Lintcode: Product of Array Exclude Itself

Given an integers array A.

Define B[i] = A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1], calculate B without divide operation.

Example
For A=[1, 2, 3], B is [6, 3, 2]

非常典型的Forward-Backward Traversal 方法：

但是第一次做的时候还是忽略了一些问题：比如A.size()==1时，答案应该是空[]

public class Solution {
    /**
     * @param A: Given an integers array A
     * @return: A Long array B and B[i]= A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1]
     */
    public ArrayList<Long> productExcludeItself(ArrayList<Integer> A) {
        // write your code
        ArrayList<Long> res = new ArrayList<Long>();
        if (A==null || A.size()==0 || A.size()==1) return res;
        long[] lProduct = new long[A.size()];
        long[] rProduct = new long[A.size()];
        lProduct[0] = 1;
        for (int i=1; i<A.size(); i++) {
            lProduct[i] = lProduct[i-1]*A.get(i-1);
        }
        rProduct[A.size()-1] = 1;
        for (int j=A.size()-2; j>=0; j--) {
            rProduct[j] = rProduct[j+1]*A.get(j+1);
        }
        for (int k=0; k<A.size(); k++) {
            res.add(lProduct[k] * rProduct[k]);
        }
        return res;
    }
}