Question

我应该在String数组上编写顺序/线性搜索。我非常接近完成，但部分任务让我感到困惑。它表示将目标项与列表的连续元素进行比较，直到目标匹配或目标小于比数组的当前元素为止。当没有数值时，String如何比其他元素更多或更少？也许我只是没有正确地思考它。到目前为止，这是我的计划：

public class SequentialSearchString {
public static boolean sequential (String[] numbers){
    //Set the target item to an arbitrary String that should return true.
    String T1 = "Frank";  

    for (int i = 0; i < numbers.length; i++){
        if (numbers[i] == T1){
            return true;    
        }
        if (numbers[i] != T1){
            numbers[i] = numbers[i+1];
        }           
    }
    return false;   
}

public static boolean sequential2 (String[] numbers){
    //Set the target key to String that should return false.
    String T2 = "Ian";
    for (int i = 0; i < numbers.length; i++){
        if (numbers[i] == T2){
            return true;    
        }
        if (numbers[i] != T2){
            numbers[i] = numbers[i+1];
        }
    }
    return false;   
}


public static void main(String[] args) {
    //Create a list of 8 Strings.
    String [] numbers = 
{"Ada", "Ben", "Carol", "Dave", "Ed", "Frank", "Gerri", "Helen", "Iggy", "Joan"};   
    //If the first target item (T1) is found, return Succuss. If not, return failure.
        if (sequential(numbers) == true){
            System.out.println("Success. 'T1' was found");
        }
        else {
            System.out.println("Failure. 'T1' was not found");  
        }
    //If the second target item (T2) is found, return Succuss. If not, return failure.
        if (sequential2(numbers) == true){
            System.out.println("Success. 'T2' was found");
        }
        else {
            System.out.println("Failure. 'T2' was not found");  
        }   
    }   
}

第一种方法工作正常，但我似乎遇到了搜索列表中不存在的元素的问题。以下是运行程序后得到的错误消息：

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 10
at SequentialSearchString.sequential2(SequentialSearchString.java:32)
at SequentialSearchString.main(SequentialSearchString.java:50)
Success. 'T1' was found

非常感谢任何理解分配和修复异常的帮助。

Answer 1

numbers[i] = numbers[i+1];

可能会导致您的ArrayIndexOutOfBoundsException。

您的条款检查i < numbers.length。所以你设置了界限。但是，如果i == numbers.length - 1，那么您将尝试访问比您的数组更大的i+1，因此它超出范围。

例如：numbers.length为4。所以i可以 3。使用i+1，您尝试访问numbers[4]，这将是第五个位置，因为数组以0开头，而numbers[3]将是最后一个位置。

Answer 2

ArrayIndexOutOfBoundsException是由于您使用的事实：

for (int i = 0; i < numbers.length; i++)

和后者：

 numbers[i] = numbers[i+1];

当i等于numbers.length-1（最后一次迭代）时，i + 1等于numbers.length。然后你尝试读取错误的数字[numbers.length]（有效索引从0到numbers.length-1）。

你必须使用：

for(int i=0;i<numbers.length-1;i++)

防止异常。现在，我不确定它会解决你的整个问题，但肯定是例外。

在Java中对String数组进行顺序搜索

2 个答案: