Creating a Connect-Four Game in Java

Problem:

Connect four is a two-player board game in which the players alternately drop colored disks into a seven-column, six-row vertically suspended grid, as shown below.

                                                 

The objective of the game is to connect four same-colored disks in a row, a column, or a diagonal before your opponent can do likewise. The program prompts two players to drop a RED or YELLOW disk alternately. Whenever a disk is dropped, the program re-displays the board on the console and determines the status of the game (win, draw, or continue).

Output:

| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
---------------
Drop a red disk at column (0–6): 
0
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
|R| | | | | | |
---------------
Drop a yellow disk at column (0–6): 
1
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
|R|Y| | | | | |
---------------
Drop a red disk at column (0–6): 
2
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
|R|Y|R| | | | |
---------------
Drop a yellow disk at column (0–6): 
3
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
|R|Y|R|Y| | | |
---------------
Drop a red disk at column (0–6): 
4
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
|R|Y|R|Y|R| | |
---------------
Drop a yellow disk at column (0–6): 
3
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | |Y| | | |
|R|Y|R|Y|R| | |
---------------
Drop a red disk at column (0–6): 
2
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | |R|Y| | | |
|R|Y|R|Y|R| | |
---------------
Drop a yellow disk at column (0–6): 
1
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | | | | | | |
| |Y|R|Y| | | |
|R|Y|R|Y|R| | |
---------------
Drop a red disk at column (0–6): 
2
| | | | | | | |
| | | | | | | |
| | | | | | | |
| | |R| | | | |
| |Y|R|Y| | | |
|R|Y|R|Y|R| | |
---------------
Drop a yellow disk at column (0–6): 
1
| | | | | | | |
| | | | | | | |
| | | | | | | |
| |Y|R| | | | |
| |Y|R|Y| | | |
|R|Y|R|Y|R| | |
---------------
Drop a red disk at column (0–6): 

1
| | | | | | | |
| | | | | | | |
| |R| | | | | |
| |Y|R| | | | |
| |Y|R|Y| | | |
|R|Y|R|Y|R| | |
---------------
Drop a yellow disk at column (0–6): 
2
| | | | | | | |
| | | | | | | |
| |R|Y| | | | |
| |Y|R| | | | |
| |Y|R|Y| | | |
|R|Y|R|Y|R| | |
---------------
Drop a red disk at column (0–6): 
4
| | | | | | | |
| | | | | | | |
| |R|Y| | | | |
| |Y|R| | | | |
| |Y|R|Y|R| | |
|R|Y|R|Y|R| | |
---------------
Drop a yellow disk at column (0–6): 
3
| | | | | | | |
| | | | | | | |
| |R|Y| | | | |
| |Y|R|Y| | | |
| |Y|R|Y|R| | |
|R|Y|R|Y|R| | |
---------------
Drop a red disk at column (0–6): 
4
| | | | | | | |
| | | | | | | |
| |R|Y| | | | |
| |Y|R|Y|R| | |
| |Y|R|Y|R| | |
|R|Y|R|Y|R| | |
---------------
Drop a yellow disk at column (0–6): 
3
| | | | | | | |
| | | | | | | |
| |R|Y|Y| | | |
| |Y|R|Y|R| | |
| |Y|R|Y|R| | |
|R|Y|R|Y|R| | |
---------------
The yello player won.

Solution:

 import java.util.Scanner;
  
//This will be.. a huge code so we will have to make
//lots of methods for each aspect of the game
public class ConnectFour
{
  //We need to first create the basic visual pattern 
  public static String[][] createPattern()
  {
    //Although the game is more like a table of 6
    //columns and 6 rows, we're going to have to make
    //a 2D array of 7 rows and 15 columns because graphically
    //there's an extra row to show the ___ at the bottom
    //and you have double the columns that show | | | 
    //between each number 
     String[][] f = new String[7][15];
    
    //Time to loop over each row from up to down
    for (int i =0;i<f.length;i++)
    {
      
       //Time to loop over each column from left to right
       for (int j =0;j<f[i].length;j++)
      {
        //Note how it is always the even column
        //that has the border and the odd column
        //between them that will be either empty or
        //have a number 
        if (j% 2 == 0) f[i][j] ="|";
        else f[i][j] = " ";
        
        //Time to make our lowest row
        if (i==6) f[i][j]= "-";
      }
      
    }
    return f;
  }
  
  //Yes, we even need to make a new method for visually
  //printing our game, but at least it's not hard to do
  public static void printPattern(String[][] f)
  {
    for (int i =0;i<f.length;i++)
    {
      for (int j=0;j<f[i].length;j++)
      {
        System.out.print(f[i][j]);
      }
      System.out.println();
    }
  }
  
  //Here's are basic move, making the lowest empty row
  //of a specific column have a Red
  public static void dropRedPattern(String[][] f)
  {
    //We need to have the user tell us what column he wants
    //to drop a red into
    //Note: the user isn't supposed to know that we have 15 columns
    //starting at index 0 till 14 but just 6 nice ones
    System.out.println("Drop a red disk at column (0–6): ");
    Scanner scan = new Scanner (System.in);
    
    //Thankfully, there's a simple formula for converting a 1-2-3-4-5-6 
    //user column number into a 1-3-5-7-9-11-13
    int c = 2*scan.nextInt()+1;
   
    //Now that we know our column, we have to loop
    //over each row from the bottom to the top
    //till we find the first  empty space, drop, and
    //then finish (i.e., break) the move
    //Note: although as coders we're used to starting from 
    //0 to the end, here that wouldn't work so well because
    //it would involve multiple if statements, but try it out
    //on your own if you want to
    for (int i =5;i>=0;i--)
    {
      if (f[i][c] == " ")
      {
        f[i][c] = "R";
        break;
      }
      
