Product of Array Except Self

先从左往右扫一遍。res 数组对应的位置是nums对应的数字的左边的乘积，然后再用一个right 变量，然后从数组右往左遍历，同时更新当前遍历数组右边的乘积，当前数组res数组的数是其左边的乘积，左右相乘就是所要结果

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.)

时间复杂度 o(n) 空间复杂度 o(n)

class Solution {
    public int[] productExceptSelf(int[] nums) {
        if(nums == null || nums.length == 0){
            return new int[0];
        }
        
        int[] res = new int[nums.length];
        
        res[0] = 1;
        
        for(int i = 1; i < res.length;i++){
            res[i] = nums[i-1] *res[i-1]; 
        }
        
        int right = 1;
        
        for(int i = res.length-1;i>= 0 ;i--){
            res[i] = right * res[i];
            right = right * nums[i];
        }
        
        return res;
    }
}