Product of Array Except Self

题目地址：
https://leetcode.com/problems/product-of-array-except-self/#/description

题目：

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：
这道题就是两个loop。第一个loop是将每一个位置的数置为自己index之前的数的乘积，第二个loop首先要有一个right的数，首先更新rst的数，然后更新right。

代码：

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