238. Product of Array Except Self.md

##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.)

解题思路： 不能用除法，又必须是线性的loop, 所以，想到左右乘法。 i 左边的数字相乘保存起来，右边的数字相乘保存起来。

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

        res[0] = 1; //不要忘了
        
        for (int i = 1; i< nums.length; i++){
        	left *= nums[i-1];
        	res[i] = left; 
        }
        
        for (int i = nums.length-2; i >= 0; i--){
        	right *= nums[i+1];
        	
        //res[i]中的值是上面一个循环计算出的。
        	res[i] = res[i] *right; 
        }
        return res;
        
    }