获得2D阵列的边缘值，同时防止越界

Question

这里有4x4个字符的数组，我需要获得数组边缘的字符的公共值。。。我尝试了其他与我的问题类似的问题的解决方案，但我仍然得到了同样的错误。，

这是我的代码。。

//arr2[][]
//      arr2[3][0] = 'H';
//      arr2[3][1] = 'E';
//      arr2[3][2] = 'L';
//      arr2[3][3] = 'P';
//arr3[][]
//      arr3[1][3] = 'T';
//      arr3[2][3] = 'O';
//      arr3[3][3] = 'P';
//I specifically need the get the 'P' at [3][3]..
for(o = 0;o<count;o++){
        char letter = out.charAt(o);                        
        for(int m = 0; m < 4; m ++){    
            for(int n = 0; n < 4; n ++){
                if(Arrays.asList(arr3[m][n]).contains(letter)){ 
                    r = m;
                    c = n;
                }
            }
        }
        right  = arr2[r][c+1];
        left  = arr2[r][c-1];
        up  = arr2[r-1][c];
        down  = arr2[r+1][c];
        if(o==0){
                if(c==0){
                    if(r==0||r==3){
                        if(right!=null){
                            l = right;
                        }
                    }else{
                        if(right!=null){
                            l = right;
                        }else if(up!=null){
                            l = up;
                        }
                    }
                }else if(c==3){
                    if(r==0||r==3){
                        if(left!=null){
                            l = left;
                        }
                    }else{
                        if(left!=null){
                            l = left;
                        }else if(up!=null){
                            l = up;
                        }
                    }
                }else{
                    if(r==0||r==3){
                        if(left!=null){
                            l = left;
                        }else if(right!=null){
                            l = right;
                        }
                    }else{
                        if(left!=null){
                            l = left;
                        }else if(right!=null){
                            l = right;
                        }else if(up!=null){
                            l = up;
                        }
                    }
                }
            }
        }else if(o==(count-1)){
            if(vertical == 1){
                if(c==0){
                    if(r==0||r==3){
                        if(right!=null){
                            l = right;
                        }
                    }else{
                        if(right!=null){
                            l = right;
                        }else if(down!=null){
                            l = down;
                        }
                    }
                }else if(c==3){
                    if(r==0||r==3){
                        if(left!=null){
                            l = left;
                        }
                    }else{
                        if(left!=null){
                            l = left;
                        }else if(down!=null){
                            l = down;
                        }
                    }
                }else{
                    if(r==0||r==3){
                        if(left!=null){
                            l = left;
                        }else if(right!=null){
                            l = right;
                        }
                    }else{
                        if(left!=null){
                            l = left;
                        }else if(right!=null){
                            l = right;
                        }else if(down!=null){
                            l = down;
                        }
                    }
                }
            }
        }else{
            if(vertical == 1){
                if(c==0){
                    if(right!=null){
                        l = right;
                    }
                }else if(c==3){                                 
                    if(left!=null){
                        l = left;
                    }
                }else{
                    if(right!=null){
                        l = right;
                    }else if(left!=null){
                        l = left;
                    }
                }
            }
        }
        k = Character.toString(letter);
        letr = Character.toString(l);

Answer 1

在您的代码中，

for(int m = 0; m < 4; m ++){    
    for(int n = 0; n < 4; n ++){
        if(Arrays.asList(arr3[m][n]).contains(letter)){ 
            r = m;
            c = n;
        }
    }
}

在这里，你分配r和c，在某种情况下可能是0或3，然后你下面的代码，

right  = arr2[r][c+1];
left  = arr2[r][c-1];
up  = arr2[r-1][c];
down  = arr2[r+1][c];

在上面的赋值中，您有c + 1、r + 1、c - 1、r - 1，如果上一个循环将r和c赋值为0或3，则这些值肯定会超出界限。

Answer 2

你能运行这个并告诉我们它是否有帮助吗？

char[][] arr2 = ... // your array
char[][] arr3 = ... // your other array
// inspect all common indeces:
for (int i = 0; i < arr2.length && i < arr3.length; i++) {
    for (int j = 0; j < arr2[i].length && j < arr3[i].length; j++) {
        // print value, if value is the same: 
        if (arr2[i][j] == arr3[i][j]) {
            // here you know that two characters on the same index are the same.
            // you have the information about the character and the indeces:
            System.out.println("Found common value at index (i,j)->"
                    + "(" + i + "," + j + "), char '" + arr2[i][j] + "'");
        }
    }
}

使用一个类似于您评论的示例中的数组，这将给出以下输出：

Found common value at index (i,j)->(3,3), char 'P'

这似乎就是你想要的。假设两个数组具有相同的维度，此解决方案将查看两个数组中的每个索引。如果一个数组比另一个大，那么只会查看常见的索引。你提到了数组的边缘。我不清楚，如果你的意思是只有边或包括边

Answer 3

感谢那些帮助我解决这个问题的人。。我能够使用模得到数组边缘的值。。我能够阻止出界。。。https://stackoverflow.com/a/12743834/2078012这个来自另一个问题的链接回答了我的问题。。再次感谢…：-）