Java Can'；t在不破坏编译的情况下反转数组

目前，我正在尝试获取一个列表，以便在测试类中向后打印。我认为Collections.reverse（discardPile）可以工作，所以我使用了它，结果是混合的，至少可以这么说

这是到目前为止我的代码

           DiscardPile<Card> discardPile = null; 
       discardPile = new DiscardPile<Card>(52, 0);
           discardPile.push(new Card(8));
        discardPile.push(new Card(32));
        discardPile.push(new Card(48));
        discardPile.push(new Card(2));
        discardPile.push(new Card(17));
        discardPile.push(new Card(20)); //removeTopCard should remove all that's above
        discardPile.push(new Card(25));
        discardPile.push(new Card(50));
        discardPile.push(new Card(19));
        discardPile.push(new Card(41)); //10 Cards that must be popped
        //Collections.reverse(discardPile); //texted it out for the meantime
        System.out.println(discardPile.toString()); 

        Card[] cardArr = discardPile.removeTopCard(new Card(50));
        for(Card co : cardArr) { //for loop to print the method removeTopCards
            System.out.println(co);
            }                
        }

然后丢弃。卸下OPCARD（新卡（50））进行打印

没有集合。反向（丢弃堆） 我的代码打印

[10 of Clubs
, 8 of Hearts
, J of Spades
, 4 of Clubs
, 6 of Diamonds
, 9 of Diamonds
, A of Diamonds
, K of Spades
, 8 of Diamonds
, 4 of Spades
]
4 of Spades

8 of Diamonds

K of Spades

public class Card {

    private int rank;

    private int suit;


    public Card() {
        rank = (int) (Math.random() * 13);
        suit = (int) (Math.random() * 4);
    }


    public Card(int n) {
        if (n >= 0 && n <= 51) {
            rank = n % 13;
            suit = n / 13;
        }
    }


    public Card(int r, int s) {
        if ((r >= 0 && r <= 12) && (s >= 0 && s <= 3)) {
            rank = r;
            suit = s;
        }
    }


    public String toString() {
        String c = getRankAsString() + " " + "of " + getSuitAsString() + "\n";
        c = c.replace(",", "").replace("[", "").replace("]", "");
        return c;
        }


    public void setRank(int r) {
        if (r >= 0 && r <= 12)
            rank = r;
    }


    public void setSuit(int s) {
        if (s >= 0 && s <= 3)
            suit = s;
    }


    public int getRank() {

        return rank;
    }


    public int getSuit() {
        return suit;
    }


    public String getRankAsString() {
        String[] ranks = { "2", "3", "4", "5", "6", "7", "8", "9", "10", "J", "Q", "K", "A" }; // 0-12
        return ranks[rank];
    }


    public String getSuitAsString() {
        String[] suits = { "Clubs", "Diamonds", "Hearts", "Spades" }; // 0-3
        return suits[suit];
    }


    public boolean equals(Card otherCard) {
        return ((rank == otherCard.rank) && (suit == otherCard.suit));
    }


    public int compareByRank(Card otherCard) {
        return rank - otherCard.rank;
    }


    public int compareBySuit(Card otherCard) {
        return suit - otherCard.suit;
    }
}

它以正确的顺序打印所有括号内的数字，但下一系列字符串出现问题。它可以打印3张以上的卡片

这是我的卡片课

[10 of Clubs
, 8 of Hearts
, J of Spades
, 4 of Clubs
, 6 of Diamonds
, 9 of Diamonds
, A of Diamonds
, K of Spades
, 8 of Diamonds
, 4 of Spades
]
4 of Spades

8 of Diamonds

K of Spades

public class Card {

    private int rank;

    private int suit;


    public Card() {
        rank = (int) (Math.random() * 13);
        suit = (int) (Math.random() * 4);
    }


    public Card(int n) {
        if (n >= 0 && n <= 51) {
            rank = n % 13;
            suit = n / 13;
        }
    }


    public Card(int r, int s) {
        if ((r >= 0 && r <= 12) && (s >= 0 && s <= 3)) {
            rank = r;
            suit = s;
        }
    }


    public String toString() {
        String c = getRankAsString() + " " + "of " + getSuitAsString() + "\n";
        c = c.replace(",", "").replace("[", "").replace("]", "");
        return c;
        }


    public void setRank(int r) {
        if (r >= 0 && r <= 12)
            rank = r;
    }


    public void setSuit(int s) {
        if (s >= 0 && s <= 3)
            suit = s;
    }


    public int getRank() {

        return rank;
    }


    public int getSuit() {
        return suit;
    }


    public String getRankAsString() {
        String[] ranks = { "2", "3", "4", "5", "6", "7", "8", "9", "10", "J", "Q", "K", "A" }; // 0-12
        return ranks[rank];
    }


    public String getSuitAsString() {
        String[] suits = { "Clubs", "Diamonds", "Hearts", "Spades" }; // 0-3
        return suits[suit];
    }


    public boolean equals(Card otherCard) {
        return ((rank == otherCard.rank) && (suit == otherCard.suit));
    }


    public int compareByRank(Card otherCard) {
        return rank - otherCard.rank;
    }


    public int compareBySuit(Card otherCard) {
        return suit - otherCard.suit;
    }
}

公共类卡{
私有整数秩；
私人诉讼；
公共卡（）{
rank=（int）（Math.random（）*13）；
suit=（int）（Math.random（）*4）；
}
公共卡（int n）{
如果（n>=0&&n=0&&r=0&&s=0&&r=0&&s据我所知，您希望使用堆栈数据结构

Stack stack = new Stack();
stack.push(discardPile.push(new Card(32))); .......

然后，您可以使用
stack.peek（）获得堆栈的顶部
和所有其他与堆栈相关的操作都可以高效地完成。如果您不了解堆栈数据结构，请按照本文档操作。
请编辑您的问题，并添加DiscardPile和Card类。@Rcordoval嘿，我刚刚添加了您请求的类。
stack
在java中已过时（
class Stack extends Vector
）。应改用
Deque
。
Stack stack = new Stack();
stack.push(discardPile.push(new Card(32))); .......