while (first <= last) {
            if (compare(array[middle], key))
                first = middle + 1;
            else if (array[middle].equals(key)) {
                System.out.println(key + " found at location " + (middle) + ".");
                break;
            } else {
                last = middle - 1;
            }
            middle = (first + last) / 2;
        }
        if (first > last)
            System.out.println(key + " is not found.\n");

    }

    private static boolean compare(String string, String key) {
        // TODO Auto-generated method stub
        for (int i = 0; i < Math.min(string.length(), key.length()); ++i)
            if (string.charAt(i) < key.charAt(i))
                return true;
        return false;

    }
于 2015-08-19T06:02:50.963 回答
0

你在这里做什么：

if (listOfBooks.get(i) == number) {

是您将数组的内容与输入数字匹配，这是无关紧要的。

您可以直接使用输入数字来获取存储在索引中的值。

例如：
txtLinearOutput.setText(listOfBooks[number-1]);

此外，int number = Integer.parseInt(numberInput);应放置在 try-catch 块中以验证输入数字解析。您可以检查输入数字是否在数组范围内，以避免出现以下异常：

try{

    int number = Integer.parseInt(numberInput);

    // Linear search to match index number  and user input number
    if (number > 0 && number <=100) {
        txtLinearOutput.setText(listOfBooks[number-1]);
    } else {
    // Display error message
    }
} catch(Exception e) {
    // Handle exception and display error message
}

而对于使用二分查找，字符串数组需要进行排序。您可以使用 Arrays.sort()方法对其进行排序。关于使用二进制搜索，您可以使用Java Arrays Binary Search 方法

于 2015-08-19T06:04:19.877 回答
0

您的线性搜索代码看起来像这样

try{
txtLinearOutput.setText(listOfBooks[yourNumber]);
}
catch(IndexOutOfBoundsException ie){
// prompt that number is not an index
}
catch(Exception e){
// if any other exception is caught
}
于 2015-08-19T06:09:47.447 回答