由用户输入引导的气泡搜索

Question

我正在编写一个程序，帮助用户制作Top X或Top favorites列表。

用户将首先输入列表中的项目数，然后用于创建相同大小的数组。然后用户填充数组。最后，用户会被问到一系列问题，程序会将列表中的每个项目与列表中的另一个项目逐一进行比较，并在两者之间询问你更喜欢哪一个。根据他们的选择，他们的分数保存在另一个存储整数值的数组中。然后按降序进行排序。

然而，当我运行程序时，它不会将所有元素与每个元素进行比较。

例如，我有一个三个元素长的数组，项目是"香草"、"巧克力"one_answers"草莓"。它会要求我将香草与巧克力进行比较，然后将香草与草莓进行比较。然而，它不会将巧克力与草莓进行比较，然后打印结果。我想知道我犯了什么逻辑错误。我从未对字符串使用过冒泡搜索。

这是我的代码供参考：

 import java.util.Scanner;
public class FavoriteListMaker {
public static void main(String[] args)
{
    Scanner inputDevice = new Scanner(System.in);
    System.out.println("This program is to help you order a favorites list. Please enter the amount of items on the list.");
    int ListSize = inputDevice.nextInt();
    inputDevice.nextLine();
    String [] TopXA = new String [ListSize];
    int [] TopXB = new int[ListSize];
    for (int x = 0; x < TopXA.length; ++ x)
    {
        System.out.println("Please enter an item to be organized on the list");
        TopXA[x] = inputDevice.nextLine();
        System.out.println("You have " + (ListSize - x - 1) + " items left to fill on the list.");
    }
    System.out.println("Now we will compare each item on the list with every item on the list one at a time.");
    System.out.println("We will ask you a series a question on whether you like item A better then item B and tally the score.");
    System.out.println("At the end the item with the most points wins.");
    int comparisonsToMake = TopXA.length - 1;
    for (int y = 0; y < TopXA.length - 1; ++ y) 
    {   
        for (int z = 0; z < comparisonsToMake; ++ z)
        {
            if(TopXA[y] != TopXA[z + 1])
            {
                String compareA = TopXA[y];
                String compareB = TopXA[z + 1];
                System.out.println("Do you prefer "  + compareA + " or " + compareB + " .");
                System.out.println("If you prefer " + compareA + " Please press 1. If you prefer " + compareB + " please press 2.");
                int choice = inputDevice.nextInt();
                inputDevice.nextLine();
                switch(choice)
                {
                case 1:
                    TopXB[y] =+ 1;
                    break;
                case 2:
                    TopXB[z + 1] =+ 1;
                    break;
                default:
                    System.out.print("I'm sorry but that is not a valid input.");
                }
            }
        }
        --comparisonsToMake;
    }
    int comparisonsToMakeB = TopXB.length - 1;
    for(int a = 0; a < TopXB.length - 1; ++ a)
    {
        for(int b = 0; b < comparisonsToMakeB; ++b)
        {
            if(TopXB[b] < TopXB[b + 1])
            {
                String temp = TopXA[b];
                TopXA[b] = TopXA[b + 1];
                TopXA[b + 1] = temp;
            }
        }
        --comparisonsToMakeB;
    }
    for(int q = 0; q < TopXA.length; ++ q)
    {
        System.out.print("Your number " + (q + 1) + " pick is "  + TopXA[q] + ".");
    }
    }
 }

Answer 1

没关系，我找到了答案。。。。其实很简单。我只是简单地将z的所有特定实例更改为y。

int comparisonsToMake = TopXA.length - 1;
for (int y = 0; y < TopXA.length - 1; ++ y) 
{   
    for (int z = 0; z < comparisonsToMake; ++ z)
    {
        if(TopXA[y] != TopXA[z + 1]) // changed TopXA[z + 1] to [y + 1]
        {
            String compareA = TopXA[y];
            String compareB = TopXA[y + 1]; //instance of change
            System.out.println("Do you prefer "  + compareA + " or " + compareB + " .");
            System.out.println("If you prefer " + compareA + " Please press 1. If you prefer " + compareB + " please press 2.");
            int choice = inputDevice.nextInt();
            inputDevice.nextLine();
            switch(choice)
            {
            case 1:
                TopXB[y] =+ 1; 
                break;
            case 2:
                TopXB[y + 1] =+ 1; //instance of change.
                break;
            default:
                System.out.print("I'm sorry but that is not a valid input.");
            }
        }
    }
    --comparisonsToMake;
}

老实说，我不知道我为什么要做出这样的改变。可能是因为我累了，昨晚正在做这件事，并且在思考一些疯狂的巨魔逻辑。