java - 对数组进行排序并获取第 3 个最大的元素

Question

我是初学者java程序员。我有这样的作业:写一个静态方法，方法中的参数是一个 Drink 数组。该方法返回 3 个最大的酒精饮料数组元素。

我的问题是:是否有我写的更短的解决方案？

    public static void alcoholMax(AlcoholDrink[] t){
    // sort the AlcoholDrink[] t
    Arrays.sort(t, new Comparator<AlcoholDrink>() { 
        // here I try sorting the elements by their alcohol value
        @Override
        public int compare(AlcoholDrink o1, AlcoholDrink o2) {
            int at1 = (int)o1.getAlcohol(); // get the alcohol value
            int at2 = (int)o2.getAlcohol(); // get the alcohol value
            if(at1>at2)
                return -1;
            else if(at1<at2 )
                return 1;
            return 0;
        }
    });
    // get the 3 largest element of the sorted array
    double t0 = t[0].getAlcohol();
    double t1 = t[1].getAlcohol();
    double t2 = t[2].getAlcohol();
    // check, 1st > 2nd > 3rd, if this not true, return with null reference
    if(t0>t1 && t1>t2)
        System.out.println(t[0] + "\n" + t[1] + "\n" + t[2]);
    else
        System.out.println("null");
}

public static void main(String[] args){
    AlcoholDrink[] t = new AlcoholDrink [8];
    // new AlcoholDrink( String Name, String Stripping, int price, double alcohol)
    // these are hungarian wines :). If somebody curious for the languange
    t[0] = new AlcoholDrink("Kék Portói", "0.75 l", 1200, 20.5);
    t[1] = new AlcoholDrink("Kék Oportó", "0.75 l", 1100, 10.5);
    t[2] = new AlcoholDrink("Tokaji Asszú", "0.75 l ", 1600, 14.5);
    t[3] = new AlcoholDrink("Egri Bikavér", "0.75 l", 1500, 23.5);      
    t[4] = new AlcoholDrink("Egri Leányka", "0.75 l", 1100, 8.5);
    t[5] = new AlcoholDrink("Egri Merlot", "0.75 l", 1700, 18.5);
    t[6] = new AlcoholDrink("Egri Medina", "0.75 l", 900, 16.5);
    t[7] = new AlcoholDrink("Törley Talisman", "0.75 l", 750, 4.5);     

    alcoholMax(t);
    // It is always return with "null"

Answer 1

如果您的 getAlcohol() 方法返回 double，那么您不应该将其转换为 int，这会导致精度损失。此外，您可以自动比较 double ，而不是自己进行比较，如下所示:

Arrays.sort(t, new Comparator<AlcoholDrink>() { 
        // here I try sorting the elements by their alcohol value
        @Override
        public int compare(AlcoholDrink o1, AlcoholDrink o2) {
           return o1.getAlcohol().compareTo(o2.getAlcohol());
        }
    });

您还可以使您的 Alcohol 类实现 Comparable 接口(interface)，如 this 所示。例子。

最后，如果您想坚持使用自己的代码，您可能需要考虑对比较方法返回的值进行更改，如下所示:

 @Override
        public int compare(AlcoholDrink o1, AlcoholDrink o2) {
            int at1 = (int)o1.getAlcohol(); // get the alcohol value
            int at2 = (int)o2.getAlcohol(); // get the alcohol value
            if(at1>at2)
                return -1;
            else if(at1<at2 )
                return 1;
            return 0;
        }

我目前无法测试代码，但您可能会以升序方式对数组进行排序。