如何在元素中搜索2d數組？

Question

你必須創建兩個獨立的程序，一個主程序和一個戰列艦類。其他人正在讓主程序詢問用戶猜測座標，您可以輸入設置變量來僞裝成用戶的猜測。

公共類MainMethod {

public static void main(String[] args) { 

    Battleship_BraydenH_R1 var = new Battleship_BraydenH_R1(); 

    //initialize variables 
    int[][] LAYOUT = new int[7][7]; 

    int ROW = 1, COLUMN = 2; 
    Battleship_BraydenH_R1.hitDetect(ROW, COLUMN); 

} 

}

公共類Battleship_BraydenH_R1 {

/** 
*Program Header 
*Program Name: Battleship 
*Program Description: This class tells the main method whether or not a ship has been hit or whether or not a ship has been sunk. 
*Program Creator: Brayden H 
*Revision 1 
*Date: May 9, 2016 
*/ 

public static void batClass() 
{ 
    // initialize variables 
    int[][] LAYOUT = { 
     {0,0,0,1,0,0,0}, 
     {0,0,1,0,0,0,0}, 
     {0,1,0,1,0,0,1}, 
     {1,0,0,1,0,1,0}, 
     {0,0,0,1,1,0,0}, 
     {0,0,0,1,0,0,0}, 
     {0,0,1,0,0,0,0}, 
    }; // layout array 

} // batClass 

public static boolean[] hitDetect (int X, int Y) 
{ 
    int[] counter = new int[3]; // counter array 
    //counter[0] = boat[0]; 
    boolean HIT = false, SUNK = false; // sets 'HIT' and 'SUNK' booleans to false 
    boolean[] STATUS = new boolean [2]; // array with 2 elements to be able to return 'HIT' and 'SUNK' 
    STATUS[0] = false; // MISS 
    STATUS[1] = true; // HIT 

    // if the user guessed the correct coordinates for boat 1 
    if ((X == 3 && Y == 4)||(X == 4 && Y == 4)||(X == 5 && Y == 4)) 
    { 
     STATUS[1] = true; // the boat was hit 
     X = 4; // turns 1 into 4 so the program knows when a ship is sunk 
     Y = 4; // turns 1 into 4 so the program knows when a ship is sunk 
    } // boat 1 

    // if the user guessed the correct coordinates for boat 2 
    else if ((X == 1 && Y == 4)||(X == 2 && Y == 3)||(X == 3 && Y == 2)||(X == 4 && Y == 1)) 
    { 
     STATUS[1] = true; // the boat was hit 
     X = 4; // turns 1 into 4 so the program knows when a ship is sunk 
     Y = 4; // turns 1 into 4 so the program knows when a ship is sunk 
    } // boat 2 

    // if the user guessed the correct coordinates for boat 3 
    else if ((X == 3 && Y == 7)||(X == 4 && Y == 6)||(X == 5 && Y == 5)||(X == 6 && Y == 4)||(X == 7 && Y == 3)) 
    { 
     STATUS[1] = true; // the boat was hit 
     X = 4; // turns 1 into 4 so the program knows when a ship is sunk 
     Y = 4; // turns 1 into 4 so the program knows when a ship is sunk 
    } // boat 3 

    // if a boat was not hit 
    else 
    { 
     STATUS[1] = false; // the boat was not hit 
    } 

    (THIS IS WHERE I WANT THE PROGRAM TO SEARCH THE LAYOUT) 

    // searches layout 
    double a[][]=new double[7][7]; 
    for(int row = 0 ; row < 3 ; row++) 
    { 
     for(int col = 0 ; col < 7 ; col++) 
     { 

     } 
    } 


    return STATUS; // returns 'STATUS' to main method 
} // hitDetect method 

} //戰艦類

Answer 1

首先我需要知道在哪裏的船在你的矩陣中。之後，在用戶輸入的情況下，您只需搜索矩陣以確定輸入是否與任何船舶匹配。

假設你有位置爲a [3] [3]的船。驗證的最佳方法是將所述位置的內容設置爲1（有船時），當沒有時爲0。

所以，當你得到一個用戶猜測如：x = 1; y = 4;

所有你必須是檢查所接收到的位置，以找出是否它是一個1（有船）或0（無船有）：

if(a[x][y] == 1) { 
    //user guesses correctly 
} else { 
    //user found no ship 
}