I have to apply linear scan to this problem, and I just don't even know where to start. I feel as if I should change my career path honestly, this is a simple intro problem and i'm just not understanding anything about it.

/**
 * Applies the linear scan strategy to counting the number of negative
 * values in an array.
 */
public class CountNegatives {

    /**
     * Returns the number of negative values in the given array.
     */
    public static int countNegatives(int[]a) {
        neg=0;
        for{i=0;i<a.length;i  }
            if(i>0)
                neg = neg 1;
        return neg;
        
    }
}

I've tried running this in VS and a tool my school uses called JGrasp. Both told me the { should be ( , the ; between length and i should be > , i<a.length is not a statement, and that } should be ;

When I change any of these things, it tells me variable neg=0 cannot be found and doubles the amount of errors in the code.

CodePudding user response：

public class CountNegatives {
/**
 * Returns the number of negative values in the given array.
 */
public static int countNegatives(int[]a) {
        int neg=0;
        for(int number : a){
            if(number<0) {
                neg=neg 1;
            }
        }
        return neg;
    }
}

CodePudding user response：

There are a few things you missed,

1. You did not initialize the neg=0; value with the proper data type; you should initialize it with the proper data type, which int his case is int, correct initialization would be int neg = 0;

2. the other problem would be in the for loop you still need to initialize i with a data type

3. the if statement is wrong, it should be 'i<0' instead cause with your code you are just counting the positive values

4. You are comparing the index value with 0 so you won't get the right value, instead, you should compare the value in the array that on that index, array[index].

The correct code would be as follows:

static int countNegatives(int[] a) {
        
        int neg = 0;
        
        for(int i = 0; i < a.length; i  ) {
            if(a[i] < 0) neg  ;
            // this will just work as neg = neg  1;
        }
        
        return neg;
    }
}