Question

我正在制作一个Java纸牌游戏，现在我尝试为电脑玩家创造一个“大脑”。条件：

有3张牌（只有价值，不适合）;
计算机播放器（CP）有6张卡;
赢得一轮CP必须击败桌上至少2张牌，或者它必须击败1张牌并与另外2张牌进行平局;
当然，卡片组必须是最佳的。

实施例：在桌子上是值为1,2,3的卡.CP具有值为1,1,3,2,5,6的卡。它必须选择值为2的卡来击败Card1，而不是选择值为3的卡来击败Card2和选择值为1的卡，因为它已经击败了2张牌。现在我编写了下一个代码，但它根本不起作用或根本不起作用。

for (i = 0, j = i++, k = j++; i < oppHand.size() - 2 && j < oppHand.size() - 1 && k < oppHand.size(); i++) {

a = oppHand.get(i).getPower(); //value of card1 from CP hand
b = oppHand.get(j).getPower(); //value of card2 from CP hand
c = oppHand.get(k).getPower(); //value of card3 from CP hand

x = oppHand.indexOf(i);        //position of card1 in a CP hand
y = oppHand.indexOf(j);        //position of card2 in a CP hand
z = oppHand.indexOf(k);        //position of card3 in a CP hand

if (a > Score1 && b > Score2 ||  a> Score1 && c > Score3) {    //Score - value of the cards on the table
choice1 = x;
choice2 = y;
choice3 = z;}

else if (a > Score1 && b > Score3 || a > Score1 && c > Score2) {
choice1 = x;
choice2 = z;
choice3 = y;} ........

// moving cards from the CP hand to the table with assignment values to the piles
    validPower5 = oppHand.get(choice1).getPower();  
    discardPile5.add(0, oppHand.get(choice1));
    oppHand.remove(choice1);

    validPower6 = oppHand.get(choice2).getPower();  
    discardPile6.add(0, oppHand.get(choice2));
    oppHand.remove(choice2);

    validPower7 = oppHand.get(choice3).getPower();  
    discardPile7.add(0, oppHand.get(choice3));
    oppHand.remove(choice3);
                    }

Answer 1

会有这样的帮助吗？

sortCardsOnTable();//now Score1 <= Score2 <= Score2
sortOpponentCards();
boolean beatScore1 = false;
boolean beatScore2 = false;
int indexFirst = 0;
int indexSecond = 0;
int indexThird = 0;

for(int i = 0;i < oppHand.size();i++){
   Card c = oppHand.get(i);
   if(!beatScore1 && c.getPower() > Score1){
      indexFirst = i;
      beatScore1 = true;
   }
   else if(beatScore1 && c.getPower() > Score2){
      indexSecond = i;
      beatScore2 = true;
      indexThird = 1;//here compare with indexFirst so you dont pull the first card twice, just pull the first available card
   }
}

你应该对你的Cardlist进行排序，以简化计算

Answer 2

我认为你是否还需要正确的条件..

不确定但请检查此代码。

if (a > Score1) 
{    //Score - value of the cards on the table
    if (b > Score2 || c > Score3)   {
        choice1 = x;
        choice2 = y;
        choice3 = z;    }
    else if (b > Score3 || c > Score2)  {
        choice1 = x;
        choice2 = y;
        choice3 = z;    }
}

我认为这应该有用......

按条件从数组列表中选择3个元素

2 个答案: