在Java中搜索2D数组行

Question

I am searching an 2D array that has 1 and 0 . If there is an adjacent 3 to a 0, the 0 gets converted to a 3. I have figured out how to search a row once, but I would like the same for loop to repeat until all the 0's have been converted to 3. Thanks in advance for the help.

Here is a visual depiction. So if I am searching the 2nd row, I will find the first 0 touching a 3, but then I want to go through the same row and find the next 0 that is touching a 3. I hope that clarifies! :)

1 1 1 1 1 3 1 1

1 0 1 1 0 0 1 1

1 1 1 1 0 1 1 1

etc.

I start the search on row with index 1 and I find the first 0 that touches a 3 and convert it to a 3. So I have the following, I then need to go back and repeat the process until the remainding 0 is converted to a 3 and then repeat for each row:

1 1 1 1 1 3 1 1

1 0 1 1 0 3 1 1

1 1 1 1 0 1 1 1

    public boolean searchRows(){
    boolean found = false;
    for (int c = 0; c < cave[rowIndex].length; c++){

        if(cave[rowIndex][c]==0){
            if (cave[rowIndex-1][c] == 3){
                cave[rowIndex][c] = 3;
                found=true;
            }
            else if(cave[rowIndex][c-1] == 3){
                    cave[rowIndex][c] = 3;}
            else {found=false;}
        }
    }   
    return found;
}

Answer 1

It seems like your for loop is working fine to go through the one row. It looks like you aren't checking all of the possible elements that could be touching the one you are currently looking at. It looks like you are checking the element directly above it and the one directly to the left of it. You should be checking whatever other directions your problem asks for as well. You should also be checking that you aren't going to be checking an array index that doesn't exist so that you don't get an ArrayIndexOutOfBounds error. Along with that make sure you are are setting your found boolean anytime that you actually find a zero that will be turned into a 3. Currently you are only setting the found to true in one case and not the other.