Question

我试图编写一个程序来检测可以使用ArrayList中任何数字子集进行的最大总和，并且总和必须低于用户输入的目标数。到目前为止，我的程序运行良好，除了一行（没有双关语）。请注意，此代码也尚未完成。

出于某种原因，在调用我的扫描仪时，我在下面标记的行上不断出现异常的InputMismatchException。当我运行程序并要求它应用默认列表时（正如您将在代码中看到的那样），我得到一个不寻常的输出

Enter integers (one at a time) to use, then type "done" 
OR
You can just type "done" to use the default set
done   //User Input here
Done
Enter your target number

在此行之后，抛出异常并且我被踢出程序（在BlueJ中编写并运行）。你们认为你可以帮助我吗？我假设它只是一个＆＃34;语法＆＃34;错误，如果你知道我的意思。代码如下：

import java.util.*;
/** This program will provide the largest sum of numbers within a set that is less than a user-implied target number. This is done by asking for a user input of integers (or none, for a default set), sorting the list, and calulating the highest sum starting with the largest number and moving downward on the set
   @param input A Scanner used to detect user input
   @param user A placeholder for user-implied integers
   @param target The user-implied target number
   @param index A reference to the currently "targeted" index on "list"
   @param splitIndex Used if the list needs to be checked in non-adjacent indexes
   @param finalOutput The final product of this program
   @param list An ArrayList of Integers used as the set*/
class LargestSum{
    public static void main(String[] args){
        Scanner input = new Scanner(System.in);
        int user = 0, target = 0, index = 0, finalOutput = 0, splitIndex = 0;
        ArrayList<Integer> list = new ArrayList<Integer>();
        System.out.println("Enter integers (one at a time) to use, then type \"done\" \nOR\n You can just type \"done\" to use the default set");
        /** A try-catch block is used to detect when the user has finished ading integers to the set*/
        try{
            user = input.nextInt();
            list.add(user);
        }
        catch(InputMismatchException e){
            System.out.print("Done");
            if (list.size() == 0){
                list.add(1);
                list.add(2);
                list.add(4);
                list.add(5);
                list.add(8);
                list.add(12);
                list.add(15);
                list.add(21);
            }
            if (list.size() > 0){
            Collections.sort(list);
            System.out.println("\nEnter your target number");
            target = input.nextInt();           //EXCEPTION IS THROWN HERE
            index = list.size() - 1;
            while (list.get(index) > target){                
                list.remove(index);
                index = index - 1;
                if (index == -1){
                    System.out.println("No sum can be made that is lower than the target number");
                    System.exit(0);
                }
            }        
            while (finalOutput < target){
                if(list.get(index) + list.get(index - 1) < target){
                   finalOutput = list.get(index) + list.get(index - 1);
                   list.remove(index);
                   list.remove(index - 1);
                   list.add(finalOutput);
                   index = list.size() - 1;
                }
                else{
                    for (int i = index; i >= 0; i--){
                        if (list.get(i) + list.get(index) < target){
                            finalOutput = list.get(i) + list.get(index);
                            list.remove(i);
                            list.remove(index);
                            list.add(finalOutput);
                            index = list.size() - 1;
                            i = index;
                        }
                    }
                }
            } 
        }
            System.out.println("Done!");
            System.out.println("Largest Sum is: " + finalOutput);
        }
    }           
}

谢谢大家，你可以忽略文档评论。正如我所说，该计划尚未完成，我计划增加更多。

〜AndrewM

Answer 1

您需要致电$('#panel').ready(function () { $("#stbutton").click(); });以允许扫描仪接收另一行输入。目前input.nextLine();正在尝试将input.nextInt()转换为导致Done的{{1}}，因为它使用您之前输入的相同输入行。

{{2} }

Answer 2

问题出在这里 - user = input.nextInt（）; 直到你通过nextInt（）捕获整数，你的代码才行。但是当你提供＆＃34;完成＆＃34;时，你会得到inputMistmathexception。这个函数显然是有帮助的InputMismatchException.Hope。

为什么在这里输入InputMismatchException？（Java）的

2 个答案:

为什么在这里输入InputMismatchException？ （Java）的

2 个答案:

为什么在这里输入InputMismatchException？（Java）的