Searching Squares in 2D Array

Question

Hi, I am new to Java and I am currently trying to learn how to do multiple row searching in 2D array..
I have an input of character array that starts like this.
private static final char[][] bigSquare =     {
          { 'F', 'G', '3', '4'},
          { 'F', 'G', 'C', 'D'},
          { 'E', 'F', 'G', 'H'},
          { 'E', 'F', 'G', 'H'}};
private static final char[][] smallSquare ={
          { 'F', 'G'},
          { 'F', 'G'}};I am suppose to find the number of occurence of the small squares in the bigSquare in this 2D character array.. but I have no idea how to start,
I currently have this code
//b is the big square, r is the small square, bSize is the width of the big square, rSize is the width of the small square
             public static int checkOccurence(char[][] b, char[][] r, int bSize, int rSize) {
          int n = 0; //to return
          // TODO Search for square occurrences..
          for (int i = 0; i < b[0].length; i++) {
               for (int j = 0; j < b.length; j++) {
               System.out.println();
          return n;
     }The above example should return 2
Can someone point me in the right direction? :(
Edited by: 873045 on Jul 17, 2011 2:01 PM

Answer 1

mGx-Alander wrote:
Thanks Winston and Ram
I got it :DWell done. And isn't it much more satisfying when you get there by yourself?
Now to the critique - and don't worry, these are all things that you'll learn as you go on.
First: I don't like those 'bS' and 'rS' names (what is rS?). The 'bigSquare' and 'smallSquare' that you had originally were much better, and there's nothing to stop you using them in your checkOccurrence() method too. Just rename the parameters.
Second, your code assumes that your matrices are squares. In this case, you'll get the right answer (if the code is right; and at first glance it appears to be), but it makes the solution "brittle". What if the next assignment is to find a small rectangle inside a big one? Wouldn't it be nice to reuse the same code?
Third: I rather like your idea of banging the entire bigSquare into a 'line' of Locations - quite clever - but I suspect you can go even further. Have a think about it.
Fourth: You've got quite a lot of 'loops inside loops', which usually means that you haven't broken the problem down enough, and makes the code quite hard to read.
Now that you've got some working code, I'll give you my "pseudo description" of the solution:set 'numberFound' to 0.
for each cell in 'bigSquare', starting from the top-left:
   if its value matches the top-left corner of 'smallSquare'
      compare the 'internal square' starting here with the contents of 'smallSquare'.
      if the contents match
         add 1 to 'numberFound'.
end-for-each.Obviously, it's not detailed enough - the "compare the 'internal square'" line needs some fleshing out - but basically it describes the major steps.
So what about "compare the internal square"?
Get another piece of paper and describe how that works (which is basically all the inner stuff starting from "for (int r = 0; r < rS.length; r++) {...").
Now, most importantly, when you do your translation into Java, make all your pieces of paper separate methods. It's not a hard and fast rule; but it's not a bad one to start out with.
I think I might define the method something like:public static boolean completeMatch(char[][] bigSquare,
   int atRow, int atColumn, char[][] smallSquare) {...and then your outer loop might look something like:for (int i = 0; i < loc.length; i++) {
   int r = loc.getRow();
int c = loc[i].getCol();
boolean found = false;
if (bS[r][c] == rS[0][0])
found = completeMatch(bS, r, c, rS);
if (found)
++n;
}a bit neater, don't you think?
There are many other ways to do it (for example, you could make the 'top left corner' check the first thing in your 'completeMatch()' method; I probably would), but the main idea is: break things down.
If you find yourself writing reams of code in the same method (my usual rule of thumb is about a screenfull): Stop; and think about how you could break it up.
HIH
Winston