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Александр
Asked:2020-11-30 19:21:42 +0000 UTC2020-11-30 19:21:42 +0000 UTC

对整数数组进行排序

  • 772

有 2 个带数字的数组。它们中的元素具有一定的依赖性，即 数组的 0 元素 1 “链接”到数组的 0 元素 2，依此类推。有必要以某种方式以相对于 Vi / Wi 的降序对对象进行排序，并将其写入新的集合或数组中。然后，如有必要，当从那里取出对象时，以某种方式获得它们的原始关系。

static int[] values = new int[] {10, 8, 3, 8, 8}; // цены
static int[] weights = new int[] {8, 3, 4, 3, 3}; //вес предметов

PS：我用贪心算法解决背包问题

java

2 个回答

  1. 你可以这样做：

    List<Integer> sortedValues = Stream
            .generate(new Supplier<Map.Entry<Integer, Integer>>() {
                private int index = -1;

                @Override
                public Map.Entry<Integer, Integer> get() {
                    index++;
                    return new AbstractMap.SimpleEntry<>(values[index], weights[index]);
                }})
            .limit(values.length)
            .sorted(Map.Entry.comparingByValue())
            .map(Map.Entry::getKey)
            .collect(toList());
    • 2
  2. Best Answer
    import java.util.Collections;
import java.util.Arrays;
import java.util.List;

public class HelloWorld{

     public static void main(String []args){
        List<Entity> entities = Arrays.asList(new Entity(3,5),new Entity(5,7),new Entity(3,1),new Entity(1,10));
        Collections.sort(entities);
        entities.forEach(System.out::println);
     }
}

class Entity implements Comparable<Entity> {

        private int weight;
        private int cost;

        public Entity(int weight, int cost) {
            this.weight = weight;
            this.cost = cost;
        }

        private double getMeasure(){
            return cost * 1.0 / weight;
        }

        @Override
        public int compareTo(Entity entity) {
            double thisMeasure = getMeasure();
            double thatMeasure = entity.getMeasure();

            if(thisMeasure < thatMeasure) return -1;
            return thisMeasure > thatMeasure ? 1 : 0;
        }

        @Override
        public String toString() {
        return "Entity{" +
                "weight=" + weight +
                ", cost=" + cost +
                ", difference=" + getMeasure() +
                '}';
    }
}
    • 2

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