Question

我的代码的输出是：

The fibonacci list that smaller than 40 is:
0 1 1 2 3 5 8 13 21 34 
The prime list that smaller than 40 is:
2 3 5 7 11 13 17 19 23 29 31 37

我想在这两个列表之间建立交集。

使其变为：（当我在fibo（）和allPrime（）方法中放入变量n = 40时）

2 3 5 13

但是我不知道该怎么做。我已经搜索了论坛，并且大多数交集问题是在两个arraylist或两个setlist之间。

我想知道是否可以像这样在两个函数之间进行交集？

public class FiboAndPrime {

    static boolean IsPrime(int n) {
        for (int i = 2; i <= Math.sqrt(n); i++) {
            if (n % i == 0)
                return false;
        }
        return true;
    }

    // Find all the prime numbers that are less than or equal to n
    static void allPrime(int n) {
        for(int i=2; i<=n; i++) {
            if(IsPrime(i)) System.out.print(i+ " ");
        }
    }

    //Find the Fibonacci numbers that are less than or equal to n
    static void fibo(int n) {
        int fibo[] = new int[n];
        fibo[0] = 0;
        fibo[1] = 1;
        System.out.print(fibo[0]+" "+fibo[1]+" ");
        for (int i = 2; i <= n; i++) {
            fibo[i] = fibo[i - 1] + fibo[i - 2];
            if (n >= fibo[i]) {
                System.out.print(fibo[i]+ " ");
            } else {
                break;
            }
        }
    }

    public static void main(String[] args) {
        int k = 40;
        System.out.println("The fibonacci list that smaller than " + k + " is:");
        fibo(k);
        System.out.println();
        System.out.println("The prime list that smaller than " + k + " is:");
        allPrime(k);
    }

}

我试图将代码更改为使用ArrayList，但是我陷入了fibo（）方法的困境。

输出为：

The final intersection that are both fabonacci and prime is:
0 1true true true true true true true true 
The prime list that smaller than 40 is:
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37]

为什么它在我的fibo列表中变成布尔类型？

static void allPrime(int n) {
    List<Integer> primes = new ArrayList<Integer>(n);
    for(int i=2; i<=n; i++) {
        if(IsPrime(i)) {
            primes.add(i);
        }
    }
    System.out.print(primes);
}

static void fibo(int n) {
    List <Integer> fibos = new ArrayList<>(n);
    int fibo[] = new int[n];
    fibo[0] = 0;
    fibo[1] = 1;
    System.out.print(fibo[0]+" " + fibo[1]);
    for (int i = 2; i <= n; i++) {
        fibo[i] = fibo[i - 1] + fibo[i - 2];
        if (n >= fibo[i]) {
           int in =fibo[i];
           System.out.print(fibos.add(in)+ " ");
        } else {
            break;
        }
    }
}

Answer 1

您将需要使用HashSet或ArrayList之类的数据结构来执行此操作，然后找到它们之间的交集。

使用ArrayList的解决方案：

import java.util.List;
import java.util.ArrayList;

public class FiboAndPrime {


    static boolean IsPrime(int n) {
        for (int i = 2; i <= Math.sqrt(n); i++) {
            if (n % i == 0)
                return false;
        }
        return true;
    }
    // Find all the prime numbers that are less than or equal to n
    static void allPrime(int n, List<Integer> prime_set ) {
        for(int i=2; i<=n; i++) {
            if(IsPrime(i)) System.out.print(i+ " ");
            prime_set.add(i);
        }
    }
    //Find the Fibonacci numbers that are less than or equal to n
    static void fibo(int n, List<Integer> fibo_set ) {
        int fibo[] = new int[n];
        fibo[0] = 0;
        fibo[1] = 1;
        System.out.print(fibo[0]+" "+fibo[1]+" ");
        for (int i = 2; i <= n; i++) {
            fibo[i] = fibo[i - 1] + fibo[i - 2];
            if (n >= fibo[i]) {
                System.out.print(fibo[i]+ " ");
                fibo_set.add(fibo[i]);
            } else {
                break;
            }
        }
    }


    public static void main(String[] args) {
        int k = 40;


