Question

我试图检测我的arraylist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中的最后3个对象如下：

King of Diamonds
Ace of Clubs
Ace of Spades

俱乐部王牌和黑桃王牌之间的共同因素是Ace值。在现实世界中，你会将黑桃王牌移到俱乐部的王牌上，在比赛期间再也不会显示俱乐部的王牌。

就ArrayList而言，这可以通过在arraylist中的位置替换黑桃王牌并完全删除俱乐部的Ace来实现，使ArrayList的大小为51（52张卡减去我们刚刚删除的那个。

在某些情况下，以下是Card类的外观，其中包含每张卡的特征：

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shuffle方法（从.txt文件中读取所有卡片＆＃34; cards.txt＆＃34;并随机化他们的订单）：

    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
}

以下是我用来删除一张卡片的方法，如果先前发出的卡片具有匹配值（套装或价值）：

    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arraylist中删除先前处理的卡片，最近处理的卡片在arraylsit中取代（并且就现实生活中的板子本身而言，最多最近发出的牌是在先前发出的牌之上，将其隐藏在游戏的剩余部分中。）

我尝试做的与上面的删除方法类似。如果满足这两个特征中的一个，我想要更换2回的卡。让我们假设arraylist现在看起来像这样，我们必须用object（0）替换object（2）：

Ace of Clubs
King of Diamonds
Ace of Spades

假设已经发出3张牌，我将如何用位置2处的对象替换位置0处的对象？

Answer 1

更好的方法：

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

或者您可以使用方法＆＃34; set（int index，E element）＆＃34;来自ArrayList。

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。

Answer 2

以下代码根据您的问题使用0和2的硬索引。你应该决定如何防范那里的例外，

像这样替换

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

如果你想删除2 ，请使用这两个

cardsList.remove(2);

用Arraylist中的另一个预先存在的对象替换Arraylist中的对象

2 个答案: