Question

我试图弄清楚如何在我自己的Java中创建一个拼图求解器但是卡住了。我能够读取文件的输入，其中填字游戏的尺寸存储在一个数组中。我在搜索实际拼图中的单词时遇到问题。这就是我到目前为止所做的：

package crossWrdNMaze2;

import java.io.File;
import java.util.Scanner;

public class crossWord {



public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(new File("puzzle.txt"));
        //r and c describe the dimensions of the puzzle array
        int r = sc.nextInt();
        int c = sc.nextInt();
        char[][] array = new char[r][c];
        //to scan puzzle chars into two-dimensional array
        for (int i = 0; i<r; i++){
            String getChar = new String(sc.next());
            for(int j=0; j<c; j++){
                array[i][j] = getChar.charAt(j);
            }//end for2
        }//end for1
        //Test to make sure array was filled
        for(int i=0; i<r; i++){
            for(int j=0;j<r;j++)
                System.out.print(array[i][j]);
        System.out.println("");
        }//end for

        //scan in the number of words to be found in array (these integers follow the lines of 
        //letters which make up the array in the text file
        int num = sc.nextInt();
        System.out.println(num);

        for(int i=0; i<num; i++){
            //scan in the first word to be found
            String getWord = new String(sc.next());
            //determine word length to find size of array to look for in puzzle
            int z = getWord.length();
            char array2[] = new char[z];
            //scan new string (word we are looking for) into an array so we can look for one letter
            //at time in puzzle
            for(int j=0; j<z;j++){
                array2[j] = getWord.charAt(j);
            }//end for2
            //Size two array for the purpose of passing by reference so we can get coordinates of
            //word in puzzle
            int[] result = new int[1];
            //pass array(puzzle), array2(word we are looking for), result(coordinates of word in array),
            //r(number of rows in puzzle), c (number of columns in puzzle), z (length of word we
            //are searching for)
            findWord(array, array2, result, r, c, z);
            //print loaction of word each time through loop
            System.out.println(getWord + " is located at position (" + result[0] + ") (" + result[1] + ")");
        }//end for1

        sc.close();
    }//end main

static void findWord(char arr1[][], char arr2[], int[] result, int a, int b, int c){
        for(int i=0; i<a;i++){
            for(int j=0; i<b;j++){
                for(int k = 0; k<c;k++){
                    if(arr1[i][j] != arr2[k]){
                        System.out.println("Word not found!");//test to see if comparing letters and giving false return
                    }else if(arr1[i][j]== arr2[k]){
                    System.out.println("progress");//test to check if returning true when correct

                }//end if
            }//end for3
        }//end fosr2
    }//end for1
}//end function
}//end class

编辑：我在代码中添加了一些评论，我也记得最近我读过的一篇文章生日快乐堆栈溢出，我是一个很长时间的读者，但这是我第一次发布。再次感谢您的时间

Answer 1

您可以构建问题的一种方法是将行，列和最终的对角条带提取到字符串（或字符数组）中，并针对该字符串执行测试。

for (int row=0; row<rows; row++) {
    String strip=extractRow(row, arr);
    if (strip.contains(word)) { ... }
}
for (int col=0; col<cols; col++) {
    String strip=extractCol(col, arr);
    if (strip.contains(word)) { ... }
}

如果你只是专注于提取你想要比较你的单词的字符序列，它将使你的问题更容易管理。

我正在尝试在java中创建一个填字游戏解算器

1 个答案: