在2D数组中搜索字符

时间:2012-09-25 19:46:45

标签: java arrays search 2d

我正在编写一个迷宫程序。用户用字符'P'表示,我需要能够找到该字符以便为我的移动命令分配值。我很困惑如何在迷宫中找到'P'。

public static void main(String[] args) throws Exception{

    //Display the maze
    char treasureMaze[][] = {{'P','.','X','X','.'},{'.','X','.','.','.'},{'.','.','.','X','.'},{'X','X','T','.','.'},{'.','.','X','.','.'}}; 
    display(treasureMaze);


    //Give Move Options
    options();

    //Get Users Decision
    Scanner moveChoice = new Scanner(System.in);
    int choice = moveChoice.nextInt();

    if(choice == 1){
        System.out.println("You chose to Move up");
    }
    else if(choice == 2){
        System.out.println("You chose to Move down");
    }
    else if(choice == 3){
        System.out.println("You chose to Move left");
    }
    else if(choice == 4){
        System.out.println("you chose to Move right");
    }
    else{
        return;
    }


    //Move the Player
    //Move Up
    if(choice == 1){
        if(treasureMaze[0-1][0] == '.'){
            treasureMaze[0-1][0] = 'P';
            treasureMaze[0-1][0] = '.';
        }
        else if(treasureMaze[0-1][0] == 'T'){
            System.out.println("Congratulations you won!");
        }
        else{
            System.out.println("Cannot move there! Try something else");
        }
    }

    //Move Down
    else if(choice == 2){
        if(treasureMaze[0+1][0] == '.'){
            treasureMaze[0+1][0] = 'P';
            treasureMaze[0][0] = '.';
        }               
        else if(treasureMaze[0+1][0] == 'T'){
            System.out.println("Congratulations you won!");
        }
        else{                   
            System.out.println("Cannot move there! Try something else");
            }
            }

    //Move Left
    else if(choice == 3){
        if(treasureMaze[0][0-1] == '.'){
            treasureMaze[0][0-1] = 'P';
            treasureMaze[0][0] = '.';
        }
        else if(treasureMaze[0][0-1] == 'T'){
            System.out.println("Congratulations you won!");
        }
        else{
            System.out.println("Cannot move there! Try something else");
        }
    }

    //Move Right
    else if(choice == 4){
    if(treasureMaze[0][0+1] == '.'){
        treasureMaze[0][0+1] = 'P';
        treasureMaze[0][0] = '.';
    }
    else if(treasureMaze[0][0+1] == 'T'){
        System.out.println("Congratulations you won!");
    }
    else{
        System.out.println("Cannot move there! Try something else");
    }
    }
    else{
        return;
    }
    display(treasureMaze);
    options();
}



//Display Object: prints out the maze for the user
public static void display(char x[][]){
    for(int row = 0; row < x.length; row++){
        for(int column = 0; column < x[row].length; column++){
            System.out.print(x[row][column] + "\t");
        }
        System.out.println();
    }
}

//Options Object: gives the options menu to the user
 static void options(){
     System.out.println("You may:");
        System.out.println("\t1) Move up");
        System.out.println("\t2) Move down");
        System.out.println("\t3) Move left");
        System.out.println("\t4) Move right");
        System.out.println("\t0) Quit");

}

我已将其设置为从其起始位置移动“P”,但不知道如何在下次运行时找到它。有什么想法吗?

2 个答案:

答案 0 :(得分:3)

我建议跟踪当前位置,这样您就不必找到它。为玩家的水平和垂直位置声明两个变量。设置电路板时初始化它们(看起来应该是(0,0))并在播放器移动时更新它们。

答案 1 :(得分:1)

Ted的回答可能最适合您当前设置的程序。

作为替代方案,您可以用树替换2D数组。这种方法的另一个好处是你不必担心数组索引越界。每个房间都会引用其连接的其他房间,无效的指示可能只是null,您可以在玩家的当前房间周围留下参考。