    }
  }
  
  //Same as the above step, just yellow
  public static void dropYellowPattern(String[][] f)
  {
    System.out.println("Drop a yellow disk at column (0–6): ");
    Scanner scan = new Scanner (System.in);
    int c = 2*scan.nextInt()+1;
    for (int i =5;i>=0;i--)
    {
      if (f[i][c] == " ")
      {
        f[i][c] = "Y";
        break;
      }
      
    }
  }
  

  //Here's where it gets hard.
  //That's because there are basically four patterns
  //of Reds or Yellows that can win the game
  //One pattern is a horizontal line of four Rs or Ys,
  //another is a vertical line, another is a left-up to right-down
  //diagonal line, and the last is left-down to right-up diagonal,
  //We thus need to code for each type of line
  //and the various places where the line can be
  public static String checkWinner(String[][] f)
  {
    
    //Time to look for the first type of winning line,
    //a horizontal line
    //This line can be on any row, so let's loop over 
    //each row starting from 0 to 5 (since 6 is just ___)
    for (int i =0;i<6;i++)
    {
      //On every row, the four-dotted line can look like
      //----_ _,  _----_, or _ _----
      //Here, _ can be an empty space or one of the colors 
      //and - is not empty space  AND every - has the same
      // color (R or Y)
      //Note: since our R/Y/Empty's can only be in odd places,
      //because of how we created the pattern in the first
      //method, then our count has to be incremented by 2 
      //and will start from 0 (which will be 1, ----_ _) 
      //and stop at 6 (which will be 7, _ _----)
      for (int j=0;j<7;j+=2)
      {
        if ((f[i][j+1] != " ")
        && (f[i][j+3] != " ")
        && (f[i][j+5] != " ")
        && (f[i][j+7] != " ")
        && ((f[i][j+1] == f[i][j+3])
        && (f[i][j+3] == f[i][j+5])
        && (f[i][j+5] == f[i][j+7])))

        //If we found a same-colored pattern, we'll return 
        //the color so that we will know who won
          return f[i][j+1];  
      }
    }
    
    //For a vertical line, let's first loop over each
    //odd-numbered column by incrementing with 2
    //and check for consecutive boxes in the same column
    //that are the same color
    //Note: make sure you understand the horizontal line's
    //codes first or else everything below this point will 
    //make no sense to you
    for (int i=1;i<15;i+=2)
    {
      //Of course, our lines will look like ----__ but reversed
      //and there is need to our rows by 2 but just one
      //and we have to start at the vertical version of ----__ and 
      //and stop at _ _ ---- so it's from 0 to 2
      for (int j =0;j<3;j++)
      {
            if((f[j][i] != " ")
            && (f[j+1][i] != " ")
            && (f[j+2][i] != " ")
            && (f[j+3][i] != " ")
            && ((f[j][i] == f[j+1][i])
            && (f[j+1][i] == f[j+2][i])
            && (f[j+2][i] == f[j+3][i])))
              return f[j][i];  
      }  
    }
    
    //For the left-up to right-down diagonal line
    //We'll have to loop over the 3 uppermost
    //rows and then go from left to right column-wise
    for (int i=0;i<3;i++)
    {
      
      //As expected, our uppermost box will start from 1
      //and increase by 2 until it becomes 7 (the 3rd box
      //on a row)
      //Note how we used 1 instead 0 for the count here
      //There's no real reason to use 1 instead of 0 or 
      //vice versa, since we're still using an odd index
      //for the columns and incrementing by 2
      for (int j=1;j<9;j+=2)
      {
            if((f[i][j] != " ")
            && (f[i+1][j+2] != " ")
            && (f[i+2][j+4] != " ")
            && (f[i+3][j+6] != " ")
            && ((f[i][j] == f[i+1][j+2])
            && (f[i+1][j+2] == f[i+2][j+4])
            && (f[i+2][j+4] == f[i+3][j+6])))
              return f[i][j];  
      }  
    }
    
    //Similar to the method above, but we're just reversing our
    //trajectory, i.e. we're starting from the rightmost column
    //instead of the leftmost like we did above
    for (int i=0;i<3;i++)
    {
      for (int j=7;j<15;j+=2)
      {
            if((f[i][j] != " ")
            && (f[i+1][j-2] != " ")
            && (f[i+2][j-4] != " ")
            && (f[i+3][j-6] != " ")
            && ((f[i][j] == f[i+1][j-2])
            && (f[i+1][j-2] == f[i+2][j-4])
            && (f[i+2][j-4] == f[i+3][j-6])))
              return f[i][j];  
      }  
    }
    
    //If after going over the table and we find no
    //same colored lines, then we have to return something
    //that says that we didn't find a winning color :P
    return null;
  }
  
  //The easy part: using these methods
  public static void main (String[] args)
  {
    //Time to make a pattern
    String[][] f = createPattern();
    //Time to make a condition for our game to keep on
    //playing
    boolean loop = true;
    //We need something to keep track of whose turn it is
    int count = 0;
    printPattern(f);
    while(loop)
    {
       //Let's say that Red gets the first turn and thus
       //every other turn 
       if (count % 2 == 0) dropRedPattern(f);
       else dropYellowPattern(f);
       count++;//We need to keep track of the turns
       printPattern(f);
       //Let's say we want to check for a winner during every
       //turn made and say who it is
       if (checkWinner(f) != null)
       {
          if (checkWinner(f) == "R")
             System.out.println("The red player won.");
          else if (checkWinner(f)== "Y")
            System.out.println("The yello player won.");
         //Well, if someone one, then the game has to end
         loop = false;
    }
  }
}
}