Question

我正在使用NetBeans编写一个带参考编号的图书馆数据库来查找图书。我正在使用线性和二进制搜索来确定引用号是否在库中，但二进制搜索在应该为真时返回false。这就是我的整体代码：

package u3a3_bookslist;

import java.util.*;
import java.io.*;

public class bookslist {

//Define the default variables to use in all other methods of the program
public static int index, numSearches;

public static void main(String[] args) {

    //Define the ArrayList to store the values of 'BookList.txt'
    ArrayList <String> books = new ArrayList <String>();

    //Define the default values to use later in the code
    BufferedReader br = null;
    String referenceNumber;

    //Use a try statement to analyze the entire 'BookList.txt' file and add
    //each value on a new line into the arrayList 'books'
    try {
        br = new BufferedReader(new FileReader("BookList.txt"));
        String word;
        while ((word = br.readLine()) != null ){
            books.add(word);
        }
    } catch (IOException e) {
        e.printStackTrace();
    } finally {
        try {
            br.close();
        } catch (IOException ex) {
            ex.printStackTrace();
        }
    }

    //Create a new Array called 'bookList' to store and convert all the values
    //from the 'books' arrayList
    String [] bookList = new String[books.size()];
    books.toArray(bookList);

    //Create a scanner input to ask the user to enter a reference number
    Scanner input = new Scanner(System.in);
    System.out.println("Enter the reference number of a book to determine"
            + " if it is in the library or not: ");

    //Set the variable 'referenceNumber' to whatever value that the user inputted
    //into the Scanner
    referenceNumber = input.next();

    //Obtain the boolean result of either true or false from the Binary and
    //Linear search methods
    Boolean resultLinear = linearSearch(bookList, referenceNumber);
    Boolean resultBinary = binarySearch(bookList, 0, bookList.length-1, referenceNumber);

    //Analyze each element that is contained in the 'bookList' Array
    for (int i = 0; i < bookList.length; i++) {

        //Determine if the value of 'i' is equal to the 'referenceNumber'
        //converted into an int format
        if (i == Integer.parseInt(referenceNumber)) {

            //Determine if the 'i' index of the 'bookList' Array is equal
            //to the user inputted reference number
            if (bookList[i].equals(referenceNumber)) {

                //If the previous statement were true, the 'index' variable
                //was set to equal to current value for 'i'
                index = i;
            }
        }
    }

    //Determine the message to display to the user depending on if the reference
    //number was found in the Array using a Linear Search
    if (resultLinear == true) {
        System.out.println("Linear Search: Reference Number " + referenceNumber +
                " was found in the library. The book with that number is: " + bookList[index+1]);
    } else {
        System.out.println("Linear Search: Reference Number " + referenceNumber
                + " not in the library. No book with that number.");
    }

    //Determine the message to display to the user depending on if the reference
    //number was found in the Array using a Binary search
    if (resultBinary != false) {
        System.out.println("Binary Search: Reference Number " + referenceNumber +
                " was found in the library. The book with that number is: " + bookList[index+1]);
    } else {
        System.out.println("Binary Search: Reference Number " + referenceNumber
                + " not in the library. No book with that number.");
    }


}

//Execute a linear search to determine if the user inputted reference number
//is contained in the bookList Array
static public Boolean linearSearch(String[] A, String B) {
    for (int k = 0; k < A.length; k++) {
        if (A[k].equals(B)) {
            return true;
        }
    }
    return false;
}

//Execute a binary search to determine if the user inputted reference number
//is contained in the bookList Array
public static Boolean binarySearch(String[] A, int left, int right, String V) {

    int middle;
     numSearches ++;
     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);
     }

}

}

我遇到问题的程序部分是这部分：

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

    int middle;
     numSearches ++;
     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);
     }

}

我只是不明白为什么二进制搜索在假定为True时返回false。当线性搜索成立时，它甚至会返回true。

Answer 1

我看到你使用相同的数组进行线性搜索和二进制搜索你以前忘记了对数组进行排序吗？

使二进制搜索成为可能的原因是给定数组已排序 - 值只以一种方式进行（大多数变大）。
我在代码中看不到任何类型的排序 - 二进制搜索工作的最重要条件是排序数组。如果你考虑一下，会发生什么：每次给定值和找到的值不相等时，数组被“切割”成2个部分（没有定义每个部分的任何内容，不像分类的数组，其中一个部分大于中间，另一个部分小于*）。
因为数组没有排序，所以获得正确值的机会非常小，搜索无法根据他“指向”的值找到所需的值。如果希望的值在中间，或者如果它所采用的“路线”会以某种方式导致它（通常不太可能发生），那么你的代码就可以工作了。

*也可能是平等的。只是提出这个想法。

当返回值为true时，二进制搜索会一直返回false

1 个答案: