Is there an efficient algorithm for calculating which tiles are within a set walking distance of your character in a 2d grid? - javafx

Question

Lets say that on a 2d grid of 20x20, you have your character at position (5,5).
He is able to walk up to 4 tiles in his move. However, there may be obstacles blocking your path such as a wall.
Is there any efficient / easy way for calculating exactly which tiles he would be able to walk to without checking every single possible move ( e.g. move up 0 and right 0 then move up 0 and right 1 e.t.c )?
At the moment I'm calculating the places that you can walk through with this horrific thing:
int playerx = GridPane.getRowIndex(button);
int playery = GridPane.getColumnIndex(button);
int position = playery*8+playerx;
for (int i = 0; i < 5; i++)
{
for (int j = i-4; j < 5-i; j++)
{
try
{
int expectedCollumn = playerx+j;
int actualCollumn = ((position+i+j*8)-((position+i+j*8)%8))/8;
if(expectedCollumn==actualCollumn)
{
Button temp = (Button)gridPane.getChildren()
.get(position+i+j*8);
if (!temp.getText().equals("W") &&
!temp.getText().equals("P"))
{
temp.setText("T");
}
}
actualCollumn = ((position-i+j*8)-((position-i+j*8)%8))/8;
if(expectedCollumn==actualCollumn)
{
Button temp2 = (Button)
gridPane.getChildren().get(position-i+j*8);
if (!temp2.getText().equals("W") &&
!temp2.getText().equals("P"))
{
temp2.setText("T");
}
}
}
}
}
However, its showing as if you are able to walk to the otherside of the wall and I'm not sure how I would go about fixing this.
Many thanks in advance.

Answer 1

For path finding, you should figure out how this works:
https://en.wikipedia.org/wiki/Dijkstra%27s_algorithm
and then move on to A* or something more efficient.

Answer 2

Thanks to everyone that answered but the solution was simple
incase somehow someone finds this post and is interested it was a simple recursive call
void getReachableTiles(Tile current, Int stamina, List<Tile> visited, List<Tile> reachable) {
if (stamina <= 0) return;
List<Tile> neighbours = new List<>(current + up, current + left, ..)
for (Tile t in neighbours) {
if (!visited.contains(t)) {
visited.append(t);
if (!t.isWall()) {
reachable.append(t);
getReachableTiles(t, stamina - 1, visited, reachable);
}
}
}
}