Question

我在txt文件中有一个排序的整数列表（在线公开托管），我需要让用户通过二进制搜索给出一个在整数列表中搜索的数字。

我参加算法课程的介绍，我是Java的新手。

现在我已经得到了这些错误：

Exception in thread "main" java.lang.IndexOutOfBoundsException: Index: 52, Size: 22
    at java.util.ArrayList.rangeCheck(Unknown Source)
    at java.util.ArrayList.get(Unknown Source)
    at BinSearch4.main(BinSearch4.java:67)

* while循环（下面列出）是第67行

似乎无法弄清问题在这里。我也相信，即使我工作的一些事情也可以更有效地实施。任何人都可以提供的帮助将不胜感激。

/*  Binary Search Algorithm
 * 
 * - Import a sorted list of integers of an unknown length
 * 
 * - Ask the user for a number to search for (searchValue)
 * 
 * - Run a Binary Search on the list of integers without simply using binarySearch
 * 
 * - Return the index of an occurrence of the search number (searchValue)
 * 
 * - Or -1 if the target is not found.
 * 
 * 
 */


import java.io.FileNotFoundException;
import java.io.IOException;
import java.net.URL;
import java.util.Scanner;
import java.util.ArrayList;

public class BinSearch4 {

public static void main(String[] args) throws IOException {

    URL url = new URL("http://m.uploadedit.com/ba3a/1423978916244.txt"); 
    ArrayList<Integer> arrList = new ArrayList<Integer>(); 
    int searchValue;
    int low;
    int high;
    int mid;
    int NOT_FOUND = -1;
    Scanner in = new Scanner(System.in);

    System.out.println("Enter your number to search for:");
    searchValue = in.nextInt(); 
    System.out.println(searchValue);

    try {

        Scanner scr = new Scanner(url.openStream()); 

        while (scr.hasNextLine()) {
            int i = scr.nextInt();
            arrList.add(i);
        }
        scr.close();
        System.out.println(arrList);


    }

    catch (FileNotFoundException e) {
        e.printStackTrace();
    }




    low = arrList.get(0);                      
    System.out.println(low);
    high = arrList.get(arrList.size() - 1);
    System.out.println(high);
    mid = (low + high) / 2;

    while (low <= high && arrList.get(mid) != searchValue) {
        if (arrList.get(mid) < searchValue) {
            low = mid + 1;
        } else  {
            high = mid - 1;
        }
        mid = (low + high) / 2;
    }
    if (low > high) {
        mid = NOT_FOUND;
    }
    System.out.println(mid);

 }
}

文本文件中的数字列表是： [-2,3,6,9,11,22,24,31,35,43,48,52,62,65,69,70,73,83,86,90,100,107]

Answer 1

更改您的行：

low = arrList.get(0);                      
System.out.println(low);
high = arrList.get(arrList.size() - 1);

要

low = 0;                      
System.out.println(low);
high = arrList.size() - 1;

你获得异常的原因是，如果你的列表中有10个元素，并且你在列表中有100 - 110个元素，那么你试图按索引访问列表元素，即列表中不存在的100个元素因为你的列表只包含10个元素，即访问索引9之外的元素（因为数组索引从0开始到n-1），你将获得IndexOutOfBoundsException。

Answer 2

二进制搜索

public class BinarySearch {

    public int binarySearch(int[] a) {
        int start = 0;
        int end = a.length - 1;
        int searchedValue = 70;

        for (int i = 0; i <= a.length; i++) {
            int mid = (start + end) / 2;

            if (searchedValue == a[mid]) {
                return 1;
            }

            if (searchedValue < a[mid]) {
                end = mid - 1;
            } else {
                start = mid + 1;
            }
        }
        return -1;
    }

    public static void main(String[] args) {
        int a[] = { 2, 5, 40, 50, 60, 70 };
        BinarySearch binarySearch = new BinarySearch();
        System.out.println(binarySearch.binarySearch(a));
    }
}

start必须为0，end将为arr.length-1

Answer 3

感谢android_dev和almas，他们都是正确的。

此处我的原始代码在此处发布：

low = arrList.get(0);                      
System.out.println(low);
high = arrList.get(arrList.size() - 1);

我正在引用该值，我需要引用索引。因此，“低”＆＃39;对于像这样的二进制搜索程序，index将始终需要为零。

我现在正常运作的代码是：

low = 0;                          
System.out.println(low);
high = arrList.size() - 1;

Java中文本文件的二进制搜索整数

3 个答案: