用于基于网格的运动的多维数组

Question

I am practicing my programming skill using Unity3D. I have a grid set up that has coordinates in an x,y type of setup.

[0,0] to [10,10]

With all the numbers in between (IE 5,5 would be close to center of the map).

The thing that I am trying to do now is figure out a mathematical formula to calculate the coordinates my character can move. If the character is at position 5,5 and has a movement radius of 2 what is the most efficient way to return a list or an array of coordinates my character can move to? Every single grid square is its own object and have public variables for its X and Y so once I have the available results actually using them in the code isn't hard.

Here's what I'm trying so far (I'm ignoring the out of range possibility for right now, that's an easy fix):

for(int x = currentGridSquare.xCoord - myMovementRange; 
x <= currentGridSquare.xCoord + myMovementRange; x++){
for(int y = currentGridSquare.yCoord - myMovementRange;
    y <= currentGridSquare + myMovementRange; y++)
{
    //Starting at 5,5 with a movement range of 2 should
    //start this process at the value of 3,3 which
    // is incorrect
}
}

I may be too tired to actually calculate a formula for this but I've been searching and haven't come across anything so if anyone's had experience with this and knows a quick way to do it I would be greatly appreciative.

Update: The values that I am expecting this to return would be coordinates. In this example starting at 5,5, the values I'd want back would be [3,5],[4,4],[4,5],[4,6],[3,5],[4,5],[5,5],[6,5],[4,6],[5,6],[6,6] and [5,7]

Answer 1

So I used formula Ben provided and came up with:

foreach(var gridSquare in allGridSquares)
{
   if( (Mathf.Abs(myX - gridSquare.Xcoord) + (Mathf.Abs(myY - gridSquare.Ycoor) >= myMovementValue)
   {
        gridSquare.activate();
   }

}

I know that if the grid starts getting bigger than the distance formula will be more complicated and I will update this later if that's the case but for the size of my grid this works wonderfully.