Question

我的程序应该计算数组中最长的行以及最长的行。我的计数停滞不前，我的眼睛因试图找到我的OBO（一个一个）错误而受伤。请帮忙:)谢谢！方法horizontalPath

中的错误

＆＃13;

public class game {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        // TODO code application logic here
         // TODO code application logic here
        Scanner kbd = new Scanner(System.in);
        System.out.println("ENTER A SINGLE INTEGER: ");
        int n = kbd.nextInt();
        char[][] mazeValue = new char[n+2][n+2];
        System.out.println("ENTER A PATH: ");
        for(int i = 0; i < mazeValue.length; i++ ){
            for(int j = 0; j< mazeValue[i].length; j++){
                if (i==0 || j==0 || i==n+1 || j == n+1) mazeValue[i][j] = 'X'; 
                else mazeValue[i][j]= kbd.next().charAt(0);
        }
     }
    printMaze(mazeValue);
    horizontalPath(mazeValue,n);
  }
    public static void printMaze(char newArray[][])
    {
        System.out.println("MAZE");
        for(int i = 0; i < newArray.length-2; i ++)
     {
            for (int j = 0; j < newArray[i].length-2; j++)
            {
            System.out.printf("%5c",newArray[i+1][j+1]);
            }
         System.out.printf("\n");
     }
   }
   public static void horizontalPath(char mazeValue[][], int n)
   {
       int[] totalRow = new int[n];
       
       //int horizontalPath=0;
       int count=0;
       int i;
       int j;
       for(i= 0; i<mazeValue.length-2; i++){
           for(j = 0; j<mazeValue[i].length-2; j++){
                if(mazeValue[i][j]== 'O'){
                 count++;
           }
           else{
               if(totalRow[i] < count)
                   totalRow[i]=count;
              count = 0;
           }          
        }
        if(totalRow[i] < count)
                totalRow[i]=count;
        count = 0;
    }
           int biggestRow = totalRow[0];
           //int longestRow=0;
           int finalLongestRow =0;
           for(int x =0; x <n; x++){
               if(biggestRow < totalRow[x]){
                   biggestRow = totalRow[x];
                   finalLongestRow  = x;
            }
       }
       System.out.printf("Longest horizontal path row %d length %d",finalLongestRow+1,biggestRow);

   }

＆＃13;

Answer 1

在您的主要内容System.out.println("ENTER A PATH: ");之后，在您的for循环中，它等待用户输入，但不会再次询问该路径。将System.out.println("ENTER A PATH: ");移动到该巢中的其他地方。

修改

以下是我的代码：

/**
 * @param args
 *            the command line arguments
 */
public static void main(String[] args) {
    // TODO code application logic here
    // TODO code application logic here
    Scanner kbd = new Scanner(System.in);
    System.out.println("ENTER A SINGLE INTEGER: ");
    int n = kbd.nextInt();
    char[][] mazeValue = new char[n + 1][n + 1];
    for (int i = 0; i < mazeValue.length; i++) {
        for (int j = 0; j < mazeValue[i].length; j++) {
            if (i == 0 || j == 0 || i == n + 1 || j == n + 1)
                mazeValue[i][j] = 'X';
            else {
                System.out.println("ENTER A PATH: ");
                // You know there are better ways... you are throwing away the entire line -- is that intended?
                mazeValue[i][j] = kbd.next().charAt(0);
            }
        }
    }
    printMaze(mazeValue);
    horizontalPath(mazeValue, n);
}

public static void printMaze(char newArray[][]) {
    System.out.println("MAZE");
    for (int i = 1; i < newArray.length; i++) {
        for (int j = 1; j < newArray[i].length; j++) {
            System.out.printf("%5c", newArray[i][j]);
        }
        System.out.printf("\n");
    }
}

public static void horizontalPath(char mazeValue[][], int n) {
    int[] totalRow = new int[n];

    // int horizontalPath=0;
    int count = 0;
    for (int i = 1; i < mazeValue.length; i++) {
        for (int j = 1; j < mazeValue[i].length; j++) {
            if (mazeValue[i][j] == 'O') {
                count++;
            }
        }
        if (totalRow[i - 1] < count)
            totalRow[i - 1] = count;
        count = 0;
    }
    int biggestRow = totalRow[0];
    // int longestRow=0;
    int finalLongestRow = 0;
    for (int x = 0; x < n; x++) {
        if (biggestRow < totalRow[x]) {
            biggestRow = totalRow[x];
            finalLongestRow = x;
        }
    }
    System.out.printf("Longest horizontal path row %d length %d", finalLongestRow + 1, biggestRow);
}

所以，有些事情需要注意 - 我们在每个维度（x和y）上添加一个。这是因为您的第一行是边框，第一列是边框。因此，在迭代时，我们从i开始为1，为j开始为1。

现在，记住数组从0开始，int[] totalRow = new int[n];从索引0开始是n长.n是用户输入的，但我们的二维数组是n + 1高，所以我们有一个边框。我们的totalRow数组是不同的大小，因为我们不包括第一行，因为它是一个边框。这意味着mazeValue中的行 a 是totalRow中的行 a-1 。为了从mazeValue到totalRow交叉引用，我们减去1.然而，实际上，不需要第二个数组。我们需要的只是跟踪当前最高计数的行。为此，我们可以将horizontalPath更改为：

public static void horizontalPath(char mazeValue[][], int n) {
    int largestRow = 0;
    int largestCount = 0;

    // int horizontalPath=0;
    int count = 0;
    for (int i = 1; i < mazeValue.length; i++) {
        for (int j = 1; j < mazeValue[i].length; j++) {
            if (mazeValue[i][j] == 'O') {
                count++;
            }
        }
        if (largestCount < count) {
            largestCount = count;
            largestRow = i - 1;
        }
        count = 0;
    }

    System.out.println("Longest horizontal path row " + largestRow + " length " + largestCount);
}

请注意，我修复了内部for循环中的错误。

计数器未指示为预期的2D阵列

1 个答案:

修改