Reputation: 405
I currently have following code:
int[][] legalForBlack = {{0,1},{1,0},{2,3},{3,2}};
for (int x=0;x<boardSize;x++) {
for (int y=0;y<boardSize;y++) {
if (x,y) in legalForBlack
methodA()
else
methodB()
}
}
Of course this code won't compile. I am looking for a fancy and compact way to check when (x,y) are in the given list. I can do this with 4 if-statements or a loop, but this is not a proper way imo. I am looking for something that does this in constant time.
EDIT:
I think I found a way. What do you think of this?
int[][] legalForBlack = {{0,1},{1,0},{2,3},{3,2}}; // keep in order!
int cur = 0;
for (int x=0;x<boardSize;x++) {
for (int y=0;y<boardSize;y++) {
int[] buffer = legalForBlack[cur];
if (x==buffer[0] && y==buffer[1]) {
cur++;
methodA();
} else {
methodB();
}
}
}
Upvotes: 1
Views: 96
Reputation: 2297
Heres pseudocode for arrays:
input data in array
find x with for to match first column (legalForBlack[i][0])
if x matches legalForBlack[i][0] check if legalForBlack[i][1] matches y
if yes, count it
But there is a better way, when you just want to check if they are in array. Create object Pair
with variable x
and y
, create equals()
and hashCode()
functions to have unique for each pair (like get hashCode
from string xy
), place all inputs in Set
and then check if given Pair(x,y)
is in Set
.
Upvotes: 2
Reputation: 76
I have an alternative solution for your question. Judging from your code, I assume that you are writing something chess-related and your list of legalForBlack is a series of coordinates that the player is allowed to move to. Best way IMO would be to code each square on the board with an index (0 to maximum 63) and store all of these in a Map of type <Integer, Coordinate>, where Coordinate has int x and int y. If you don't have any particular use for the coordinates, you can also skip and convert your Map to a simple Array of allowed-squares. This would only require you to check whether a given value is in your allowed-square list. I hope this gives you a better approach to your problem. Good luck!
Upvotes: 2
Reputation: 4707
boolean isLegal = false;
for(int[] coord: legalForBlack)
if(Arrays.equals(coord, someXYArray)) {
isLegal = true;
break; //Credit to Adnan Isajbegovic
}
if(isLegal)
methodA();
else
methodB();
Upvotes: 2