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.)

 //不让我们用除法。如果让用除法的话，那这道题就应该属于Easy，因为可以先遍历一遍数组求出所有数字之积，然后除以对应位置的上的数字。但是这道题禁止我们使用除法，那么我们只能另辟蹊径。我们想，对于某一个数字，如果我们知道其前面所有数字的乘积，同时也知道后面所有的数乘积，那么二者相乘就是我们要的结果，所以我们只要分别创建出这两个数组即可，分别从数组的两个方向遍历就可以分别创建出乘积累积数组
//  Given numbers [2, 3, 4, 5], regarding the third number 4, the product of array except 4 is 2*3*5 which consists of two parts: left 2*3 and right 5. The product is left*right. We can get lefts and rights:

// Numbers:     2    3    4     5
// Lefts:            2  2*3 2*3*4
// Rights:  3*4*5  4*5    5      
// Let’s fill the empty with 1:

// Numbers:     2    3    4     5
// Lefts:       1    2  2*3 2*3*4
// Rights:  3*4*5  4*5    5     1
// We can calculate lefts and rights in 2 loops. The time complexity is O(n).

// We store lefts in result array. If we allocate a new array for rights. The space complexity is O(n). To make it O(1), we just need to store it in a variable which is right in @lycjava3’s code.
    public int[] productExceptSelf(int[] nums) {
        int n = nums.length;
        int[] res = new int[n];
        // Calculate lefts and store in res.
        int left = 1;
        for (int i = 0; i < n; i++) {
            if (i > 0)
                left = left * nums[i - 1];
            res[i] = left;
        }
        // Calculate rights and the product from the end of the array.
        int right = 1;
        for (int i = n - 1; i >= 0; i--) {
            if (i < n - 1)
                right = right * nums[i + 1];
            res[i] *= right;
        }
        return res;
    }