238. Product of Array Except Self

题意： 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.)

思路：题目禁止使用除法，所以两遍扫描的解法肯定就不行了，但是可以使用一个数组记录前n项的乘积，再从右向左扫描，把每个数左边的乘积和右边的乘积乘一下，就OK了。

class Solution:    def productExceptSelf(self, nums):        if not nums:            return []        left_product = [1 for i in xrange(len(nums))]        for i in xrange(1, len(nums)):            left_product[i] = left_product[i - 1] * nums[i - 1]        right_product = 1        for i in xrange(len(nums) - 2, -1, -1):            right_product *= nums[i + 1]            left_product[i] = left_product[i] * right_product        return left_product