traversing next element of 2D array

Question

    //add adj nodes
    try{
        for(int r2 = 0; r2 < rows; r2++){
            for(int c2 = 0; c2 < cols; c2++){
                Node currentNode = nodeGrid[r2][c2];
                Node rightNode = nodeGrid[r2][c2+1];
                Node bottomNode = nodeGrid[r2+1][c2];

                if(!rightNode.isWall()){
                    currentNode.addNeighbor(rightNode);
                }
                if(!bottomNode.isWall()){
                    currentNode.addNeighbor(bottomNode);
                }
            }
        }
    }catch(ArrayIndexOutOfBoundsException e){
        System.out.println("no next node");
    }

Hello, given that i have this 2D array, and i would like to access the next element i would eventually reach a index out of bounds, is there any other workaround for accessing the next element?

Test if there is a right element before accessing it: if (c2 + 1 < cols) — JB Nizet
– JB Nizet, Commented Jan 25, 2015 at 8:52

Eran · Accepted Answer · 2015-01-25 08:52:02Z

1

You must check your indices more carefully :

    for(int r2 = 0; r2 < rows; r2++){
        for(int c2 = 0; c2 < cols; c2++){
            Node currentNode = nodeGrid[r2][c2];
            Node rightNode = null;
            if (c2 < cols - 1) // the condition for the existence of a right node
                rightNode = nodeGrid[r2][c2+1];
            Node bottomNode = null;
            if (r2 < rows - 1) // the condition for the existence of a bottom node
                bottomNode = nodeGrid[r2+1][c2];

            if(rightNode != null && !rightNode.isWall()){
                currentNode.addNeighbor(rightNode);
            }
            if(bottomNode != null && !bottomNode.isWall()){
                currentNode.addNeighbor(bottomNode);
            }
        }
    }

answered Jan 25, 2015 at 8:52

Eran

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