Question

这是一个单词搜索程序。它搜索的文本被输入并转换为另一个类中的2D数组。

这是该计划正在搜索的文字：

10 //rows
15 //columns
fqexfecmxdvjlgu
cxomfslieyitqtz
nucatfakuxofegk
hfytpnsdlhcorey
pgrhdqsypyscped
ckadhyudtioapje
yerjodxnqzztfmf
hypmmgoronkzhuo
hdskymmpkzokaao
amuewqvtmrlglad

由于某种原因，即使我输入了终止字符串end，它也会进入我的checkDown()方法并创建一个越界错误。如果我注释掉该方法并只执行checkRight()和checkDiagonal()方法，那么一切似乎都能正常工作。这是我的代码：

import java.util.Scanner;

public class WordSearch
{
    private char[][] array;
    private String targetWord;
    private int rowLocation;
    private int colLocation;

    public WordSearch(char[][] inArray)
    {
        array = inArray;
    }

    public void play()
    {
        do{
            for (int row = 0; row < array.length; row++)
            {
                for (int col = 0; col < array[row].length; col++)
                {
                    System.out.print(array[row][col]);
                }
                System.out.println();
            }

            System.out.println();
            Scanner input = new Scanner(System.in); 
            System.out.println("What word would you like to search for? Type end to quit: ");
            targetWord = input.nextLine();
            System.out.println("Typed in: " + targetWord);
            System.out.println();

            compareFirst(targetWord);
        } while (!targetWord.equals("end"));

    }

    public void compareFirst(String inWord)
    {
        for (int row = 0; row < array.length; row++)
        {
            for (int col = 0; col < array[row].length; col++)
            {
                if(array[row][col] == inWord.charAt(0))
                {

                    rowLocation = row;
                    colLocation = col;

                    suspectAnalysis();
                }
            }
        }
    }

    public void suspectAnalysis()
    {
        checkRight();
        checkDown();
        checkDiagonal();
    }


    public void checkRight()
    {
        for(int i = 1; i < (targetWord.length()); i++)
        {
            if(colLocation + i > array[0].length - 1)
            {
                return;
            }

            else if(array[rowLocation][colLocation + i] != targetWord.charAt(i))
            {
               return;
            }
        }
        System.out.println(targetWord + " found horizontally at row " + rowLocation + " and column " + colLocation);
        System.out.println();

        return;

    }


    public void checkDown()
    {
        for(int i = 1; i < (targetWord.length()); i++)
        {
            if(rowLocation + i > array.length - 1 && colLocation + i > array[0].length - 1)
            {
                return;
            }
            else if(array[rowLocation + i][colLocation] != targetWord.charAt(i))
            {
                return;
            }
        }
        System.out.println(targetWord + " found vertically at row " + rowLocation + " and column " + colLocation);
        System.out.println();          
    }

    public void checkDiagonal()
    {
        for(int i = 1; i < (targetWord.length()); i++)
        {
            if(colLocation + i > array[0].length - 1 || rowLocation + i > array.length - 1)
            {
                return;
            }

            else if(array[rowLocation + i][colLocation + i] != targetWord.charAt(i))
            {
                return;
            }
        }
        System.out.println(targetWord + " found diagonally at row " + rowLocation + " and column " + colLocation);
        System.out.println();
    }
}

当我发表checkDown()方法时，为什么不会发生这种情况？我该如何解决？

我很感激任何帮助。谢谢！

Answer 1

有三种情况导致失败：

因为你使用了do-while-loop，它在循环结束时检查他的状态，程序正在寻找单词＆＃34;在结束前结束
你的最后一行有一个＆＃34; e＆＃34; （那就是你的运气;-)）。所以你的分析开始了第9行，第3行。
这个条件：

if（rowLocation + i＆gt; array.length - 1＆amp;＆amp; colLocation + i＆gt; array [0] .length - 1）

又名if(9 + 1 > 9 && 3 + 1 > 14)

返回true＆amp;＆amp; false =＆gt;假在下一个条件下导致反弹：

else if(array[rowLocation + i][colLocation] != targetWord.charAt(i))

又名

else if(array[9 + 1][3] != targetWord.charAt(i))

所以改变＆amp;＆amp;到|| ...

如果你不想查看单词＆＃34; end＆＃34;用一段时间（真实）交换你的do-while并通过

if(!targetword.equals("end"))
    break;

。

Java：即使条件为假，也执行Do-while循环

1 个答案: