Given an array nums of n integers where n > 1, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].
Example:

Input:  [1,2,3,4]
Output: [24,12,8,6]

Note: Please solve it without division and in O(n).

Follow up:
Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.)

难度: medium

题目：给定一整数组其长度大于1, 返回输出数组其元素由除当前输入元素外所有元素的乘积组成。

思路：先用输出数组从右向左计算累积元素的乘积，然后从左向左计算nums[i] = product(0, i-1) * result(i + 1)

Runtime: 1 ms, faster than 100.00% of Java online submissions for Product of Array Except Self.
Memory Usage: 40.8 MB, less than 84.97% of Java online submissions for Product of Array Except Self.

class Solution {
    public int[] productExceptSelf(int[] nums) {
        if (null == nums || nums.length < 2) {
            return new int[] {1};
        }
        int n = nums.length;
        int[] result = new int[n];
        int product = 1;
        for (int i = n - 1; i >= 0; i--) {
            result[i] = product * nums[i];
            product = result[i];
        }
        
        result[0] = result[1];
        product = nums[0];
        for (int i = 1; i < n - 1; i++) {
            result[i] = product * result[i + 1];
            product *= nums[i];
        }
        result[n - 1] = product;
        
        return result;
    }
}