用 Arraylist 中的另一个预先存在的对象替换 Arraylist 中的对象

Question

我正在尝试检测我的数组列表中 2 个对象（卡片）之间的 2 个特征之一是否相同，并将其中一个对象替换为另一个。

例如ArrayList中最后3个对象如下：

King of Diamonds
Ace of Clubs
Ace of Spades

梅花 A 和黑桃 A 的共同因素是 A 值。在现实世界中，您会将黑桃 A 移到梅花 A 上，在游戏中不再显示梅花 A。

就 ArrayList 而言，这可以通过在 arraylist 中的位置用黑桃 A 替换梅花 A 并完全删除梅花 A 来实现，ArrayList 的大小为 51（52 张牌减去我们刚刚删除了一个。

在某些情况下，下面是 Card 类的外观，它包含每张卡片的特征：

package uk.ac.aber.dcs.cs12320.cards;

import java.util.ArrayList;


public class Card {

protected String value;
protected String suit;
 ArrayList<Card> cardsList = new ArrayList<>();

public Card(String v, String s){
    this.value = v;
    this.suit = s;
}

public Card() {

}

public  Card[] getAll(){
    Card[] brb = new Card[cardsList.size()];
    int tempCount = -1;
    for(Card c : cardsList){
        tempCount++;
        brb[tempCount] = c;
    }
    return brb;
}

public  void deleteAll(){
    cardsList.clear();
}

public  String getValue() {
    return value;
}

public void setValue(String value) {
    this.value = value;
}

public  String getSuit() {
    return suit;
}

public void setSuit(String suit) {
    this.suit = suit;
}

public void addCard(Card card){
    cardsList.add(card);
}

}

这是我的洗牌方法（从 .txt 文件“cards.txt”中读取所有卡片并随机排列它们的顺序）：

    private void dealCard(){
    //TODO
    int totalLeftOver = 0; // used to count the cards left in the shuffled-but-not-dealt pack
    Card topCard = shuffledPack.get(0);
    shuffledPack.remove(0);
    theFlop.add(topCard);
    System.out.print("Cards on the flop: ");
    for(Card dealt : theFlop){
        String definitelyDealt = dealt.getValue() + dealt.getSuit() + " ";
        System.out.print(definitelyDealt);
    }
    System.out.println("\n");
    for(Card card : shuffledPack){ // for loop to count how cards haven't been dealt
        totalLeftOver++;
    }
    System.out.println("Total number of cards left to deal: " + totalLeftOver); // show how many cards haven't been dealt to the player
}

如果之前发的一张牌具有匹配的价值（花色或价值），这是我用来移除一张牌的方法：

    private void makeMovePreviousPile(){
    int lastDealtCardPos = theFlop.size() - 1; //allows us to see how many cards have been dealt, are you even trying to challenge us Chris?
    int previouslyDealtCardPos = lastDealtCardPos - 1;

    if(lastDealtCardPos != 0){ // check that the deck has been shuffled and at least 1 card has been dealt.

        String lastDealtCardValue = theFlop.get(lastDealtCardPos).getValue(); // fetches the value of the last dealt card
        String lastDealtCardSuit = theFlop.get(lastDealtCardPos).getSuit(); // fetches the suit of the last dealt card
        String previouslyDealtCardValue = theFlop.get(previouslyDealtCardPos).getValue(); // fetches the 2nd to last dealt card's value
        String previouslyDealtCardSuit = theFlop.get(previouslyDealtCardPos).getSuit(); // fetches the 2nd to last dealt card's suit

        if(lastDealtCardValue.equals(previouslyDealtCardValue)){
            theFlop.remove(previouslyDealtCardPos);
        }
        else if(lastDealtCardSuit.equals(previouslyDealtCardSuit)) {
            theFlop.remove(previouslyDealtCardPos);
        }
        else {
            System.out.println("Cannot make a move. Are you sure you know the rules?");
        }
        System.out.println("\n");
        printCardsFromFlop();
    }
    else { // if it hasn't been shuffled we shun the user.
        System.out.println("Are you sure you shuffled the deck and dealt a card before trying to make a move?");
    }
    System.out.print("Total cards on the flop: " + lastDealtCardPos + "\n"); // checking to see that its working as intended
}

上述方法检测两个值中的一个是否匹配，然后从数组列表中删除之前发过的牌，而最近发过的牌在数组中取而代之（就现实生活中的棋盘本身而言，最最近发过的牌在之前发过的牌之上，在接下来的游戏中隐藏）。

我正在尝试做的类似于上面的删除方法。如果满足两个特征之一，我正在寻找替换发到 2 回合的卡片。假设数组列表现在看起来像这样，我们必须将对象 (2) 替换为对象 (0)：

Ace of Clubs
King of Diamonds
Ace of Spades

假设已经发了 3 张牌，我将如何将位置 0 的对象替换为位置 2 的对象？

Answer 1

根据您的问题，下面的代码使用 0 和 2 的硬索引。由您决定如何防止出现异常，

像这样替换

cardsList.set(0, cardsList.get(2));

如果你想去掉 2 那就用这两个

cardsList.remove(2);

Answer 2

更好的方法：

Collections.swap(list, list.indexOf(firstCardToBeReplaced), list.indexOf(secondCardToBeReplaced));

或者你可以使用ArrayList中的“set(int index, E element)”方法。

int firstPosition = list.indexOf(firstCardToBeReplaced);
int secondPosition = list.indexOf(secondCardToBeReplaced);
list.set(firstPosition, secondCardToBeReplaced);
list.set(secondPosition, firstCardToBeReplaced);

它将用新元素替换对象的位置。