连接四个绘制条件java

Question

此代码适用于连接四个游戏，除了绘制条件。我认为对于绘制条件我必须检查所有单元格并检查它们是否已满。我不知道如何循环所有单元格并检查它们是否为空，如果它们是完全返回＆＃34;绘制＆＃34 ;.我在这里写的只是检查板的第一个原始。这听起来很容易，但我不知道该怎么做。任何评论都赞赏。

 import  java.util.Scanner;

 public class conncetFour {
 public static String[][] createpattern(){
     String [][] f = new String[7][15];
    for (int i=0;i<f.length ; i++){
        for (int j=0 ; j<f[i].length ; j++){
            if (j%2 == 0) f[i][j] = "|";
            else f[i][j] = " " ;
            if (i == 6 ) f[i][j] = "-";
        }
    }
    return f;
}

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();
    }

}
public static void dropRedpattern(String [][] f){
    System.out.println("Drop a red disk at column (0-6): ");
    Scanner input = new Scanner(System.in);
    int c = 2*input.nextInt()+1;
    for (int i=5;i>= 0;i--){
        if (f[i][c] == " "){
            f[i][c] = "R";
            break;
        }
    }
}
public static void dropyellowpattern(String[][] f){
    System.out.println("Drop a yellow disk at column (0-6): ");
    Scanner input = new Scanner(System.in);
    int c = 2*input.nextInt()+1;
    for(int i =5;i>= 0 ; i--){
        if (f[i][c] == " "){
            f[i][c] = "Y";
            break;
        }
    }
}
public static String checkwinner(String[][] f){
    for(int i =0;i<6;i++){
        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])))
                return f[j][j+1];
        }
    }
    for (int i=1;i<15;i+=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(int i=0;i<3;i++){
        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];
        }
    }
    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];
        }
    }
 //checking for draw condition 

    for(int i=0;i<6;i++){
        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] != " "))
                return "draw";
        }
    }

    return null;
}
public static void main(String[] args){
    String[][] f = createpattern();
    boolean repeat = true;
    int count = 0;
    printpattern(f);
    while(repeat){
        if (count % 2 == 0) dropRedpattern(f);
        else dropyellowpattern(f);
        count++;
        printpattern(f);
        if(checkwinner(f) != null ){
            if (checkwinner(f) == "R")
                System.out.println("red palyer won!");
            else if (checkwinner(f)== "Y")
                System.out.println("yellow player won!");
            else if (checkwinner(f) == "draw")
                System.out.println("Tha game draw!");
            repeat = false;
          }

      }
  }

 }