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Java 无法解析方法';比较(int)和#x27 ;;_Java_Arrays_Sorting_Comparison_Bubble Sort - Fatal编程技术网

Java 无法解析方法';比较(int)和#x27 ;;

Java 无法解析方法';比较(int)和#x27 ;;,java,arrays,sorting,comparison,bubble-sort,Java,Arrays,Sorting,Comparison,Bubble Sort,我正在尝试使用冒泡排序和compareTo方法对对象数组进行排序。“。IF条件中的compareTo不起作用,并表示无法解析方法“compareTo(int)”,如何解决此问题 我的排序方法: public static void sort(Card[] hand) { for (int i = 1; i < hand.length; i++) { for (int j = 0; j < hand.length - i; j++) {

我正在尝试使用冒泡排序和compareTo方法对对象数组进行排序。“。IF条件中的compareTo不起作用,并表示无法解析方法“compareTo(int)”,如何解决此问题

我的排序方法:

public static void sort(Card[] hand) {
    
    for (int i = 1; i < hand.length; i++) {
        for (int j = 0; j < hand.length - i; j++) {
            if (((hand[j].getRankValue()).compareTo((hand[j + 1].getRankValue()))) > 0) {
                Card temp = hand[j];
                hand[j] = hand[j + 1];
                hand[j + 1] = temp;
            }
        }
    }
}
}您的
compareTo()
函数比较两张卡,因此在
if
语句中不需要任何
getRankValues()
。如果你摆脱了这些,你的代码应该可以工作

public final int suit;
public final int rank;
public int rankValue = 0;

public static final String[] SUIT = {
        "Hearts", "Clubs", "Diamonds", "Spades"};

public static final String[] RANK = {
        "2", "3", "4", "5", "6", "7",
        "8", "9", "10", "Jack", "Queen", "King", "Ace"};

public Card(int rank, int suit) {
    this.rank = rank;
    this.suit = suit;
    this.rankValue = getRankValue();
}

public int compareTo(Card other){
    if(getRankValue() > other.getRankValue()) return 1;
    else if(getRankValue() < other.getRankValue()) return -1;
    else return 0;
}

public int getRank(){
    return this.rank;
}

public int getSuit(){
    return this.suit;
}

public int getRankValue() {
    return this.suit * 13 + this.rank - 1;
}

public String toString() {
    return RANK[this.rank] + " of " + SUIT[this.suit];
}