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

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.)

LeetCode：連結

首先想到的思路是計算全部數字的乘積，然後分別除以num陣列中的每一個數（需要排除數字0）。然而，題目要求不能使用除法。

由於output[i] = (x0 * x1 * ... * xi-1) * (xi+1 * .... * xn-1)

因此執行兩趟迴圈：第一趟正向遍歷陣列，計算x0 ~ xi-1的乘積；第二趟反向遍歷陣列，計算xi+1 ~ xn-1的乘積。

class Solution(object):
    def productExceptSelf(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        tmp = 1
        res = [0] * len(nums)
        for i in range(len(nums)):
            res[i] = tmp
            tmp *= nums[i]
        tmp = 1
        for i in range(len(nums)-1, -1, -1):
            res[i] *= tmp 
            tmp *= nums[i] 
        return res