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;
     }
 }