        System.out.println("The fibonacci list that smaller than " + k + " is:");
        List<Integer> fibo_set = new ArrayList<Integer>();
        fibo_set.add(0);
        fibo_set.add(1);
        List<Integer> prime_set = new ArrayList<Integer>();
        fibo(k,fibo_set);
        System.out.println();
        System.out.println("The prime list that smaller than " + k + " is:");
        allPrime(k,prime_set);

    fibo_set.retainAll(prime_set); // fibo_set now contains only elements in both sets

    System.out.println();
    System.out.println("intersection between the fibo and prime set:");
    for (Integer intersection : fibo_set) {
    System.out.println(intersection);
    }

    }
}

使用HashSet的解决方案：

import java.util.Set;
import java.util.HashSet;

public class FiboAndPrime {


    static boolean IsPrime(int n) {
        for (int i = 2; i <= Math.sqrt(n); i++) {
            if (n % i == 0)
                return false;
        }
        return true;
    }
    // Find all the prime numbers that are less than or equal to n
    static void allPrime(int n, Set<Integer> prime_set ) {
        for(int i=2; i<=n; i++) {
            if(IsPrime(i)) System.out.print(i+ " ");
            prime_set.add(i);
        }
    }
    //Find the Fibonacci numbers that are less than or equal to n
    static void fibo(int n, Set<Integer> fibo_set ) {
        int fibo[] = new int[n];
        fibo[0] = 0;
        fibo[1] = 1;
        System.out.print(fibo[0]+" "+fibo[1]+" ");
        for (int i = 2; i <= n; i++) {
            fibo[i] = fibo[i - 1] + fibo[i - 2];
            if (n >= fibo[i]) {
                System.out.print(fibo[i]+ " ");
                fibo_set.add(fibo[i]);
            } else {
                break;
            }
        }
    }


    public static void main(String[] args) {
        int k = 40;


        System.out.println("The fibonacci list that smaller than " + k + " is:");
        Set<Integer> fibo_set = new HashSet<Integer>();
        fibo_set.add(0);
        fibo_set.add(1);
        Set<Integer> prime_set = new HashSet<Integer>();
        fibo(k,fibo_set);
        System.out.println();
        System.out.println("The prime list that smaller than " + k + " is:");
        allPrime(k,prime_set);

    fibo_set.retainAll(prime_set); // fibo_set now contains only elements in both sets

    System.out.println();
    System.out.println("intersection between the fibo and prime set:");
    for (Integer intersection : fibo_set) {
    System.out.println(intersection);
    }

    }
}

Answer 2

绝对有可能。我不会为您编写代码，但是我可以建议您应该怎么做。您应该将它们存储在2个不同的数组中并尝试查找它们之间的共同点，而不是打印40以下的斐波那契数和素数。但是我认为您不熟悉arraylist和set，我建议您使用另一种方法。

当斐波那契数在40以下时，请检查该数是否为素数。如果是这样，则打印它，否则就不打印。

Answer 3

为什么在Java中使用array而不是list。您可以通过retainAll来实现，如下所示：

    List<Integer> ret = new ArrayList<>(primes);
    ret.retainAll(fibos);

对于本地测试：

public static void main(String... args) {
    List<Integer> primes = new ArrayList<>(Arrays.asList(2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37));
    List<Integer> fibos = new ArrayList<>(Arrays.asList(0, 1, 1, 2, 3, 5, 8, 13, 21, 34));
    List<Integer> ret = new ArrayList<>(primes);
    ret.retainAll(fibos);
    System.out.println(ret);
}

输出将是：

[2, 3, 5, 13]

已更新

根据OP的要求，我将其添加为参考：

static void fibo(int n) {
    List <Integer> fibos = new ArrayList<>(n);
    fibos.add(0);
    fibos.add(1);
    System.out.print(fibos.get(0) + "," + fibos.get(1));
    for (int i = 2; i <= n; i++) {
        int a = fibos.get(i-1) + fibos.get(i-2);
        if (a <= n) {
            fibos.add(a);
            System.out.print("," + a);
        } else break;
    }
}

如何使方法之间相交

3 个答案:

已更新