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], return [6, 3, 2].

分析：

其实这题的限制条件挺脑残的，但是感觉实现的方式还是可以用在其它地方的。

 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      * cnblogs.com/beiyeqingteng/
 6      */
 7     public ArrayList<Long> productExcludeItself(ArrayList<Integer> A) {
 8          ArrayList<Long> result = new ArrayList<Long>();
 9         if (A.size() <= 1) {
10             result.add((long)1);
11             return result;
12         }
13         long[] left = new long[A.size()];
14         long[] right = new long[A.size()];
15         left[0] = 1;
16         for (int i = 1; i < A.size(); i++) {
17             left[i] = left[i-1] * A.get(i-1);
18         }
19         right[A.size()-1] = 1;
20         for (int i = A.size()-2; i >= 0; i--) {
21             right[i] = right[i+1] * A.get(i+1);
22         }
23         for (int i = 0; i < A.size(); i++) {
24             result.add(left[i] * right[i]);
25         }
26         return result;
27     }
28 }

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