LeetCode-238.Product of Array Except Self

https://leetcode.com/problems/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.)

如果不考虑空间，可以声明两个数组存放左边乘积和右边乘积

public int[] ProductExceptSelf(int[] nums)        {            int n = nums.Length;            int[] left = new int[n - 1];            int[] right = new int[n - 1];            int[] res = new int[n];            left[0] = nums[0];            right[0] = nums[n - 1];            for (int i = 1; i < n-1; i++)            {                left[i] = left[i - 1] * nums[i];                right[i] = right[i - 1] * nums[n - 1 - i];            }            res[0] = right[n - 2];            res[n - 1] = left[n - 2];            for (int i = 1; i < n-1; i++)                res[i] = left[i - 1] * right[n - 2 - i];            return res;        }

优化一下，省去两个数组

public int[] ProductExceptSelf(int[] nums)     {        int n = nums.Length,right=1;        int[] res = new int[n];        res[0] = nums[0];        for (int i = 1; i < n; i++)            res[i] = res[i - 1] * nums[i];        for (int i = n-1; i >0; i--)        {            res[i] = res[i - 1] * right;            right *= nums[i];        }        res[0] = right;        return res;    }