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.)

题意：给定一个数组（每个元素都大于1） 返回一个数组，每位的值都是原数组中除本身以外的所有元素乘积

    public int[] productExceptSelf(int[] nums) {
        int n = nums.length;
        int[] res = new int[n];
        res[0] = 1;

        for (int i = 1; i < nums.length; i++) {
            res[i] = res[i - 1] * nums[i - 1];//res每一位都保存了他前面所有数的乘积
        }

        int right = 1;
        for (int i = nums.length - 1; i >= 0; i--) {
            res[i] *= right; //right此时保存的是右边一位到末尾的数乘积
            right *= nums[i];//right保存了从结尾开始到i截止所有数的乘积
        }
        return res;
    }