![](/img/trans.png)
[英]Implementing a second battleship into a one dimensional game “River Battleship”
[英]Implementing a second ship into one dimensional battleship game Java
我將戰艦游戲設計為只隱藏一艘船,現在我必須在游戲中實現另一艘船。 我是 Java 新手,想知道是否有人可以為我提供一種簡單的方法來解決這個問題。 我需要創建另一種方法還是只調整顯示河流方法?
public static void displayRiver(int[] river, boolean showShip) {
System.out.println();
System.out.print("|");
for (int val : river) {
switch (val) {
case -1: // No Ship
System.out.print("x");
break;
case 0: // Unknown
System.out.print(" ");
break;
case 1: // Ship Found
System.out.print(showShip ? "Y" : " ");
break;
}//switch
System.out.print("|");
}//for
System.out.println();
System.out.println();
}//displayRiver
// main method
public static void main(String[] arg) {
int riverLength = promptForInt("Please, enter the river lenght");
int [] shipArray = new int[riverLength];
int randomBattleshipLocation = new Random().nextInt(riverLength);
shipArray[randomBattleshipLocation] = 1;
boolean showShip = false ;
int userGuess;
do
{
displayRiver (shipArray, false);
userGuess = promptForInt(String.format("Guess, enter a location from 1 to " + riverLength));
userGuess = userGuess -1;
if(shipArray[userGuess] == 1)
{
System.out.println("Boom! ");
showShip = true;
displayRiver(shipArray, true);
}
else if(shipArray[userGuess] == -1)
{
System.out.println("Location was already hit, try again! ");
}
else if(shipArray[userGuess] == 0)
{
System.out.println("Splash...");
shipArray[userGuess] = -1 ;
}
} while(!showShip);
System.exit(0);
}
}
您的邏輯似乎是數組中的1
表示一艘船,而您的船的寬度顯然永遠不會超過 1。
您當前使用以下內容來創建一艘船
int randomBattleshipLocation = new Random().nextInt(riverLength);
shipArray[randomBattleshipLocation] = 1;
所以你可以把它變成一個創建戰艦的方法,然后為多艘船調用任意多次。 只要確保您沒有將一艘船分配到另一艘船的頂部,或者犯另一個邏輯錯誤(例如嘗試將 5 艘船放入大小為 4 的河流中,它會一直循環嘗試為船找到空間)。
偽代碼和非偽代碼:
for(int i = 0;i < shipsToAdd; i++) {
addShip(shipArray);
}
// Use a shared class-level Random object, don't do new Random().nextInt();
private static Random rnd = new Random();
private static addShip(int[] array) {
// Here you should loop to check if the array is already full of ships
// otherwise it's a design flaw that will result in an infinite loop with bad input
// loop until we find a shipless array index
int index = rnd.nextInt(array);
while(array[index] == 1)
index = rnd.nextInt(array);
array[index] = 1;
}
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.