给定一个包含（-100 到 100）之间的 N 个整数的数组 A Java

Question

I'm practicing Java and my prof has given me this problem:

You're given an array A of N integers (between -100 and 100), calculate the multiplied value of all elements inside the array and return (-1, 0, 1) based on the output.

For instance:

A[1,2,3] will return 1 because the output is (6), 
A[-1,2,3] will return -1 because the output is (-6), 
A[1,2,0] will return 0 because the output is (0)

Method Definition:

public static int solution(int[] A) {
}

My approach: I was thinking this would be a recursive process because I will multiple have inputs.

Process one block A[1,2,3], store the value (1) into a newArr[], process another block A[-1,2,3], store the value (-1) into newArr[] and so on... at the end return newArr[] with the proper values in it.

What I have so far, now I am kind of stuck..

public static void main(String[] args) {
        // TODO Auto-generated method stub
         int arr[] = {6,-7,8,9};
         int arr1[] = {-6,-7,8,9};
         int arr2[] = {0,-7,8,9};
         System.out.println("[ " + solution(arr) + " ]  => Needs to be -1" );
         System.out.println("[ " + solution(arr1) + " ] => Needs to be  1");
         System.out.println("[ " + solution(arr2) + " ] => Needs to be 0");
         System.out.println("Final return should be (-1, 1, 0)");
    }
    
    public static int solution(int[] A) {
        int temp = 1;
        for (int i = 0; i < A.length; i++){
            temp *= A[i];
        }
        return temp;
    }

I don't know if this is the right way to proceed, any ideas as to how I should go about doing this?

Thanks in advance!

Answer 1

No need to do any multiplication.

• if any number is 0 return 0 immediately since the product will be zero.
• if the count of negative numbers is even, return 1 since the product of an even number of negatives is a positive.
• else return -1

public static int solution(int[] ints) {
    int count = 0;
    for (int i : ints) {
        if (i == 0) {
            return 0;
        }
        if (i < 0) {
            count++;
        }
    }
    return count % 2 == 0 ? 1 : -1;
}

Answer 2

Array multiplication left to OP.

Look at Integer.signum() - which will convert the result to 1, 0, -1 for >0, 0, <0.

Answer 3

Implementing with a wrapper, along with @Mr R's suggestion

public class test {
    public static void main(String[] args) {
        // TODO Auto-generated method stub
         int arr[] = {6,-7,8,9};
         int arr1[] = {-6,-7,8,9};
         int arr2[] = {0,-7,8,9};

         int[][] array_of_arrays = {arr, arr1, arr2};
         int[] results_array;
         results_array = solution_wrapper(array_of_arrays); 
         
         // then do what you neeed to do with the results_array
         
         System.out.println(results_array[0]);
         System.out.println(results_array[1]);
         System.out.println(results_array[2]);
         System.out.println("[ " + solution(arr) + " ]  => Needs to be -1" );
         System.out.println("[ " + solution(arr1) + " ] => Needs to be  1");
         System.out.println("[ " + solution(arr2) + " ] => Needs to be 0");
         System.out.println("Final return should be (-1, 1, 0)");
    }
    
    public static int solution(int[] A) {
        int temp = 1;
        for (int i = 0; i < A.length; i++){
            temp *= A[i];
        }
        return Integer.signum(temp);
    }

    public static int[] solution_wrapper(int[][] allA) {
        int[] rv = new int[allA.length];
        for (int j = 0; j < allA.length; j++) {
            rv[j] = solution(allA[j]);
        }
        return rv;
    }
}

Answer 4

You can declare and initialize a final array by using the method you created:

int[] finalArr = {solution(arr), solution(arr1), solution(arr2)};

However, I'm not sure where you want to return the final array from. You have your main method which doesn't have a return type and your solution method. Are you certain that you're supposed to return the final array, or is simply having one enough?