Question

我必须创建一个接受用户输入（一个数字）的程序，然后程序应该具有该数字并将搜索应用于数组并通过匹配索引和用户输入的数字输出相应的标题。但是在运行期间，没有任何反应。我在代码中设置了断路器，并注意到for循环（搜索算法）存在问题。请帮帮我，让我知道我的搜索算法有什么问题。我想要做的是使用用户输入的数字来匹配索引，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存在问题。此外，我需要应用二进制搜索，只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进制搜索。

如何制作一个判断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。变量都是例如：

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);
     }
 }

Answer 1

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

删除你的代码

// 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; 
}

与

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
 }

Answer 2

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

Answer 3

这不是二元搜索答案。只是HashMap的实现。看看吧。

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"));

如何使用二分查找比较Strings。

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;

    }

Answer 4

你在这做什么：

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，以避免出现例外情况：

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
}

对于使用二进制搜索，需要对字符串数组进行排序。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 method

Answer 5

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
}

将线性和二进制搜索应用于数组

5 个答案: