题目描述

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.)

分析

可以定义两个辅助数组，left和right。

left[i]：存放nums[i]之前的乘积
right[i]：存放nums[i]之后的乘积

但是分析一下，空间可以优化，因为我们先从右向左算，再从左向右算，只需要返回的那一个数组就够了。

代码

    public int[] productExceptSelf(int[] nums) {

        int[] rt = new int[nums.length];
        rt[nums.length - 1] = 1;

        for (int i = nums.length - 2; i >= 0; i--) {
            rt[i] = rt[i + 1] * nums[i+1];
        }

        int left = 1;
        for (int i = 0; i < nums.length; i++) {
            rt[i] *= left;
            left *= nums[i];
        }

        return rt;
    }