我如何在数组中找到两个值相乘的数字

Question

What I'm trying to do is search through an array and find out if there's two numbers that multiply to 225. Here's what I have right now:

    int n = A.length;

    for(int i = 0; i >= n; i++){
        for(int j = 0; j >= n; j++){
            if(i == j){
            }
            else if(A[i] * A[j] == 225){
                return true;
            }
        }
    }
    return false;
}

What happens is it finds the first value in the array and multiplys that with every other value and once it finds two numbers that are 225 then it returns true. I also made it so that if i and j are the same number then it does nothing because I don't want it comparing values in the same position. The problem is it keeps on returning false even on array's which do have two numbers that multiply to 225 (like 15, 45, 60, 15). So what's wrong with my code.

Answer 1

you have 1 problem with for loop:

it should look like this:

int n = A.length;
for(int i = 0; i < n; i++)
    for(int j = 0; j < n; j++)
        if(i != j)
            if(A[i] * A[j] == 225)
                return true;

also, you could optimize this loop a little bit to:

int n = A.length;
for(int i = 0; i < (n-1); i++)
    for(int j = (i+1); j < n; j++)
        if (A[i] * A[j] == 225)
            return true;

here is a little bit different version, which has average complexity O(N):

    final Map<Integer, Boolean> set = new HashMap<>(A.length);
    for (final int i : A) {
        set.put(i, set.containsKey(i)); // store number from array, set to true if there are multiple occurrences of 2 numbers
    }
    for (final int i : A) {
        final int mult = 225 / i;
        if ((mult * i) == 225) { // test that integral multiplication leads to 225
            final Boolean m = set.get(mult);
            if (m != null) { // we have both numbers in set, now we should filter double counting
                if ((i != mult) || ((i == mult) && m.booleanValue()))
                    return true;
            }

        }
    }
    return false;

Answer 2

Here Arrays.sort() use quicksort internally

public class FindMultiplicationValueUsingpointer 
{
    private void operation(int [] arr, int required)
    {//2, 3, 4, 5, 6, 7
        Arrays.sort(arr);
        int right =arr.length-1;
        int left= 0, val;

            while(left<right)
            {
                System.out.println("left is "+arr[left]);
                System.out.println("right is "+arr[right]);
                val = arr[left]*arr[right];
                System.out.println("value is " +val);
                if(val<required)
                {
                    left = left+1;

                }
                else if(val>required)
                {
                    right=right-1;
                }

                else if(val==required)
                {
                    System.out.println("value matched");
                    break;

                }
            }   

    }

    public static void main(String[] args) 
    {
        int arr[] = {3, 2, 6, 7, 4, 5};
        int required = 20;
        FindMultiplicationValueUsingpointer  up = new FindMultiplicationValueUsingpointer();
        up.operation(arr, required);

    }

Answer 3

It can also be done for any number with one for loop using hashset which will also minimize the number of iteration . Best case is o(1) and worst case o(n):

   public boolean  hasAnyPair(int[] arr, int n) {


        boolean hasPair=false;
        HashSet<Integer> set= new HashSet<Integer>();

        for(int i=0; i<arr.length; i++) {

            if(n>=arr[i] && n%arr[i]==0) {
                set.add(arr[i]);
                if(set.contains(n/arr[i])) {
                    hasPair=true;
                    break;
                }
            }

        }

        return hasPair;

    }