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[]

 1 public class Solution {

 2     /**

 3      * @param A: Given an integers array A

 4      * @return: A Long array B and B[i]= A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1]

 5      */

 6     public ArrayList<Long> productExcludeItself(ArrayList<Integer> A) {

 7         // write your code

 8         ArrayList<Long> res = new ArrayList<Long>();

 9         if (A==null || A.size()==0 || A.size()==1) return res;

10         long[] lProduct = new long[A.size()];

11         long[] rProduct = new long[A.size()];

12         lProduct[0] = 1;

13         for (int i=1; i<A.size(); i++) {

14             lProduct[i] = lProduct[i-1]*A.get(i-1);

15         }

16         rProduct[A.size()-1] = 1;

17         for (int j=A.size()-2; j>=0; j--) {

18             rProduct[j] = rProduct[j+1]*A.get(j+1);

19         }

20         for (int k=0; k<A.size(); k++) {

21             res.add(lProduct[k] * rProduct[k]);

22         }

23         return res;

24     }

25 }