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.)

思路：不能用除法，只能用乘法。對於每個位置，最後的結果是他左邊的所有數的積乘上右邊所有數的積，

比如，對於 3 這個數， 他這個位置上最後的結果為 8 。8 是左邊 [ 1, 2 ] 和 右邊 [ 3 ] 的乘積。那麽我們可

以先反向遍歷數組，得到每個位置上相對應的右邊的乘積。先初始化一個結果數組[ 1, 1, 1, 1 ]。執行一次

的結果為 [ 24, 12, 4, 1]，再正向遍歷一次，得到最後的結果為 [24,12,8,6].

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

        product = 1;
        for (int i = n-1; i>=0; i--){
            res[i] 
 
*= product;
            product *= nums[i];
        }
        return res;
    }
}

[Leetcode]238. Product of Array Except Self