238. Product of Array Except Self

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

Solve it without division and in O(n).

For example, given [1,2,3,4], return [24,12,8,6].

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

假设n = 5，那么组成output[i] 的num[i] 如下：

output  1  2  3  4  5

num     x  1  1  1  1

            2  x  2  2  2 

            3  3  x  3  3 

            4  4  4  x  4

            5  5  5  5  x

那么把上表分为右上三角和左下三角，可以发现规律，右上三角是从num[1]向后的的累加，左下三角是之后的num[length - 1]向前的累加。代码如下：

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        int[] res = new int[nums.length];
        res[0] = 1;
        for (int i = 1; i < res.length; i ++) {
            res[i] = res[i - 1] * nums[i - 1];
        }
        int right = 1;
        for (int i = res.length - 1; i > 0; i --) {
            right = right * nums[i];
            res[i - 1] = res[i - 1] * right;
        }
        return res;
    }
}