Multiplication of numbers

Q:

There is an array A[N] of N numbers. You have to compose an array Output[N] such that Output[i] will be equal to multiplication of all the elements of A[N] except A[i]. Solve it without division operator and in O(n).
For example Output[0] will be multiplication of A[1] to A[N-1] and Output[1] will be multiplication of A[0] and from A[2] to A[N-1].
Example:
A: {4, 3, 2, 1, 2}
OUTPUT: {12, 16, 24, 48, 24}

A:

Tiger's own solution is to use another two array to store the left and right muliplication result.

    ,  
    private static int[] multi(int[] A) {
        // use another matrix to help record its left multiplication result, and right
        int n = A.length;
        if(n==0){
            System.err.print("array should have at least 1");
            System.exit(0);
        }
        int[] left = new int[n];
        int[] right = new int[n];
        left[0] =1;
        for(int i =1;i<n;i++) // calculate the left part
            left[i]=left[i-1] * A[i-1];
        
        right[n-1] = 1;
        for(int i = n-2;i>=0;i--)
            right[i] = right[i+1] * A[i+1];
        
        int[] output = new int[n];
        for(int i =0;i<n;i++)
            output[i] = left[i] * right[i];
        return output;
    }

But above would require O(n) space

To save space ,we can use the right matrix to replace the output matrix,
as follows

private static int[] multi(int[] A) {
        // use another matrix to help record its left multiplication result, and right
        int n = A.length;
        if(n==0){
            System.err.print("array should have at least 1");
            System.exit(0);
        }
        int[] left = new int[n];
        int[] right = new int[n];
        left[0] =1;
        for(int i =1;i<n;i++) // calculate the left part
            left[i]=left[i-1] * A[i-1];
        
        right[n-1] = 1;
        for(int i = n-2;i>=0;i--)
            right[i] = right[i+1] * A[i+1];
        
        for(int i =0;i<n;i++)
            right[i] = left[i] * right[i];
        return right;
    }

Further, let's try to eliminate the left array.
After analyzing, we saw that the left value could be calculated iteratively in the loop. thus, we got following.

    private static int[] multi(int[] A) {
        // use another matrix to help record its left multiplication result, and right
        int n = A.length;
        if(n==0){
            System.err.print("array should have at least 1");
            System.exit(0);
        }
        int leftTillNow = 1;
        // record the multiplication result of its right sided, from i+1 to n-1
        int [] output = new int[n];
        output[n-1] = 1;
        for(int i = n-2;i>=0;i--)
            output[i] = output[i+1] * A[i+1];
        
        for(int i =1;i<n;i++){
            output[i] =  leftTillNow * output[i];
            leftTillNow *= A[i-1];
        }
        return output;
    }